Matrices: Class 12 Maths Chapter 3 Hinglish mein

Mathematics Part 1 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

120 min
1 simulation
61 practice questions
31 MCQs

Is chapter mein aap kya seekhenge

Matrix ka order padhna, koi bhi entry a_ij dhoondhna, aur data ya kisi rule se matrix banana
Row, column, square, diagonal, scalar, identity aur zero matrices pehchanna, aur equality of matrices se unknowns nikaalna
Matrices ko add, subtract aur scale karna, aur in operations ke laws se matrix equations solve karna
Row-by-column rule se matrices multiply karna, tay karna ki product kab defined hai, aur samjhana ki AB aur BA alag kyun ho sakte hain
Transpose nikaalna, aur sum, multiple aur product ke transpose ke rules lagaana
Symmetric aur skew symmetric matrices pehchanna, aur kisi bhi square matrix ko in do types mein todna
Check karna ki do matrices ek doosre ke inverse hain ya nahi, aur inverse ki uniqueness aur product ke inverse ka rule use karna

NCERT Mathematics Class 12, Chapter 3, pp. 34–74 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

3.1 Introduction

Cricket scoresheet, railway reservation chart aur ghar ka budget, teeno numbers ko rows aur columns mein rakhte hain. Jab sab maan lein ki har row aur har column kis cheez ke liye hai, to labels hata sakte hain aur numbers ke us block ko hi add, scale aur multiply kar sakte hain. Aise block ko matrix (आव्यूह) kehte hain, aur iske saath calculation ke rules milkar matrix algebra banate hain.

Matrices ki shuruaat linear equations ke system ko saaf-suthre tareeke se likhne aur ek fixed routine se solve karne ke liye hui thi. Aaj yahi idea computer graphics chalata hai (jab bhi tumhare phone par photo rotate ya zoom hoti hai, kaam ek matrix karta hai), saath hi web pages ki ranking, weather models, bank messages ko protect karne wale codes, aur woh input–output tables jinse economists production plan karte hain. Yeh chapter vocabulary aur arithmetic banata hai; Chapter 4 determinants ke saath inhe kaam mein lagata hai.

Fig. 3.1 us routine ka pehla step dikhata hai. Jab positions tay ho jaati hain, to unknowns aur plus signs koi information nahi dete, isliye equations ka system numbers ke do blocks mein simat jaata hai, aur poora system ek single matrix equation ban jaata hai.

3/3System ek single matrix equation ban jaata hai (Chapter 4 mein solve hoga).

Fig. 3.1Numbers ki positions tay hote hi linear equations ka system ek matrix equation ban jaata hai.NCERT 3.1 ka idea · p. 34

Step 1 mein equations 2x + 3y = 13 aur 4x minus y = 5 aligned columns mein likhi hain. Step 2 mein letters aur signs fade ho jaate hain, aur brackets mein coefficient block 2, 3 ke neeche 4, minus 1 bachta hai, aur ek column mein constants 13 ke neeche 5. Step 3 poore system ko ek matrix equation ki tarah likhta hai: coefficient matrix times column x, y barabar column 13, 5, aur neeche solution x = 2, y = 3 likha hai.

Quick check

2 mein se 1

System

5 x - 2 y = 7

x + 4 y = 3

ke coefficients kaun sa matrix rakhta hai?

[\begin{array}{cc} 5 & - 2 \\ 1 & 4 \end{array}]

[\begin{array}{cc} 5 & 2 \\ 1 & 4 \end{array}]

[\begin{array}{cc} 5 & 1 \\ - 2 & 4 \end{array}]

[\begin{array}{c} 7 \\ 3 \end{array}]

Answer dekho

Answer (A) $[\begin{array}{cc} 5 & - 2 \\ 1 & 4 \end{array}]$

Har equation ek row deti hai, pehle

x

phir

y

ke order mein, aur minus sign apne coefficient ka hissa hai:

- 2

. Equations ko columns mein neeche-neeche likhne se teesra option milta hai, jo ek alag matrix hai.

3.2 Matrix

Yeh ek school canteen ka record hai. Monday ko 42 samosas aur idli ki 30 plates biki, Tuesday ko 35 samosas aur idli ki 48 plates, aur Wednesday ko 50 samosas aur idli ki 26 plates. Table se record padhna aasaan ho jaata hai. Agar hum maan lein ki rows din hain aur columns dishes, to labels hata kar sirf numbers ka rectangular block brackets mein rakh sakte hain (Fig. 3.2).

5/5Row 2 aur column 1 entry $a_{21} = 35$ par milte hain.

Fig. 3.2Labels hata do to numbers ki table ek matrix hai; har entry apni row aur apne column se pehchani jaati hai.NCERT p. 35 ki tables ka idea · p. 35

Step 1 mein ek table hai jiske row labels Mon, Tue aur Wed hain aur column labels Samosa aur Idli, aur usmein 42 aur 30, 35 aur 48, 50 aur 26 hain. Step 2 mein labels aur grid fade ho jaate hain aur chhe numbers ke chaaron taraf square brackets lag jaate hain, jise 3 by 2 matrix likha gaya hai. Step 3 doosri row, 35 aur 48, ko row 2 ki tarah highlight karta hai. Step 4 pehle column, 42, 35 aur 50, ko column 1 ki tarah highlight karta hai. Step 5 number 35 par gol ghera banata hai jahan row 2 aur column 1 milte hain, aur use a sub 2 1 barabar 35 likhta hai.

Yeh arrangement zaroori nahi hai. Hum utni hi aasaani se dishes ko rows mein aur dino ko columns mein rakh sakte the, jisse 2 rows aur 3 columns ka block milta. Dono blocks mein same information hai; zaroori bas yeh hai ki numbers use karne se pehle convention fix ho. Section 3.5 mein hum dekhenge ki dono arrangements ek doosre ke transposes hain.

Definition

Matrix

Numbers ya functions ka rows aur columns mein ek definite order mein rectangular arrangement. Iski har item matrix ka ek element (ya entry) hai.

Example $[\begin{array}{cc} 42 & 30 \\ 35 & 48 \\ 50 & 26 \end{array}]$ , canteen ka record.

Page par aar-paar ek line mein rakhi entries ek row banati hain; page par neeche ki taraf rakhi entries ek column banati hain. Rows upar se gini jaati hain, columns left se, aur matrices ko capital letters milte hain. Entries ka whole numbers hona zaroori nahi, woh kisi variable ke functions bhi ho sakti hain, jaise in teen matrices mein:

\begin{array}{c} P = [\begin{array}{cc} - 1 & \frac{3}{4} \\ 0 & \sqrt{5} \end{array}] \\ Q = [\begin{array}{ccc} 2 & - 7 & 0.5 \end{array}] \\ R = [\begin{array}{ccc} \cos t & - \sin t & t \\ \sin t & \cos t & t^{2} \end{array}] \end{array}

$P$ mein 2 rows aur 2 columns hain, $Q$ mein 3 elements (अवयव) ki ek hi row hai, aur $R$ mein 2 rows aur 3 columns hain jinki entries $t$ ke saath badalti hain.

3.2.1 Matrix ka order

$m$ rows aur $n$ columns wale matrix ka order $m \times n$ (कोटि) hota hai, jise “ $m$ by $n$ ” padhte hain. Rows hamesha pehle aati hain. Canteen matrix ka order $3 \times 2$ hai, aur upar wale $P$ , $Q$ , $R$ ke orders $2 \times 2$ , $1 \times 3$ aur $2 \times 3$ hain. $m \times n$ matrix mein $m n$ entries hoti hain, row aur column ke har choice ke liye ek, isliye $P$ mein 4 entries hain aur $R$ mein 6.

Ek general m × n matrix

\begin{aligned} A & = [a_{i j}]_{m \times n} \\ = [\begin{array}{cccc} a_{11} & a_{12} & \dots & a_{1 n} \\ a_{21} & a_{22} & \dots & a_{2 n} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ a_{m 1} & a_{m 2} & \dots & a_{m n} \end{array}] \end{aligned}

jahan

$a_{i j}$: row $i$ aur column $j$ wali entry, jise $(i, j)$ th element kehte hain
$i, j$: natural numbers jahan $1 \leq i \leq m$ aur $1 \leq j \leq n$

Pehla subscript row batata hai aur doosra column. Isliye row $i$ mein $a_{i 1}, a_{i 2}, \dots, a_{i n}$ aate hain aur column $j$ mein $a_{1 j}, a_{2 j}, \dots, a_{m j}$ . Canteen matrix mein $a_{21} = 35$ hai, kyunki row 2 Tuesday hai aur column 1 samosas, jabki $a_{12} = 30$ Monday ki idli hai. Is chapter mein har entry ek real number ya real-valued function hai, aur $A = [a_{i j}]_{m \times n}$ hamara shorthand hai “ $A$ ek $m \times n$ matrix hai jiski entries $a_{i j}$ hain” ke liye.

Plane ka ek point khud ek chhota matrix hai: $(x, y)$ ko column $[\begin{array}{c} x \\ y \end{array}]$ ya row $[x y]$ ki tarah likh sakte hain. Kai points ke columns ko saath-saath khada karo to ek matrix poori figure store kar leta hai. Fig. 3.3 ke quadrilateral $K L M N$ ke liye, jahan $K (- 2, 1)$ , $L (3, 0)$ , $M (4, 4)$ aur $N (0, 5)$ hain, vertex matrix mein har vertex ka ek column hai, order $K, L, M, N$ mein, isliye yeh $2 \times 4$ hai:

V = [\begin{array}{cccc} - 2 & 3 & 4 & 0 \\ 1 & 0 & 4 & 5 \end{array}]

SimulationMatrix ke columns ki tarah vertices

Dono directions mein minus 5 se 7 tak ka coordinate grid, jismein quadrilateral KLMN hai jiske vertices K (minus 2, 1) par, L (3, 0) par, M (4, 4) par aur N (0, 5) par hain, aur sides unki lengths ke saath judi hain. Tum kisi bhi vertex ko kisi grid point tak drag karke uske naye coordinates padh sakte ho. Ek mirror control x-axis par shuru hota hai aur reflected quadrilateral K prime L prime M prime N prime dikhata hai; ise y-axis, origin ya off par switch kar sakte ho.

Fig. 3.3Reflection vertex matrix

V

ke har column ko ek hi rule se badalta hai; ek vertex drag karne se sirf ek column badalta hai.

Try karo

Sim $x$ -axis mein mirror image $K^{'} L^{'} M^{'} N^{'}$ ke saath khulta hai. Iska vertex matrix likho: $V$ ki pehli row ka kya hua, aur doosri ka?
Mirror ko origin par set karo. $V$ ko kis ek number se multiply karke naya vertex matrix milega?
Mirror off karo aur $M$ ko $(5, 2)$ tak drag karo. $V$ ka kaun sa column badalta hai, aur woh kya ban jaata hai?

Example 1NCERT Example 1 · p. 37

Teen trains ki seats

Ek booking clerk teen evening trains mein abhi khaali berths note karta hai. Train P mein 64 sleeper berths aur 12 AC-3 berths khaali hain, train Q mein 9 sleeper aur 27 AC-3, aur train R mein 0 sleeper aur 41 AC-3.

(a) Is information ko ek $2 \times 3$ matrix $B$ ki tarah likho jismein classes rows hon. (b) $b_{23}$ kis cheez ke liye hai, aur uski value kya hai? (c) Agar trains rows hoti to order kya hota?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 2NCERT Example 2 · p. 38

Baarah entries ke saare shapes

Ek matrix mein 12 elements hain. Uske saare possible orders list karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

3/6Chaar-chaar ki teen rows: order $3 \times 4$ .

Fig. 3.4Baarah entries chhe rectangular shapes mein fit hoti hain, 12 ke har ordered factor pair ke liye ek.NCERT Example 2 ka idea · p. 38

Har step mein wahi baarah dots dobara arrange hote hain. Step 1 mein 12 ki ek row hai, order 1 by 12. Step 2 mein 6-6 ki 2 rows. Step 3 mein 4-4 ki 3 rows. Step 4 mein 3-3 ki 4 rows. Step 5 mein 2-2 ki 6 rows. Step 6 mein 12 dots ka ek column, order 12 by 1. Har step order aur rows aur columns ki sankhya likhta hai.

Jab elements ki sankhya ek prime $p$ ho, to sirf do shapes possible hain, $1 \times p$ aur $p \times 1$ . Jab woh 16 jaisa perfect square ho, to ek shape, $4 \times 4$ , dono taraf se same hai, isliye orders ki ginti odd hoti hai.

Example 3NCERT Example 3 · p. 38

Rule se matrix

$3 \times 3$ matrix $A = [a_{i j}]$ banao jismein $a_{i j} = i^{j}$ ho, aur entry 9 ki position batao.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

3.3 Matrices ke Types

Entries ke kuch shapes aur patterns itni baar aate hain ki unke naam rakh diye gaye hain. Pehle teen sirf order par depend karte hain.

Column matrix: ek hi column, order $m \times 1$ . Jaise $[\begin{array}{c} 4 \\ - 1 \\ \frac{2}{3} \end{array}]$ order $3 \times 1$ ka column matrix hai.
Row matrix: ek hi row, order $1 \times n$ . Jaise $[0 \sqrt{3} - 6 1]$ order $1 \times 4$ ka row matrix hai.
Square matrix: jitni rows utne columns. $n \times n$ matrix ko order $n$ ka square matrix kehte hain; $[\begin{array}{ccc} 7 & 0 & - 2 \\ 1 & - 5 & 4 \\ 9 & 3 & 6 \end{array}]$ order 3 ka square hai.

Square matrix $A = [a_{i j}]_{n \times n}$ mein entries $a_{11}, a_{22}, \dots, a_{n n}$ , jo top-left corner se bottom-right corner tak jaati hain, main diagonal (ya bas diagonal) banati hain. Upar wale order-3 example mein diagonal entries $7$ , $- 5$ aur $6$ hain. Square matrix ke teen important types is baat se define hote hain ki diagonal par aur uske bahar kya hota hai, aur har type pichhle wale ko aur narrow karta hai (Fig. 3.5).

Diagonal matrix: diagonal ke bahar ki har entry 0 hai, yaani jab bhi $i \neq j$ ho to $a_{i j} = 0$ . Diagonal entries kuch bhi ho sakti hain, 0 bhi, isliye $[\begin{array}{cc} 3 & 0 \\ 0 & 0 \end{array}]$ diagonal hai.
Scalar matrix: aisa diagonal matrix jiski saari diagonal entries ek hi number $k$ ke barabar hon, jaise $[\begin{array}{ccc} - 2 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & - 2 \end{array}]$ . Isse multiply karna bas $k$ se scale karna hai, isi se naam aaya (Section 3.4.5).
Identity matrix: $k = 1$ wala scalar matrix. Order $n$ ki identity ko $I_{n}$ likhte hain, ya order saaf ho to bas $I$ : $I_{1} = [1]$ , $I_{2} = [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}]$ , aur aise hi aage. Multiplication mein yeh number 1 ka role nibhata hai.

3/4Barabar diagonal entries ( $k = 4$ ): ek scalar matrix.

Fig. 3.5Har special type ek square matrix hai jismein ek condition aur judti hai: har identity matrix scalar hai, har scalar matrix diagonal hai, har diagonal matrix square hai.NCERT types (iv)–(vi) · p. 40

Left mein ek 3 by 3 matrix hai jiska main diagonal shaded hai; right mein chaar nested boxes hain jin par square, diagonal, scalar aur identity likha hai. Step 1 entries 5, minus 2, 7, 1, 3, 0, 4, 8, 6 wala ek general square matrix dikhata hai aur bahar wala box roshan hota hai. Step 2 diagonal ke bahar ki har entry ko 0 bana deta hai, diagonal par 5, 3, 6 bachte hain, aur diagonal box roshan hota hai. Step 3 saari diagonal entries 4 kar deta hai aur scalar box roshan karta hai. Step 4 sabko 1 kar deta hai aur sabse andar wala identity box roshan karta hai.

To har identity matrix scalar hai, har scalar matrix diagonal hai aur har diagonal matrix square hai, lekin in mein se koi bhi statement ulta nahi kiya ja sakta. $1 \times 1$ matrix $[c]$ ek saath square, diagonal aur scalar hai, aur yeh identity tabhi hai jab $c = 1$ ho.

Aakhir mein, zero matrix (शून्य आव्यूह), jise O likhte hain, ki har entry 0 hoti hai. Iska koi bhi order ho sakta hai, jaise $[0 0 0]$ ya $[\begin{array}{cc} 0 & 0 \\ 0 & 0 \\ 0 & 0 \end{array}]$ ; order context se samajh aata hai. Square zero matrix ko scalar matrix bhi maana jaata hai, $k = 0$ ke saath.

3.3.1 Matrices ki equality

Do matrices ko entry by entry tabhi compare kar sakte hain jab unki entries line up hon, isliye equality ek saath do cheezein maangti hai.

Definition

Equal matrices

Matrices $A = [a_{i j}]$ aur $B = [b_{i j}]$ equal hain, likhte hain $A = B$ , jab (i) unka order same ho aur (ii) har $i$ aur $j$ ke liye $a_{i j} = b_{i j}$ ho.

Example $[\begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array}] \neq [\begin{array}{cc} 1 & 3 \\ 2 & 4 \end{array}]$ : same order aur wahi chaar numbers, lekin alag jagahon par.

Koi bhi condition chhodi nahi ja sakti. Row $[0 0]$ aur column $[\begin{array}{c} 0 \\ 0 \end{array}]$ mein same numbers hain phir bhi woh equal nahi hain, kyunki unke orders alag hain. Doosri taraf, ek matrix equation mein har position ke liye ek ordinary equation chhupi hoti hai. Agar $[\begin{array}{cc} x & 4 \\ - 1 & y + 2 \end{array}] = [\begin{array}{cc} 3 & 4 \\ - 1 & 0 \end{array}]$ , to $x = 3$ aur $y = - 2$ , jabki baaki do positions waise hi match karti hain.

Example 4NCERT Example 4 · p. 41

Chhe positions, paanch unknowns

\begin{array}{c} [\begin{array}{ccc} x - 2 & 3 y + 1 & 4 \\ a + 5 & 2 b - 1 & c / 2 \end{array}] \\ = [\begin{array}{ccc} 5 & 2 y + 4 & 4 \\ - 1 & b + 6 & 3 \end{array}] \end{array}

$a$ , $b$ , $c$ , $x$ aur $y$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 5NCERT Example 5 · p. 41

Equality se ek system milta hai

\begin{array}{c} [\begin{array}{cc} p + q & 2 p - q \\ 3 r + s & r - 2 s \end{array}] \\ = [\begin{array}{cc} 6 & 9 \\ 5 & - 3 \end{array}] \end{array}

$p$ , $q$ , $r$ aur $s$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

3.4 Matrices par Operations

Numbers ko add, subtract aur multiply kar sakte hain, aur matrices ko bhi, lekin har operation ke saath orders par ek condition aati hai. Hum in sab mein ek chhote business ko follow karenge. Ek juice wala do kiosks chalata hai, ek railway station par aur ek college gate par, aur teen juices bechta hai: mango, sugarcane aur mosambi (sweet lime). Neeche har matrix mein row 1 station hai, row 2 college, aur columns isi order mein teen juices hain.

3.4.1 Matrices ka addition

Ek din morning shift aur evening shift ne itne glasses beche:

\begin{array}{c} M = [\begin{array}{ccc} 40 & 55 & 30 \\ 25 & 20 & 35 \end{array}] \\ E = [\begin{array}{ccc} 35 & 60 & 20 \\ 30 & 15 & 25 \end{array}] \end{array}

Station par din ki mango sales $40 + 35 = 75$ glasses hai, jo do (1, 1) entries jodne se mili. Station ka sugarcane do (1, 2) entries se aata hai, aur isi tarah chhe ki chhe positions ke liye (Fig. 3.6). Result phir se ek $2 \times 3$ matrix hai, $M$ aur $E$ ka sum:

\begin{aligned} D & = M + E \\ = [\begin{array}{ccc} 40 + 35 & 55 + 60 & 30 + 20 \\ 25 + 30 & 20 + 15 & 35 + 25 \end{array}] \\ = [\begin{array}{ccc} 75 & 115 & 50 \\ 55 & 35 & 60 \end{array}] \end{aligned}

3/3Same rate par ek hafta: $D$ ki har entry 7 se multiply hoti hai.

Fig. 3.6Same order ke matrices position by position add hote hain; scalar har entry ko multiply karta hai.NCERT 3.4.1–3.4.2 ka idea · p. 43

Step 1 morning matrix 40, 55, 30 ke neeche 25, 20, 35 aur evening matrix 35, 60, 20 ke neeche 30, 15, 25 dikhata hai, saath mein unke sum ke liye khaali bracket. Step 2 sum 75, 115, 50 ke neeche 55, 35, 60 bharta hai aur same position ki matching entries 55, 60 aur 115 par ghera banata hai. Step 3 ek doosri line jodta hai jahan 7 times day matrix se week matrix 525, 805, 350 ke neeche 385, 245, 420 milta hai, aur 115 aur 805 par ghera banata hai.

General rule yeh hai. Agar $A = [a_{i j}]$ aur $B = [b_{i j}]$ ka same order $m \times n$ ho, to unka sum woh $m \times n$ matrix hai jiski entries corresponding entries ke sums hain:

Do matrices ka sum

A + B = [a_{i j} + b_{i j}]_{m \times n}

Alag orders ke matrices ke liye addition defined nahi hai. $[\begin{array}{cc} 1 & 0 \\ 2 & 5 \end{array}]$ aur $[3 4]$ ko jodne ka koi sensible tareeka nahi hai, kyunki ek ki zyaadatar positions ka doosre mein koi partner nahi hai. Lekin saare $m \times n$ matrices ke set mein koi bhi do add ho sakte hain aur answer usi set mein rehta hai: us set par addition ek binary operation hai.

Example 6NCERT Example 6 · p. 44

Surds aur fractions ko jodna

$A + B$ nikaalo agar $A = [\begin{array}{ccc} \sqrt{2} & - 1 & \frac{3}{2} \\ 0 & 4 & - 3 \end{array}]$ aur $B = [\begin{array}{ccc} 2 \sqrt{2} & 5 & \frac{1}{2} \\ - 2 & - 4 & 1 \end{array}]$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

3.4.2 Matrix ka scalar se multiplication

Maan lo kiosks ek hafte tak har din bilkul same amounts bechte hain. Tab har position mein hafte ka figure din ke figure ka saat guna hai:

\begin{aligned} 7 D & = 7 [\begin{array}{ccc} 75 & 115 & 50 \\ 55 & 35 & 60 \end{array}] \\ = [\begin{array}{ccc} 525 & 805 & 350 \\ 385 & 245 & 420 \end{array}] \end{aligned}

General roop se, matrix $A = [a_{i j}]_{m \times n}$ aur ek number $k$ ke liye (jise matrix se alag dikhane ke liye scalar kehte hain), scalar multiple $k A$ woh $m \times n$ matrix hai jo $A$ ki har entry ko $k$ se multiply karke milta hai. Jaise, $- \frac{1}{2} [\begin{array}{cc} 4 & - 6 \\ 1 & 0 \end{array}] = [\begin{array}{cc} - 2 & 3 \\ - \frac{1}{2} & 0 \end{array}]$ .

Scalar multiple

k A = k [a_{i j}]_{m \times n} = [k a_{i j}]_{m \times n}

Do special cases ke naam hain. $A$ ka negative $- A = (- 1) A$ hai: har entry ka sign badal do. Same order ke do matrices ka difference $A - B = A + (- 1) B$ hai, isliye yeh corresponding entries ko subtract karke milta hai, $A - B = [a_{i j} - b_{i j}]$ . Kiosks ke liye,

M - E = [\begin{array}{ccc} 5 & - 5 & 10 \\ - 5 & 5 & 10 \end{array}]

Positive entry ka matlab hai ki us kiosk par us juice ko morning shift ne zyaada becha, aur negative entry ka matlab evening shift ne. Matrix mein negative entries ka bhi matlab hota hai, bilkul negative numbers ki tarah.

Example 7NCERT Example 7 · p. 46

Do matrices ka combination

Agar $A = [\begin{array}{cc} 2 & - 1 \\ 0 & 3 \\ 1 & 4 \end{array}]$ aur $B = [\begin{array}{cc} 1 & 2 \\ - 3 & 5 \\ 4 & - 2 \end{array}]$ , to $3 A - 2 B$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

3.4.3 Matrix addition ki properties

Matrix addition har position mein ek saath kiya gaya ordinary addition hai, isliye ise numbers ke jaane-pehchaane laws mil jaate hain. Maan lo $A$ , $B$ aur $C$ same order $m \times n$ ke matrices hain.

PropertyMatrix addition ke laws

(i) Commutative: $A + B = B + A$ . (ii) Associative: $(A + B) + C = A + (B + C)$ . (iii) Additive identity: $m \times n$ zero matrix O ke liye $A + O = O + A = A$ . (iv) Additive inverse: $- A = [- a_{i j}]$ ke liye $A + (- A) = (- A) + A = O$ .

3.4.4 Matrix ke scalar multiplication ki properties

PropertyScalar multiplication ke laws

Same order ke matrices $A$ aur $B$ aur scalars $k$ aur $l$ ke liye: (i) $k (A + B) = k A + k B$ ; (ii) $(k + l) A = k A + l A$ ; (iii) $k (l A) = (k l) A$ ; (iv) $1 \cdot A = A$ .

In laws se hum simple matrix equations bilkul waise hi solve kar sakte hain jaise number $x$ ke liye $3 a - 2 x = b$ solve karte hain: terms ko doosri taraf le jao, phir kisi non-zero number se divide karo. Agle examples har step ke peeche ka law batate hain.

Example 8NCERT Example 8 · p. 47

Unknown matrix nikaalna

$A = [\begin{array}{cc} 2 & 0 \\ - 1 & 3 \end{array}]$ aur $B = [\begin{array}{cc} 4 & - 2 \\ 1 & 5 \end{array}]$ diye hain; woh matrix $X$ nikaalo jiske liye $3 A - 2 X = B$ ho.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 9NCERT Example 9 · p. 48

Do equations, do unknown matrices

$X$ aur $Y$ nikaalo agar $2 X + Y = [\begin{array}{cc} 7 & 2 \\ - 1 & 8 \end{array}]$ aur $X - Y = [\begin{array}{cc} 2 & 4 \\ - 2 & 1 \end{array}]$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 10NCERT Example 10 · p. 48

Combination ke andar unknown entries

$x$ aur $y$ nikaalo agar $3 [\begin{array}{cc} x & - 1 \\ 2 & y \end{array}] - [\begin{array}{cc} 4 & 2 \\ 6 & 1 \end{array}] = [\begin{array}{cc} 11 & - 5 \\ 0 & 11 \end{array}]$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 11NCERT Example 11 · p. 49

Crafts fair ki kamaai

Ek crafts fair mein ek pottery stall (row 1) aur ek bamboo stall (row 2) small, medium aur large items (columns, isi order mein) bechte hain. Rupees mein unki kamaai Saturday ko $A$ aur Sunday ko $B$ thi:

A = [\begin{array}{ccc} 4000 & 2500 & 1800 \\ 3200 & 4500 & 2000 \end{array}]

B = [\begin{array}{ccc} 5200 & 3100 & 2400 \\ 3000 & 6000 & 3500 \end{array}]

(a) Har stall ki har size ki weekend kamaai nikaalo. (b) Saturday se Sunday tak kamaai mein badlaav nikaalo. (c) Organisers Sunday ki kamaai par 5 % fee rakhte hain. Har stall aur har size ki fee nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

3.4.5 Matrices ka multiplication

Adding aur scaling position by position hote hain. Multiplication alag hai, aur kiosks dikhate hain kyun. Maan lo mango juice ka ek glass ₹40 ka hai, sugarcane ₹20 ka aur mosambi ₹30 ka. Station ki din ki kamaai mein glasses ki har sankhya ko uski price ke saath jodte hain, multiply karte hain, aur add karte hain: $75 \times 40 + 115 \times 20 + 50 \times 30 = 3000 + 2300 + 1500 = 6800$ , yaani ₹6800. College kiosk ki kamaai $55 \times 40 + 35 \times 20 + 60 \times 30 = 4700$ hai, yaani ₹4700.

Har total sales matrix $D$ ki ek row ko prices ke column ke saath jodta hai. Ab summer prices (₹50, ₹25 aur ₹35) ka ek doosra column jodo. Kiosk row aur price column ki har jodi answer ki ek entry banati hai:

\begin{aligned} D P & = \underset{2 \times 3}{\underset{⏟}{[\begin{array}{ccc} 75 & 115 & 50 \\ 55 & 35 & 60 \end{array}]}} \underset{3 \times 2}{\underset{⏟}{[\begin{array}{cc} 40 & 50 \\ 20 & 25 \\ 30 & 35 \end{array}]}} \\ = [\begin{array}{cc} 6800 & 8375 \\ 4700 & 5725 \end{array}] \end{aligned}

Answer ko table ki tarah padho: row 1 station hai aur row 2 college; column 1 usual prices use karta hai aur column 2 summer prices. To summer prices par station din mein ₹8375 kamaayega. Shapes par dhyaan do: $D$ ki ek row mein 3 entries hain aur $P$ ke ek column mein 3 entries, isliye woh bilkul jodi-jodi mil jaati hain.

Definition

Matrices ka product

Agar $A = [a_{i j}]$ $m \times n$ hai aur $B = [b_{j k}]$ $n \times p$ hai, to unka product $A B$ woh $m \times p$ matrix hai jiski $(i, k)$ entry $A$ ki row $i$ ko $B$ ke column $k$ ke saath entry by entry multiply karke aur add karke milti hai. Agar $A$ ke columns ki sankhya $B$ ki rows ki sankhya ke barabar nahi hai, to $A B$ defined nahi hai.

Example $[\begin{array}{cc} 1 & 2 \end{array}] [\begin{array}{c} 3 \\ 4 \end{array}] = [1 \cdot 3 + 2 \cdot 4] = [11]$

Product AB ki entry (i, k)

\begin{aligned} c_{i k} & = a_{i 1} b_{1 k} + a_{i 2} b_{2 k} \\ + \dots + a_{i n} b_{n k} \\ = \sum_{j = 1}^{n} a_{i j} b_{j k} \end{aligned}

jahan

$j$: running index: yeh $A$ ki row $i$ ke saath aur $B$ ke column $k$ mein neeche ek saath chalta hai
$C = [c_{i k}]_{m \times p}$: product $A B$

Product exist karta hai ya nahi, iska quick test hai orders ko saath-saath likhna, $(m \times n) (n \times p)$ . Andar wale do numbers match hone chahiye, aur bahar wale do numbers product ka order dete hain, $m \times p$ . Fig. 3.7 ek chhota product ek-ek entry karke nikaalta hai, aur uske baad wala lab tumhe apne matrices try karne deta hai.

3/5Row 1 × column 2 se: $2 \cdot 2 + 1 \cdot 0 + 3 \cdot 5 = 19$ .

Fig. 3.7

A B

ki entry

(i, k)

hai

A

ki row

i

times

B

ka column

k

: jodiyon mein multiply karo, phir add karo.NCERT p. 51 par product CD ka idea · p. 51

Rows 2, 1, 3 aur 0, 4, 1 wale matrix ko rows 1, 2 aur 3, 0 aur 1, 5 wale matrix se multiply kiya jaata hai. Step 1 orders 2 by 3 aur 3 by 2 dikhata hai, andar ke matching 3s aur product ka order 2 by 2 ke saath, aur answer khaali hai. Step 2 mein row 1 aur column 1 highlight hote hain aur working 2 times 1 plus 1 times 3 plus 3 times 1 barabar 8 pehli entry deti hai. Step 3 row 1 aur column 2 se 19 nikaalta hai. Step 4 row 2 aur column 1 se 13 nikaalta hai. Step 5 row 2 aur column 2 se 5 nikaalta hai, aur product 8, 19 ke neeche 13, 5 poora ho jaata hai.

InteractiveRow-by-column lab

Ek interactive multiplication lab. Ek menu ready-made jodiyan deta hai: kiosk sales aur prices, do square matrices, column times row, row times column, aur ek jodi jiske orders fit nahi hote; buttons rows aur columns ki sankhya badalte hain. A aur B ki har entry edit ho sakti hai. Panel dono orders batata hai aur yeh bhi ki andar ke numbers match karte hain ya nahi. Product ke kisi cell par tap karne se A ki matching row aur B ka matching column highlight hote hain aur products ka sum likha jaata hai; Next entry button product ko ek-ek cell karke bharta hai, Fill all use poora karta hai, aur Swap button comparison ke liye B times A dikhata hai.

Fig. 3.8Matrices ki ek jodi chuno (ya apni entries type karo), phir product ke kisi bhi cell par tap karke dekho ki use kaun si row aur column banate hain.

Example 12NCERT Example 12 · p. 52

Square matrix times ek chauda matrix

$A = [\begin{array}{cc} 3 & - 2 \\ 1 & 4 \end{array}]$ aur $B = [\begin{array}{ccc} 2 & 0 & - 1 \\ 5 & 1 & 3 \end{array}]$ ke liye $A B$ nikaalo. Kya $B A$ defined hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

To $A B$ exist kar sakta hai jabki $B A$ nahi. Agar $A$ $m \times n$ hai aur $B$ $k \times l$ , to $A B$ ke liye $n = k$ chahiye aur $B A$ ke liye $l = m$ , aur dono products tabhi exist karte hain jab dono conditions sahi hon. Khaas taur par, same order ke do square matrices ke liye dono products hamesha exist karte hain.

Order matter karta hai: AB aur BA

Numbers ke liye $3 \times 5 = 5 \times 3$ . Matrices ke liye, jab $A B$ aur $B A$ dono exist karte hain tab bhi woh aksar alag hote hain, aur unka order bhi same nahi ho sakta.

Example 13NCERT Example 13 · p. 53

Alag orders ke products

$A = [\begin{array}{ccc} 1 & 0 & 2 \\ - 1 & 3 & 1 \end{array}]$ aur $B = [\begin{array}{cc} 2 & 1 \\ 0 & - 1 \\ 1 & 4 \end{array}]$ ke liye $A B$ aur $B A$ nikaalo aur dikhao ki $A B \neq B A$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Same orders se bhi rule nahi bachta. Geometry dikhati hai kyun. $2 \times 2$ matrix $A$ plane par kaam kar sakta hai, har point $[\begin{array}{c} x \\ y \end{array}]$ ko $A [\begin{array}{c} x \\ y \end{array}]$ par bhej kar, isliye product $R S$ ka matlab hai “pehle $S$ lagao, phir $R$ ”. Fig. 3.9 mein $S$ ek flag ko $y$ -axis mein reflect karta hai aur $R$ use $90 °$ anticlockwise ghumata hai. Dono possible orders mein karne par moves flag ko do alag jagahon par chhodte hain.

3/3Phir doosra move. Flags alag jagahon par khatam hote hain, isliye $R S \neq S R$ .

Fig. 3.9Reflect phir turn karna, turn phir reflect karne jaisa nahi hai:

R S \neq S R

.NCERT Example 14 ka idea · p. 53

Teen chhote coordinate panels ki do rows. Upar wali row: pehle S, phir R. Neeche wali row: pehle R, phir S. Step 1 dono rows mein same starting flag dikhata hai: y-axis ke thoda right mein khada ek pole jiska pennant right ki taraf hai. Step 2 pehla move dikhata hai: upar wali row mein flag y-axis mein reflect hota hai, isliye pennant left ki taraf ho jaata hai; neeche wali row mein use quarter turn anticlockwise ghumaya jaata hai, isliye pole negative x-axis ke thoda upar let jaata hai. Step 3 doosra move dikhata hai: upar wali row ke ant mein pole negative x-axis ke neeche hai aur pennant neeche latakta hai, neeche wali row ke ant mein pole positive x-axis ke upar hai aur pennant upar ki taraf hai. Results ke paas matrices RS, rows 0, minus 1 aur minus 1, 0 ke saath, aur SR, rows 0, 1 aur 1, 0 ke saath, likhe hain, jo alag hain.

Example 14NCERT Example 14 · p. 53

Same order, alag products

Maan lo $A = [\begin{array}{cc} 2 & 1 \\ 0 & 3 \end{array}]$ aur $B = [\begin{array}{cc} 1 & 0 \\ 4 & 1 \end{array}]$ . Dikhao ki $A B \neq B A$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Iska matlab yeh nahi ki $A B$ aur $B A$ hamesha alag hote hain. Same order ke diagonal matrices hamesha commute karte hain, kyunki har product bas diagonal entries ko apni jagah multiply karta hai: $[\begin{array}{cc} 2 & 0 \\ 0 & - 1 \end{array}] [\begin{array}{cc} 5 & 0 \\ 0 & 3 \end{array}] = [\begin{array}{cc} 10 & 0 \\ 0 & - 3 \end{array}]$ , kisi bhi order mein. Har square matrix $I$ ke saath, O ke saath aur khud ke saath bhi commute karta hai.

Non-zero matrices se zero product

Real numbers ke liye $a b = 0$ ka matlab hai $a = 0$ ya $b = 0$ . Matrices yeh rule bhi tod dete hain.

Example 15NCERT Example 15 · p. 54

Do non-zero factors, product zero

$A = [\begin{array}{cc} 1 & 2 \\ 2 & 4 \end{array}]$ aur $B = [\begin{array}{cc} 2 & - 4 \\ - 1 & 2 \end{array}]$ ke liye $A B$ aur $B A$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

To $A B = O$ se hum yeh conclude nahi kar sakte ki $A = O$ ya $B = O$ . Yahi example dikhata hai ki $A B = O$ se $B A = O$ bhi zaroori nahi.

3.4.6 Matrices ke multiplication ki properties

Commutativity fail hoti hai, lekin baaki laws jin par hum numbers ke liye bharosa karte hain, bache rehte hain. Hum unhe state karenge, short mein samjhayenge ki woh kyun sahi hain, aur phir examples par test karenge.

PropertyMatrix multiplication ke laws

Jab bhi involved products defined hon: (i) Associative: $(A B) C = A (B C)$ . (ii) Distributive: $A (B + C) = A B + A C$ aur $(A + B) C = A C + B C$ . (iii) Multiplicative identity: order $n$ ke square matrix $A$ ke liye $I_{n} A = A I_{n} = A$ .

(iii) $A I$ ki $(i, k)$ entry $\sum_{j} a_{i j} e_{j k}$ hai, jahan $e_{j k}$ $j = k$ hone par 1 hai aur warna 0. Sirf $j = k$ wala term bachta hai, aur $a_{i k}$ rehta hai. Yahi argument $I A$ ke liye bhi chalta hai.

(ii) $A (B + C)$ ki $(i, k)$ entry $\sum_{j} a_{i j} (b_{j k} + c_{j k}) = \sum_{j} a_{i j} b_{j k} + \sum_{j} a_{i j} c_{j k}$ hai, jo $A B + A C$ ki $(i, k)$ entry hai. Doosra law bhi aisa hi hai.

(i) $(A B) C$ aur $A (B C)$ dono ki $(i, l)$ entry $\sum_{j} \sum_{k} a_{i j} b_{j k} c_{k l}$ hai; dono sides bas same products ko alag tareeke se group karti hain.

Associativity ka matlab hai ki $A B C$ jaise product ko brackets ki zaroorat nahi, aur isse square matrix ki powers ka matlab banta hai: $A^{2} = A A$ , $A^{3} = A A A$ , aur aise hi aage. Matrix mein polynomials, jaise $A^{2} - 3 A + 2 I$ , bhi ban jaate hain, jahan constant term ki jagah $I$ aata hai.

Example 16NCERT Example 16 · p. 54

Product ko do tareekon se group karna

Maan lo $A = [\begin{array}{cc} 1 & - 1 \\ 2 & 3 \end{array}]$ , $B = [\begin{array}{ccc} 2 & 0 & 1 \\ - 1 & 4 & 2 \end{array}]$ aur $C = [\begin{array}{c} 3 \\ 1 \\ - 2 \end{array}]$ . $(A B) C$ aur $A (B C)$ nikaalo, aur check karo ki woh barabar hain.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 17NCERT Example 17 · p. 56

Distributive law check karna

$A = [\begin{array}{cc} 1 & 2 \\ 0 & - 1 \end{array}]$ , $B = [\begin{array}{cc} 3 & - 1 \\ 2 & 2 \end{array}]$ aur $C = [\begin{array}{cc} 2 & 1 \\ - 1 & 4 \end{array}]$ ke liye verify karo ki $A (B + C) = A B + A C$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 18NCERT Example 18 · p. 56

Ek quadratic satisfy karne wala matrix

Maan lo $A = [\begin{array}{cc} 2 & 1 \\ 1 & 3 \end{array}]$ . Dikhao ki $A^{2} - 5 A + 5 I = O$ , aur iska use karke $A^{3}$ ko $p A + q I$ ki tarah likho.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 19NCERT Example 19 · p. 57

Awareness campaign ka kharcha

Ek health NGO do districts mein posters (₹12 each), radio spots (₹1500 each) aur street plays (₹4000 each) se awareness campaign chalata hai. District P ko 2000 posters, 30 radio spots aur 12 street plays milte hain; district Q ko 3500 posters, 20 radio spots aur 18 street plays. Matrix multiplication se har district ka kharcha nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Practice · 22 questions

Practice 3.2: answers ke saath

Section 3.4 ki practice (NCERT Exercise 3.2 se match karti hai).

1
Short answerEasyNCERT Exercise 3.2, Q1
Maan lo $A = [\begin{array}{cc} 3 & - 1 \\ 2 & 4 \end{array}]$ , $B = [\begin{array}{cc} 1 & 5 \\ - 2 & 0 \end{array}]$ aur $C = [\begin{array}{cc} - 3 & 2 \\ 1 & 6 \end{array}]$ . Nikaalo:
1. $A + B$
  Solution dikhao
  Answer $[\begin{array}{cc} 4 & 4 \\ 0 & 4 \end{array}]$
  $A + B = [\begin{array}{cc} 4 & 4 \\ 0 & 4 \end{array}]$ .
2. $A - B$
  Solution dikhao
  Answer $[\begin{array}{cc} 2 & - 6 \\ 4 & 4 \end{array}]$
  $A - B = [\begin{array}{cc} 2 & - 6 \\ 4 & 4 \end{array}]$ .
3. $2 A - 3 C$
  Solution dikhao
  Answer $[\begin{array}{cc} 15 & - 8 \\ 1 & - 10 \end{array}]$
  $2 A - 3 C = [\begin{array}{cc} 6 & - 2 \\ 4 & 8 \end{array}] - [\begin{array}{cc} - 9 & 6 \\ 3 & 18 \end{array}] = [\begin{array}{cc} 15 & - 8 \\ 1 & - 10 \end{array}]$ .
4. $A B$
  Solution dikhao
  Answer $[\begin{array}{cc} 5 & 15 \\ - 6 & 10 \end{array}]$
  $A B = [\begin{array}{cc} 3 + 2 & 15 + 0 \\ 2 - 8 & 10 + 0 \end{array}] = [\begin{array}{cc} 5 & 15 \\ - 6 & 10 \end{array}]$ .
5. $B A$
  Solution dikhao
  Answer $[\begin{array}{cc} 13 & 19 \\ - 6 & 2 \end{array}]$
  $B A = [\begin{array}{cc} 3 + 10 & - 1 + 20 \\ - 6 + 0 & 2 + 0 \end{array}] = [\begin{array}{cc} 13 & 19 \\ - 6 & 2 \end{array}]$ , jo $A B$ se alag hai.
2
Short answerMediumNCERT Exercise 3.2, Q2
Har sum ko simplify karo.
1. $[\begin{array}{cc} p & - q \\ q & p \end{array}] + [\begin{array}{cc} p & q \\ - q & p \end{array}]$
  Solution dikhao
  Answer $[\begin{array}{cc} 2 p & 0 \\ 0 & 2 p \end{array}]$ , ek scalar matrix.
  Off-diagonal entries cancel ho jaati hain: $- q + q = 0$ aur $q - q = 0$ .
2. $[\begin{array}{cc} m^{2} + n^{2} & m - n \\ 2 m n & 1 \end{array}] + [\begin{array}{cc} - 2 m n & m + n \\ m^{2} + n^{2} & 3 \end{array}]$
  Solution dikhao
  Answer $[\begin{array}{cc} (m - n)^{2} & 2 m \\ (m + n)^{2} & 4 \end{array}]$
  $m^{2} + n^{2} - 2 m n = (m - n)^{2}$ , $(m - n) + (m + n) = 2 m$ , $2 m n + m^{2} + n^{2} = (m + n)^{2}$ aur $1 + 3 = 4$ .
3. $[\begin{array}{ccc} 2 & - 5 & 7 \\ 0 & 4 & - 1 \\ 3 & 6 & 8 \end{array}] + [\begin{array}{ccc} - 2 & 3 & 1 \\ 5 & - 4 & 9 \\ 1 & 0 & - 8 \end{array}]$
  Solution dikhao
  Answer $[\begin{array}{ccc} 0 & - 2 & 8 \\ 5 & 0 & 8 \\ 4 & 6 & 0 \end{array}]$
  Position by position add karne par $[\begin{array}{ccc} 0 & - 2 & 8 \\ 5 & 0 & 8 \\ 4 & 6 & 0 \end{array}]$ milta hai; zero diagonal par dhyaan do.
4. $[\begin{array}{cc} \sec^{2} x & 1 \\ \tan^{2} x & {cosec}^{2} x \end{array}] + [\begin{array}{cc} - \tan^{2} x & - \sec^{2} x \\ 1 & - \cot^{2} x \end{array}]$
  Solution dikhao
  Answer $[\begin{array}{cc} 1 & - \tan^{2} x \\ \sec^{2} x & 1 \end{array}]$
  $\sec^{2} x - \tan^{2} x = 1$ , $1 - \sec^{2} x = - \tan^{2} x$ , $\tan^{2} x + 1 = \sec^{2} x$ aur ${cosec}^{2} x - \cot^{2} x = 1$ use karo.
3
Short answerMediumNCERT Exercise 3.2, Q3
Har product nikaalo.
1. $[\begin{array}{cc} p & q \\ q & p \end{array}] [\begin{array}{cc} p & - q \\ - q & p \end{array}]$
  Solution dikhao
  Answer $[\begin{array}{cc} p^{2} - q^{2} & 0 \\ 0 & p^{2} - q^{2} \end{array}] = (p^{2} - q^{2}) I$
  Entry (1, 1): $p^{2} - q^{2}$ ; entry (1, 2): $- p q + q p = 0$ ; entry (2, 1): $q p - p q = 0$ ; entry (2, 2): $- q^{2} + p^{2}$ .
2. $[\begin{array}{c} 2 \\ - 1 \\ 3 \end{array}] [\begin{array}{ccc} 1 & 4 & - 2 \end{array}]$
  Solution dikhao
  Answer $[\begin{array}{ccc} 2 & 8 & - 4 \\ - 1 & - 4 & 2 \\ 3 & 12 & - 6 \end{array}]$
  $(3 \times 1) (1 \times 3)$ product $3 \times 3$ hota hai; har entry ek single product hai, row ka number times column ka number.
  $[\begin{array}{ccc} 2 & 8 & - 4 \\ - 1 & - 4 & 2 \\ 3 & 12 & - 6 \end{array}]$
3. $[\begin{array}{cc} 2 & - 1 \\ 1 & 3 \end{array}] [\begin{array}{ccc} 1 & 0 & 2 \\ - 2 & 1 & 3 \end{array}]$
  Solution dikhao
  Answer $[\begin{array}{ccc} 4 & - 1 & 1 \\ - 5 & 3 & 11 \end{array}]$
  $[\begin{array}{ccc} 2 + 2 & 0 - 1 & 4 - 3 \\ 1 - 6 & 0 + 3 & 2 + 9 \end{array}] = [\begin{array}{ccc} 4 & - 1 & 1 \\ - 5 & 3 & 11 \end{array}]$ .
4. $[\begin{array}{ccc} 1 & 2 & 0 \\ 0 & - 1 & 3 \\ 2 & 1 & 1 \end{array}] [\begin{array}{ccc} 2 & - 1 & 1 \\ 1 & 0 & 3 \\ 0 & 2 & - 1 \end{array}]$
  Solution dikhao
  Answer $[\begin{array}{ccc} 4 & - 1 & 7 \\ - 1 & 6 & - 6 \\ 5 & 0 & 4 \end{array}]$
  Row 1: $(2 + 2 + 0, - 1 + 0 + 0, 1 + 6 + 0)$ . Row 2: $(0 - 1 + 0, 0 + 0 + 6, 0 - 3 - 3)$ . Row 3: $(4 + 1 + 0, - 2 + 0 + 2, 2 + 3 - 1)$ .
  $[\begin{array}{ccc} 4 & - 1 & 7 \\ - 1 & 6 & - 6 \\ 5 & 0 & 4 \end{array}]$
5. $[\begin{array}{cc} 1 & 3 \\ 0 & - 2 \\ 4 & 1 \end{array}] [\begin{array}{ccc} 2 & 0 & - 1 \\ 1 & 5 & 2 \end{array}]$
  Solution dikhao
  Answer $[\begin{array}{ccc} 5 & 15 & 5 \\ - 2 & - 10 & - 4 \\ 9 & 5 & - 2 \end{array}]$
  $(3 \times 2) (2 \times 3)$ se $3 \times 3$ milta hai: row 1 $(2 + 3, 0 + 15, - 1 + 6)$ , row 2 $(0 - 2, 0 - 10, 0 - 4)$ , row 3 $(8 + 1, 0 + 5, - 4 + 2)$ .
  $[\begin{array}{ccc} 5 & 15 & 5 \\ - 2 & - 10 & - 4 \\ 9 & 5 & - 2 \end{array}]$
6. $[\begin{array}{ccc} 3 & 0 & - 1 \\ 1 & 2 & 4 \end{array}] [\begin{array}{cc} 1 & 2 \\ 0 & - 3 \\ 5 & 1 \end{array}]$
  Solution dikhao
  Answer $[\begin{array}{cc} - 2 & 5 \\ 21 & 0 \end{array}]$
  $[\begin{array}{cc} 3 + 0 - 5 & 6 + 0 - 1 \\ 1 + 0 + 20 & 2 - 6 + 4 \end{array}] = [\begin{array}{cc} - 2 & 5 \\ 21 & 0 \end{array}]$ .
Short answerMediumNCERT Exercise 3.2, Q4
Maan lo $A = [\begin{array}{ccc} 2 & - 1 & 0 \\ 3 & 4 & - 2 \\ 1 & 0 & 5 \end{array}]$ , $B = [\begin{array}{ccc} 1 & 3 & - 1 \\ 0 & - 2 & 4 \\ 2 & 1 & 1 \end{array}]$ aur $C = [\begin{array}{ccc} - 1 & 2 & 3 \\ 4 & 0 & 1 \\ 2 & - 3 & 0 \end{array}]$ . $A + B$ aur $B - C$ nikaalo, aur verify karo ki $A + (B - C) = (A + B) - C$ .
Solution dikhao
AnswerDono sides $[\begin{array}{ccc} 4 & 0 & - 4 \\ - 1 & 2 & 1 \\ 1 & 4 & 6 \end{array}]$ ke barabar hain.
$\begin{array}{c} A + B = [\begin{array}{ccc} 3 & 2 & - 1 \\ 3 & 2 & 2 \\ 3 & 1 & 6 \end{array}] \\ B - C = [\begin{array}{ccc} 2 & 1 & - 4 \\ - 4 & - 2 & 3 \\ 0 & 4 & 1 \end{array}] \end{array}$
$\begin{array}{c} A + (B - C) = [\begin{array}{ccc} 4 & 0 & - 4 \\ - 1 & 2 & 1 \\ 1 & 4 & 6 \end{array}] \\ (A + B) - C = [\begin{array}{ccc} 4 & 0 & - 4 \\ - 1 & 2 & 1 \\ 1 & 4 & 6 \end{array}] \end{array}$
Results match karte hain, jaisa associative law ( $B - C = B + (- C)$ ke saath) predict karta hai.
Short answerMediumNCERT Exercise 3.2, Q5
Agar $A = [\begin{array}{ccc} \frac{1}{2} & \frac{3}{4} & - 1 \\ \frac{5}{4} & 0 & \frac{3}{2} \end{array}]$ aur $B = [\begin{array}{ccc} \frac{1}{3} & \frac{1}{6} & - \frac{2}{3} \\ \frac{1}{2} & \frac{5}{6} & 0 \end{array}]$ , to $4 A - 6 B$ nikaalo.
Solution dikhao
Answer $[\begin{array}{ccc} 0 & 2 & 0 \\ 2 & - 5 & 6 \end{array}]$
$\begin{array}{c} 4 A = [\begin{array}{ccc} 2 & 3 & - 4 \\ 5 & 0 & 6 \end{array}] \\ 6 B = [\begin{array}{ccc} 2 & 1 & - 4 \\ 3 & 5 & 0 \end{array}] \end{array}$
Subtract karne par $4 A - 6 B = [\begin{array}{ccc} 0 & 2 & 0 \\ 2 & - 5 & 6 \end{array}]$ .
Short answerMediumNCERT Exercise 3.2, Q6
$\sec θ [\begin{array}{cc} \sec θ & \tan θ \\ \tan θ & \sec θ \end{array}] - \tan θ [\begin{array}{cc} \tan θ & \sec θ \\ \sec θ & \tan θ \end{array}]$ ko simplify karo.
Solution dikhao
Answer $I_{2} = [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}]$
$\begin{array}{c} [\begin{array}{cc} \sec^{2} θ - \tan^{2} θ & \sec θ \tan θ - \tan θ \sec θ \\ \sec θ \tan θ - \tan θ \sec θ & \sec^{2} θ - \tan^{2} θ \end{array}] \\ = [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}] \end{array}$
7
Short answerMediumNCERT Exercise 3.2, Q7
$X$ aur $Y$ nikaalo.
1. $X + Y = [\begin{array}{cc} 6 & - 2 \\ 4 & 8 \end{array}]$ aur $X - Y = [\begin{array}{cc} 2 & 4 \\ - 2 & 0 \end{array}]$
  Solution dikhao
  Answer $X = [\begin{array}{cc} 4 & 1 \\ 1 & 4 \end{array}]$ , $Y = [\begin{array}{cc} 2 & - 3 \\ 3 & 4 \end{array}]$
  Add karne par $2 X = [\begin{array}{cc} 8 & 2 \\ 2 & 8 \end{array}]$ , isliye $X = [\begin{array}{cc} 4 & 1 \\ 1 & 4 \end{array}]$ . Subtract karne par $2 Y = [\begin{array}{cc} 4 & - 6 \\ 6 & 8 \end{array}]$ , isliye $Y = [\begin{array}{cc} 2 & - 3 \\ 3 & 4 \end{array}]$ .
2. $3 X + 2 Y = [\begin{array}{cc} 7 & 0 \\ 4 & 11 \end{array}]$ aur $2 X + Y = [\begin{array}{cc} 4 & 1 \\ 3 & 6 \end{array}]$
  Solution dikhao
  Answer $X = [\begin{array}{cc} 1 & 2 \\ 2 & 1 \end{array}]$ , $Y = [\begin{array}{cc} 2 & - 3 \\ - 1 & 4 \end{array}]$
  Doosri equation ko double karo: $4 X + 2 Y = [\begin{array}{cc} 8 & 2 \\ 6 & 12 \end{array}]$ . Pehli ko ghataao: $X = [\begin{array}{cc} 8 - 7 & 2 - 0 \\ 6 - 4 & 12 - 11 \end{array}] = [\begin{array}{cc} 1 & 2 \\ 2 & 1 \end{array}]$ .
  Phir $Y = [\begin{array}{cc} 4 & 1 \\ 3 & 6 \end{array}] - 2 X = [\begin{array}{cc} 2 & - 3 \\ - 1 & 4 \end{array}]$ .
Short answerEasyNCERT Exercise 3.2, Q8
$X$ nikaalo agar $Y = [\begin{array}{cc} 1 & - 2 \\ 2 & 4 \end{array}]$ aur $3 X - Y = [\begin{array}{cc} 2 & 1 \\ - 4 & 5 \end{array}]$ .
Solution dikhao
Answer $[\begin{array}{cc} 1 & - \frac{1}{3} \\ - \frac{2}{3} & 3 \end{array}]$
$3 X = [\begin{array}{cc} 2 & 1 \\ - 4 & 5 \end{array}] + [\begin{array}{cc} 1 & - 2 \\ 2 & 4 \end{array}] = [\begin{array}{cc} 3 & - 1 \\ - 2 & 9 \end{array}]$ , isliye $X = [\begin{array}{cc} 1 & - \frac{1}{3} \\ - \frac{2}{3} & 3 \end{array}]$ .
Short answerEasyNCERT Exercise 3.2, Q9
$x$ aur $y$ nikaalo agar $3 [\begin{array}{cc} 1 & x \\ y & 0 \end{array}] + [\begin{array}{cc} 2 & 1 \\ - 1 & 4 \end{array}] = [\begin{array}{cc} 5 & 7 \\ 8 & 4 \end{array}]$ .
Solution dikhao
Answer $x = 2$ , $y = 3$ .
Position (1, 2): $3 x + 1 = 7$ , isliye $x = 2$ . Position (2, 1): $3 y - 1 = 8$ , isliye $y = 3$ . Baaki positions match karti hain: $3 + 2 = 5$ , $0 + 4 = 4$ .
Short answerEasyNCERT Exercise 3.2, Q10
$x$ , $y$ , $z$ , $t$ ke liye solve karo: $4 [\begin{array}{cc} x & y \\ z & t \end{array}] - 2 [\begin{array}{cc} 1 & 3 \\ - 2 & 1 \end{array}] = 2 [\begin{array}{cc} 3 & 3 \\ 4 & 5 \end{array}]$ .
Solution dikhao
Answer $x = 2$ , $y = 3$ , $z = 1$ , $t = 3$ .
$4 x - 2 = 6 \Rightarrow x = 2$ ; $4 y - 6 = 6 \Rightarrow y = 3$ ; $4 z + 4 = 8 \Rightarrow z = 1$ ; $4 t - 2 = 10 \Rightarrow t = 3$ .
Short answerEasyNCERT Exercise 3.2, Q11
Agar $x [\begin{array}{c} 3 \\ 1 \end{array}] + y [\begin{array}{c} - 1 \\ 2 \end{array}] = [\begin{array}{c} 7 \\ 7 \end{array}]$ , to $x$ aur $y$ nikaalo.
Solution dikhao
Answer $x = 3$ , $y = 2$ .
$3 x - y = 7$ aur $x + 2 y = 7$ . Pehli ko double karke jodne par: $7 x = 21$ , isliye $x = 3$ aur $y = 2$ .
Short answerMediumNCERT Exercise 3.2, Q12
Diya hai $2 [\begin{array}{cc} x & y \\ z & w \end{array}] = [\begin{array}{cc} x & 5 \\ 3 & w \end{array}] + [\begin{array}{cc} 3 & y - x \\ w + z & - 4 \end{array}]$ ; $x$ , $y$ , $z$ aur $w$ nikaalo.
Solution dikhao
Answer $x = 3$ , $y = 2$ , $z = - 1$ , $w = - 4$ .
(1, 1): $2 x = x + 3$ , isliye $x = 3$ . (1, 2): $2 y = 5 + y - x = y + 2$ , isliye $y = 2$ . (2, 2): $2 w = w - 4$ , isliye $w = - 4$ . (2, 1): $2 z = 3 + w + z$ , isliye $z = 3 + w = - 1$ .
ProofMediumNCERT Exercise 3.2, Q13
Real number $t$ ke liye maan lo $G (t) = [\begin{array}{ccc} 1 & t & \frac{1}{2} t^{2} \\ 0 & 1 & t \\ 0 & 0 & 1 \end{array}]$ . Dikhao ki $G (s) G (t) = G (s + t)$ .
Solution dikhao
$G (s) G (t) = [\begin{array}{ccc} 1 & t + s & \frac{1}{2} t^{2} + s t + \frac{1}{2} s^{2} \\ 0 & 1 & t + s \\ 0 & 0 & 1 \end{array}]$
Kyunki $\frac{1}{2} t^{2} + s t + \frac{1}{2} s^{2} = \frac{1}{2} (s + t)^{2}$ , product bilkul $G (s + t)$ hai. (Baaki entries 0 ya 1 hain kyunki dono factors upper triangular hain aur diagonal par 1s hain.)
14
Short answerMediumNCERT Exercise 3.2, Q14
Har case mein dikhao ki $A B \neq B A$ .
1. $A = [\begin{array}{cc} 2 & 1 \\ - 3 & 4 \end{array}]$ , $B = [\begin{array}{cc} 0 & 1 \\ 1 & 2 \end{array}]$
  Solution dikhao
  Answer $A B = [\begin{array}{cc} 1 & 4 \\ 4 & 5 \end{array}]$ , $B A = [\begin{array}{cc} - 3 & 4 \\ - 4 & 9 \end{array}]$
  $A B = [\begin{array}{cc} 1 & 4 \\ 4 & 5 \end{array}]$ jabki $B A = [\begin{array}{cc} - 3 & 4 \\ - 4 & 9 \end{array}]$ ; (1, 1) entries hi alag hain (1 aur $- 3$ ).
2. $A = [\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 1 & - 1 \\ 3 & 1 & 0 \end{array}]$ , $B = [\begin{array}{ccc} 0 & 1 & 1 \\ 2 & 0 & 1 \\ 1 & - 1 & 0 \end{array}]$
  Solution dikhao
  Answer $A B = [\begin{array}{ccc} 2 & - 1 & 1 \\ 1 & 1 & 1 \\ 2 & 3 & 4 \end{array}]$ , $B A = [\begin{array}{ccc} 3 & 2 & - 1 \\ 5 & 1 & 4 \\ 1 & - 1 & 3 \end{array}]$
  $\begin{array}{c} A B = [\begin{array}{ccc} 2 & - 1 & 1 \\ 1 & 1 & 1 \\ 2 & 3 & 4 \end{array}] \\ B A = [\begin{array}{ccc} 3 & 2 & - 1 \\ 5 & 1 & 4 \\ 1 & - 1 & 3 \end{array}] \end{array}$
  (1, 1) entries 2 aur 3 hain, isliye $A B \neq B A$ .
Short answerMediumNCERT Exercise 3.2, Q15
$A = [\begin{array}{ccc} 1 & 2 & 0 \\ 0 & 1 & 3 \\ 2 & 0 & 1 \end{array}]$ ke liye $A^{2} - 4 A + 3 I$ nikaalo.
Solution dikhao
Answer $[\begin{array}{ccc} 0 & - 4 & 6 \\ 6 & 0 & - 6 \\ - 4 & 4 & 0 \end{array}]$
$A^{2} = [\begin{array}{ccc} 1 & 4 & 6 \\ 6 & 1 & 6 \\ 4 & 4 & 1 \end{array}]$
$A^{2} - 4 A + 3 I = [\begin{array}{ccc} 1 & 4 & 6 \\ 6 & 1 & 6 \\ 4 & 4 & 1 \end{array}] - [\begin{array}{ccc} 4 & 8 & 0 \\ 0 & 4 & 12 \\ 8 & 0 & 4 \end{array}] + [\begin{array}{ccc} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array}] = [\begin{array}{ccc} 0 & - 4 & 6 \\ 6 & 0 & - 6 \\ - 4 & 4 & 0 \end{array}]$ .
ProofHardNCERT Exercise 3.2, Q16
Agar $A = [\begin{array}{ccc} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 0 & 1 & 2 \end{array}]$ , to prove karo ki $A^{3} - 4 A^{2} + 5 A - 4 I = O$ .
Solution dikhao
$\begin{array}{c} A^{2} = [\begin{array}{ccc} 1 & 1 & 3 \\ 4 & 1 & 2 \\ 2 & 3 & 4 \end{array}] \\ \begin{aligned} A^{3} & = A \cdot A^{2} \\ = [\begin{array}{ccc} 3 & 4 & 7 \\ 6 & 3 & 8 \\ 8 & 7 & 10 \end{array}] \end{aligned} \end{array}$
$\begin{aligned} A^{3} - 4 A^{2} + 5 A - 4 I & = [\begin{array}{ccc} 3 & 4 & 7 \\ 6 & 3 & 8 \\ 8 & 7 & 10 \end{array}] - [\begin{array}{ccc} 4 & 4 & 12 \\ 16 & 4 & 8 \\ 8 & 12 & 16 \end{array}] + [\begin{array}{ccc} 5 & 0 & 5 \\ 10 & 5 & 0 \\ 0 & 5 & 10 \end{array}] - [\begin{array}{ccc} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{array}] \\ = O \end{aligned}$
Har position cancel ho jaati hai. Jaise position (1, 1) mein $3 - 4 (1) + 5 (1) - 4 = 0$ ; position (2, 3) mein $8 - 4 (2) + 5 (0) - 0 = 0$ .
Short answerMediumNCERT Exercise 3.2, Q17
$A = [\begin{array}{cc} 4 & 2 \\ - 1 & 1 \end{array}]$ ke liye aise numbers $p$ aur $q$ nikaalo ki $A^{2} = p A + q I$ .
Solution dikhao
Answer $p = 5$ , $q = - 6$ .
$A^{2} = [\begin{array}{cc} 14 & 10 \\ - 5 & - 1 \end{array}]$ aur $p A + q I = [\begin{array}{cc} 4 p + q & 2 p \\ - p & p + q \end{array}]$ .
Position (1, 2): $2 p = 10$ , isliye $p = 5$ . Position (1, 1): $20 + q = 14$ , isliye $q = - 6$ . Check (2, 2): $p + q = - 1$ , jo match karta hai.
18
ProofHardNCERT Exercise 3.2, Q18
Maan lo $t = \tan β$ ( $\cos β \neq 0$ ke saath) aur $A = [\begin{array}{cc} 0 & - t \\ t & 0 \end{array}]$ .
1. Dikhao ki $(I + A) (I - A) = (I - A) (I + A) = \sec^{2} β I$ .
  Solution dikhao
  $A^{2} = [\begin{array}{cc} - t^{2} & 0 \\ 0 & - t^{2} \end{array}] = - t^{2} I$ . Kyunki $A$ $I$ ke saath commute karta hai, dono products $I - A^{2} = (1 + t^{2}) I = \sec^{2} β I$ ke barabar hain.
2. Dikhao ki $\cos^{2} β (I + A)^{2} = [\begin{array}{cc} \cos 2 β & - \sin 2 β \\ \sin 2 β & \cos 2 β \end{array}]$ .
  Solution dikhao
  $(I + A)^{2} = I + 2 A + A^{2} = (1 - t^{2}) I + 2 A = [\begin{array}{cc} 1 - t^{2} & - 2 t \\ 2 t & 1 - t^{2} \end{array}]$ .
  $\cos^{2} β$ se multiply karo: $\cos^{2} β (1 - t^{2}) = \cos^{2} β - \sin^{2} β = \cos 2 β$ aur $2 t \cos^{2} β = 2 \sin β \cos β = \sin 2 β$ , jisse $2 β$ ka rotation matrix milta hai.
3. $β = \frac{π}{4}$ ke liye part (b) ko numbers se check karo.
  Solution dikhao
  AnswerDono sides $[\begin{array}{cc} 0 & - 1 \\ 1 & 0 \end{array}]$ ke barabar hain.
  Yahan $t = 1$ aur $\cos^{2} β = \frac{1}{2}$ , isliye $(I + A)^{2} = {[\begin{array}{cc} 1 & - 1 \\ 1 & 1 \end{array}]}^{2} = [\begin{array}{cc} 0 & - 2 \\ 2 & 0 \end{array}]$ aur iska aadha $[\begin{array}{cc} 0 & - 1 \\ 1 & 0 \end{array}]$ hai, yaani $\frac{π}{2}$ ka rotation.
19
NumericalMediumNCERT Exercise 3.2, Q19
Ek cooperative society ke paas ₹50,000 hain jo use do schemes mein deposit karne hain, jo saal ka 6 % aur 8 % simple interest deti hain. Amounts ko row $[x y]$ aur rates ko column $[\begin{array}{c} 0.06 \\ 0.08 \end{array}]$ ki tarah likh kar nikaalo ki paisa kaise baantein taaki saal ka interest (a) ₹3400, (b) ₹3700 mile.
1. Interest ₹3400: 8 % par amount (₹ mein)?
  ₹
  Answer dekho
  Answer 20000 ₹
  $0.06 x + 0.08 y = 3400$ aur $x = 50000 - y$ se $3000 + 0.02 y = 3400$ milta hai, isliye 8 % par $y = 20000$ aur 6 % par $x = 30000$ .
2. Interest ₹3700: 8 % par amount (₹ mein)?
  ₹
  Answer dekho
  Answer 35000 ₹
  $3000 + 0.02 y = 3700$ se 8 % par $y = 35000$ aur 6 % par $x = 15000$ milta hai.
NumericalEasyNCERT Exercise 3.2, Q20
Naye term se pehle ek stationery shop apna poora stock bech deti hai: 10 geometry boxes ke 7 packs ₹120 per box, 10 notebooks ke 12 packs ₹45 per notebook aur 10 water bottles ke 5 packs ₹150 per bottle. Quantities ka row matrix aur prices ka column matrix use karke uski total kamaai (₹ mein) nikaalo.
₹
Answer dekho
Answer 21300 ₹
$[70 120 50] [\begin{array}{c} 120 \\ 45 \\ 150 \end{array}] = [8400 + 5400 + 7500] = [21300]$ , yaani ₹21,300.
Quick answerMediumNCERT Exercise 3.2, Q21
$A$ $2 \times 3$ hai, $B$ $3 \times n$ hai aur $C$ $m \times 2$ hai. Product $C A B$ defined aur square tabhi hai jab
$m = 2$ aur $n = 3$ $m = n$ (koi bhi value) $m = 3$ aur $n = 2$ kabhi nahi
Answer dekho
Answer (B) $m = n$ (koi bhi value)
$C A$ $(m \times 2) (2 \times 3) = m \times 3$ hai, aur $(C A) B$ $(m \times 3) (3 \times n) = m \times n$ hai. Har $m$ aur $n$ ke liye dono products exist karte hain; result square tab hai jab $m = n$ ho.
Quick answerMediumNCERT Exercise 3.2, Q22
Order $2 \times 3$ ke $A$ aur order $3 \times 2$ ke $B$ ke saath kaun sa expression defined hai?
$A B + I_{2}$ $A + B$ $B A + I_{2}$ $A B + B A$
Answer dekho
Answer (A) $A B + I_{2}$
$A B$ $2 \times 2$ hai, isliye $A B + I_{2}$ defined hai. $B A$ $3 \times 3$ hai, isliye use $I_{2}$ ya $A B$ mein add nahi kar sakte, aur $A + B$ ek $2 \times 3$ ko $3 \times 2$ matrix mein jodta hai.

3.5 Matrix ka Transpose

Section 3.2 mein canteen record ko din rows ki tarah ya din columns ki tarah likh sakte the. Ek arrangement se doosre mein jaana khud ek operation hai, aur isse do special tarah ke square matrices milte hain.

Definition

Transpose

$m \times n$ matrix $A = [a_{i j}]$ ka transpose $n \times m$ matrix $A^{'} = [a_{j i}]$ hai, jise $A^{T}$ bhi likhte hain. Iski rows $A$ ke columns hain, same order mein.

Example ${[\begin{array}{ccc} 4 & - 1 & 7 \\ 0 & 5 & 2 \end{array}]}^{'} = [\begin{array}{cc} 4 & 0 \\ - 1 & 5 \\ 7 & 2 \end{array}]$

3/3Row 2 column 2 ban jaati hai: $A^{'}$ $3 \times 2$ hai.

Fig. 3.10Transpose karne se row

i

column

i

ban jaati hai:

(i, j)

entry position

(j, i)

par chali jaati hai.NCERT Definition 3 · p. 61

Left mein matrix A hai jiski row 1 mein 4, minus 1, 7 aur row 2 mein 0, 5, 2 hain, dono rows alag-alag rang mein. Right mein A prime ke liye khaali bracket hai. Step 1 sirf A dikhata hai. Step 2 mein row 1 ki teen entries khisak kar A prime ka pehla column ban jaati hain, upar se neeche. Step 3 mein row 2 ki entries khisak kar doosra column banati hain, aur 3 by 2 transpose poora hota hai jiski rows 4, 0; minus 1, 5; 7, 2 hain. A mein halki copies peeche rehti hain.

Symbols mein, $A^{'}$ ki $(i, j)$ entry $A$ ki $(j, i)$ entry hai. Row matrix transpose hokar column matrix banta hai aur column matrix row matrix, isiliye point ko dono tarah likh sakte hain. Square matrix ke transpose ka order same rehta hai, aur uska diagonal bilkul wahin rehta hai.

3.5.1 Matrices ke transpose ki properties

PropertyTransposes ke rules

Maan lo $A$ aur $B$ ke orders aise hain ki neeche ka har expression meaningful hai, aur $k$ ek scalar hai. Transpose sums se guzarta hai, $(A + B)^{'} = A^{'} + B^{'}$ , aur scalar multiples se bhi, $(k A)^{'} = k A^{'}$ . Do baar karne par kuch nahi badalta, $(A^{'})^{'} = A$ . Product par yeh order ulta kar deta hai: $(A B)^{'} = B^{'} A^{'}$ .

Pehle teen rules ke liye entries compare karo: entries ko add ya scale karna is par depend nahi karta ki woh kahan baithi hain, aur do baar transpose karne se har entry apni jagah wapas aa jaati hai.

Product rule ke liye maan lo $A$ $m \times n$ hai aur $B$ $n \times p$ , taaki $(A B)^{'}$ aur $B^{'} A^{'}$ dono $p \times m$ hon. $(A B)^{'}$ ki $(k, i)$ entry $A B$ ki $(i, k)$ entry hai, jo $\sum_{j} a_{i j} b_{j k}$ hai. $B^{'} A^{'}$ ki $(k, i)$ entry $\sum_{j} (B^{'})_{k j} (A^{'})_{j i} = \sum_{j} b_{j k} a_{i j}$ hai, wahi sum. Isliye $(A B)^{'} = B^{'} A^{'}$ .

Product rule order ulta karta hai, aur use karna hi padta hai: $A^{'} B^{'}$ mein orders $(n \times m) (p \times n)$ hain, jo aksar fit hi nahi hote. Rules lambe products par bhi lagte hain, jaise $(A B C)^{'} = C^{'} B^{'} A^{'}$ .

Example 20NCERT Example 20 · p. 61

Teen transpose rules check karna

$A = [\begin{array}{ccc} 1 & - 2 & 0 \\ 3 & 5 & - 4 \end{array}]$ aur $B = [\begin{array}{ccc} 2 & 1 & 6 \\ - 1 & 0 & 3 \end{array}]$ ke liye teen rules check karo: $A + B$ ka transpose $A^{'} + B^{'}$ hai; $- 3 A$ ka transpose $- 3 A^{'}$ hai; aur $A$ ko do baar transpose karne se $A$ wapas milta hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 21NCERT Example 21 · p. 63

Reversal rule kaam karte hue

$A = [\begin{array}{cc} 2 & - 1 \\ 0 & 3 \end{array}]$ aur $B = [\begin{array}{ccc} 1 & 4 & 0 \\ - 2 & 1 & 5 \end{array}]$ ke liye verify karo ki $(A B)^{'} = B^{'} A^{'}$ . Kya $A^{'} B^{'}$ defined hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

3.6 Symmetric aur Skew Symmetric Matrices

Socho ki square matrix ke main diagonal ke saath ek mirror rakha hai. Transpose bilkul mirror image hai: $a_{i j}$ $a_{j i}$ ke saath jagah badalta hai, aur diagonal wahin rehta hai. Do tarah ke square matrices is mirror par khaas tareeke se react karte hain (Fig. 3.11).

Definition

Symmetric matrix

Aisa square matrix $A$ jismein $A^{'} = A$ ho, yaani har $i$ aur $j$ ke liye $a_{j i} = a_{i j}$ .

Example $[\begin{array}{ccc} 2 & 5 & - 1 \\ 5 & 0 & 3 \\ - 1 & 3 & 4 \end{array}]$

Definition

Skew symmetric matrix

Aisa square matrix $A$ jismein $A^{'} = - A$ ho, yaani har $i$ aur $j$ ke liye $a_{j i} = - a_{i j}$ .

Example $[\begin{array}{ccc} 0 & 5 & - 1 \\ - 5 & 0 & 3 \\ 1 & - 3 & 0 \end{array}]$

3/3Partners ulte: $a_{j i} = - a_{i j}$ , aur har diagonal entry 0 hai.

Fig. 3.11Symmetric: mirror partners barabar hain. Skew symmetric: partners ek doosre ke negative hain, isliye diagonal zero hona hi chahiye.NCERT Definitions 4 aur 5 · p. 63

Ek 3 by 3 matrix jismein top-left se bottom-right corner tak dashed mirror line hai. Step 1 general matrix 3, 8, minus 2; 1, 6, 5; 4, 0, 7 dikhata hai, jismein curved arrows mirror partners jaise 8 aur 1 ko jodte hain. Step 2 symmetric matrix 2, 5, minus 1; 5, 0, 3; minus 1, 3, 4 dikhata hai jahan partners ki har jodi barabar hai aur ek jaisi shaded hai. Step 3 skew symmetric matrix 0, 5, minus 1; minus 5, 0, 3; 1, minus 3, 0 dikhata hai jahan partners ek doosre ke negative hain aur diagonal poora zeros hai.

Ek diagonal position dekho, jahan row aur column numbers same hain. Wahan skew condition maangti hai $a_{i i} = - a_{i i}$ , aur apne negative ke barabar sirf 0 hota hai, isliye skew symmetric matrix ka diagonal hamesha zero hota hai. Diagonal ke bahar mirror partners ek doosre ke negative hote hain. Jo matrix symmetric aur skew symmetric dono ho, uske liye $A = A^{'} = - A$ , isliye $2 A = O$ aur $A$ zero matrix hi hoga.

Symmetric matrices wahan dikhte hain jahan koi relationship dono taraf same kaam karta hai. Cities ke beech road distances ki table symmetric hoti hai, kyunki Pune se Nashik ki doori Nashik se Pune ki doori ke barabar hai; aisi hi woh table bhi jo record karti hai ki students ki kaun si jodiyan ek class share karti hain.

Theorem 1

Real entries wale har square matrix $A$ ke liye $A + A^{'}$ symmetric hai aur $A - A^{'}$ skew symmetric.

Theorem 2

Har square matrix ko ek symmetric matrix aur ek skew symmetric matrix ke sum ki tarah likha ja sakta hai, aur sirf ek hi tareeke se.

Square matrix ko split karna

A = \underset{symmetric}{\underset{⏟}{\frac{1}{2} (A + A^{'})}} + \underset{skew symmetric}{\underset{⏟}{\frac{1}{2} (A - A^{'})}}

Example 22NCERT Example 22 · p. 65

Matrix ko do halves mein split karna

$A = [\begin{array}{ccc} 3 & 5 & - 1 \\ 1 & - 2 & 4 \\ 7 & 0 & 6 \end{array}]$ ko symmetric part $P$ aur skew symmetric part $Q$ mein split karo, taaki $A = P + Q$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Practice · 12 questions

Practice 3.3: answers ke saath

Sections 3.5 aur 3.6 ki practice (NCERT Exercise 3.3 se match karti hai).

1
Short answerEasyNCERT Exercise 3.3, Q1
Har matrix ka transpose likho.
1. $[\begin{array}{cccc} - 2 & 4 & \frac{1}{3} & 7 \end{array}]$
  Solution dikhao
  Answer $[\begin{array}{c} - 2 \\ 4 \\ \frac{1}{3} \\ 7 \end{array}]$
  $1 \times 4$ row $4 \times 1$ column $[\begin{array}{c} - 2 \\ 4 \\ \frac{1}{3} \\ 7 \end{array}]$ ban jaati hai.
2. $[\begin{array}{cc} 4 & - 2 \\ 0 & 7 \end{array}]$
  Solution dikhao
  Answer $[\begin{array}{cc} 4 & 0 \\ - 2 & 7 \end{array}]$
  Diagonal 4, 7 wahin rehta hai; off-diagonal entries swap hoti hain: $[\begin{array}{cc} 4 & 0 \\ - 2 & 7 \end{array}]$ .
3. $[\begin{array}{cc} 1 & - 4 \\ 0 & 6 \\ - 5 & 2 \end{array}]$
  Solution dikhao
  Answer $[\begin{array}{ccc} 1 & 0 & - 5 \\ - 4 & 6 & 2 \end{array}]$
  $3 \times 2$ matrix se $2 \times 3$ matrix $[\begin{array}{ccc} 1 & 0 & - 5 \\ - 4 & 6 & 2 \end{array}]$ milta hai.
Short answerMediumNCERT Exercise 3.3, Q2
$A = [\begin{array}{ccc} 1 & 0 & - 3 \\ 2 & 4 & 1 \\ - 1 & 5 & 2 \end{array}]$ aur $B = [\begin{array}{ccc} 3 & - 2 & 0 \\ 1 & 1 & 4 \\ 0 & - 3 & 5 \end{array}]$ ke liye verify karo ki (a) $(A + B)^{'} = A^{'} + B^{'}$ aur (b) $(A - B)^{'} = A^{'} - B^{'}$ .
Solution dikhao
$\begin{array}{c} A + B = [\begin{array}{ccc} 4 & - 2 & - 3 \\ 3 & 5 & 5 \\ - 1 & 2 & 7 \end{array}] \\ \begin{aligned} (A + B)^{'} & = [\begin{array}{ccc} 4 & 3 & - 1 \\ - 2 & 5 & 2 \\ - 3 & 5 & 7 \end{array}] \\ = A^{'} + B^{'} \end{aligned} \end{array}$
$\begin{array}{c} A - B = [\begin{array}{ccc} - 2 & 2 & - 3 \\ 1 & 3 & - 3 \\ - 1 & 8 & - 3 \end{array}] \\ \begin{aligned} (A - B)^{'} & = [\begin{array}{ccc} - 2 & 1 & - 1 \\ 2 & 3 & 8 \\ - 3 & - 3 & - 3 \end{array}] \\ = A^{'} - B^{'} \end{aligned} \end{array}$
Yahan $A^{'} = [\begin{array}{ccc} 1 & 2 & - 1 \\ 0 & 4 & 5 \\ - 3 & 1 & 2 \end{array}]$ aur $B^{'} = [\begin{array}{ccc} 3 & 1 & 0 \\ - 2 & 1 & - 3 \\ 0 & 4 & 5 \end{array}]$ ; inhe add aur subtract karne se wahi do matrices milte hain.
Short answerMediumNCERT Exercise 3.3, Q3
Agar $A^{'} = [\begin{array}{cc} 2 & 0 \\ - 1 & 3 \\ 4 & 1 \end{array}]$ aur $B = [\begin{array}{ccc} 1 & 5 & - 2 \\ 0 & - 3 & 6 \end{array}]$ , to verify karo ki (a) $(A + B)^{'} = A^{'} + B^{'}$ aur (b) $(A - B)^{'} = A^{'} - B^{'}$ .
Solution dikhao
$A = (A^{'})^{'} = [\begin{array}{ccc} 2 & - 1 & 4 \\ 0 & 3 & 1 \end{array}]$ aur $B^{'} = [\begin{array}{cc} 1 & 0 \\ 5 & - 3 \\ - 2 & 6 \end{array}]$ .
$\begin{array}{c} \begin{aligned} (A + B)^{'} & = {[\begin{array}{ccc} 3 & 4 & 2 \\ 0 & 0 & 7 \end{array}]}^{'} \\ = [\begin{array}{cc} 3 & 0 \\ 4 & 0 \\ 2 & 7 \end{array}] \\ = A^{'} + B^{'} \end{aligned} \\ \begin{aligned} (A - B)^{'} & = [\begin{array}{cc} 1 & 0 \\ - 6 & 6 \\ 6 & - 5 \end{array}] \\ = A^{'} - B^{'} \end{aligned} \end{array}$
Short answerEasyNCERT Exercise 3.3, Q4
Agar $A^{'} = [\begin{array}{cc} 1 & - 2 \\ 3 & 0 \end{array}]$ aur $B = [\begin{array}{cc} 2 & 1 \\ - 1 & 4 \end{array}]$ , to $(2 A - B)^{'}$ nikaalo.
Solution dikhao
Answer $[\begin{array}{cc} 0 & - 3 \\ 5 & - 4 \end{array}]$
$(2 A - B)^{'} = 2 A^{'} - B^{'} = [\begin{array}{cc} 2 & - 4 \\ 6 & 0 \end{array}] - [\begin{array}{cc} 2 & - 1 \\ 1 & 4 \end{array}] = [\begin{array}{cc} 0 & - 3 \\ 5 & - 4 \end{array}]$ .
5
Short answerMediumNCERT Exercise 3.3, Q5
Har case mein verify karo ki $(A B)^{'} = B^{'} A^{'}$ .
1. $A = [\begin{array}{c} 3 \\ 0 \\ - 2 \end{array}]$ , $B = [\begin{array}{ccc} 4 & - 1 & 5 \end{array}]$
  Solution dikhao
  $\begin{array}{c} A B = [\begin{array}{ccc} 12 & - 3 & 15 \\ 0 & 0 & 0 \\ - 8 & 2 & - 10 \end{array}] \\ \begin{aligned} (A B)^{'} & = [\begin{array}{ccc} 12 & 0 & - 8 \\ - 3 & 0 & 2 \\ 15 & 0 & - 10 \end{array}] \\ = [\begin{array}{c} 4 \\ - 1 \\ 5 \end{array}] [\begin{array}{ccc} 3 & 0 & - 2 \end{array}] \\ = B^{'} A^{'} \end{aligned} \end{array}$
2. $A = [\begin{array}{cc} 1 & 2 \\ 0 & - 1 \end{array}]$ , $B = [\begin{array}{ccc} 3 & 1 & 0 \\ 2 & - 2 & 4 \end{array}]$
  Solution dikhao
  $\begin{array}{c} A B = [\begin{array}{ccc} 7 & - 3 & 8 \\ - 2 & 2 & - 4 \end{array}] \\ (A B)^{'} = [\begin{array}{cc} 7 & - 2 \\ - 3 & 2 \\ 8 & - 4 \end{array}] \end{array}$
  $\begin{aligned} B^{'} A^{'} & = [\begin{array}{cc} 3 & 2 \\ 1 & - 2 \\ 0 & 4 \end{array}] [\begin{array}{cc} 1 & 0 \\ 2 & - 1 \end{array}] \\ = [\begin{array}{cc} 7 & - 2 \\ - 3 & 2 \\ 8 & - 4 \end{array}] \end{aligned}$
6
ProofMediumNCERT Exercise 3.3, Q6
Har matrix ke liye verify karo ki $A^{'} A = I$ .
1. $A = [\begin{array}{cc} \cos φ & \sin φ \\ \sin φ & - \cos φ \end{array}]$ (ek reflection)
  Solution dikhao
  $A$ symmetric hai, isliye $A^{'} = A$ aur $A^{'} A = A^{2}$ :
  $\begin{aligned} A^{2} & = [\begin{array}{cc} \cos^{2} φ + \sin^{2} φ & \cos φ \sin φ - \sin φ \cos φ \\ \sin φ \cos φ - \cos φ \sin φ & \sin^{2} φ + \cos^{2} φ \end{array}] \\ = I \end{aligned}$
2. $A = [\begin{array}{cc} \frac{3}{5} & - \frac{4}{5} \\ \frac{4}{5} & \frac{3}{5} \end{array}]$
  Solution dikhao
  $A^{'} A = [\begin{array}{cc} \frac{3}{5} & \frac{4}{5} \\ - \frac{4}{5} & \frac{3}{5} \end{array}] [\begin{array}{cc} \frac{3}{5} & - \frac{4}{5} \\ \frac{4}{5} & \frac{3}{5} \end{array}] = [\begin{array}{cc} \frac{9 + 16}{25} & \frac{- 12 + 12}{25} \\ \frac{- 12 + 12}{25} & \frac{16 + 9}{25} \end{array}] = I$ .
7
Short answerEasyNCERT Exercise 3.3, Q7
Har matrix ko classify karo.
1. Dikhao ki $[\begin{array}{ccc} 2 & - 4 & 7 \\ - 4 & 0 & 1 \\ 7 & 1 & - 3 \end{array}]$ symmetric hai.
  Solution dikhao
  Partners match karte hain: $a_{12} = a_{21} = - 4$ , $a_{13} = a_{31} = 7$ , $a_{23} = a_{32} = 1$ . Isliye $A^{'} = A$ .
2. Dikhao ki $[\begin{array}{ccc} 0 & 2 & - 5 \\ - 2 & 0 & 3 \\ 5 & - 3 & 0 \end{array}]$ skew symmetric hai.
  Solution dikhao
  Diagonal poora 0 hai, aur partners ulte hain: $2$ aur $- 2$ , $- 5$ aur $5$ , $3$ aur $- 3$ . Isliye $A^{'} = - A$ .
Short answerEasyNCERT Exercise 3.3, Q8
$A = [\begin{array}{cc} 2 & - 3 \\ 4 & 1 \end{array}]$ ke liye verify karo ki (a) $A + A^{'}$ symmetric hai aur (b) $A - A^{'}$ skew symmetric hai.
Solution dikhao
$A^{'} = [\begin{array}{cc} 2 & 4 \\ - 3 & 1 \end{array}]$ , isliye $A + A^{'} = [\begin{array}{cc} 4 & 1 \\ 1 & 2 \end{array}]$ , jo apne transpose ke barabar hai (partners 1 aur 1).
$A - A^{'} = [\begin{array}{cc} 0 & - 7 \\ 7 & 0 \end{array}]$ : zero diagonal aur partners $- 7$ aur $7$ , isliye yeh skew symmetric hai.
Short answerMediumNCERT Exercise 3.3, Q9
General $2 \times 2$ matrix $A = [\begin{array}{cc} a & b \\ c & d \end{array}]$ ke liye $\frac{1}{2} (A + A^{'})$ aur $\frac{1}{2} (A - A^{'})$ nikaalo. Doosra kab $O$ hota hai?
Solution dikhao
Answer $[\begin{array}{cc} a & \frac{b + c}{2} \\ \frac{b + c}{2} & d \end{array}]$ aur $[\begin{array}{cc} 0 & \frac{b - c}{2} \\ \frac{c - b}{2} & 0 \end{array}]$ ; doosra O tabhi hai jab $b = c$ ho.
$A$ aur $A^{'}$ ka average diagonal ko rakhta hai aur har off-diagonal jodi ko uske mean $\frac{b + c}{2}$ se badal deta hai. Difference ke aadhe ka diagonal zero hai aur entries $\pm \frac{b - c}{2}$ hain.
Skew part tabhi zero hota hai jab $b = c$ , yaani jab $A$ pehle se symmetric ho.
10
Short answerMediumNCERT Exercise 3.3, Q10
Har matrix ko $P + Q$ ki tarah likho, jahan $P$ symmetric hai aur $Q$ skew symmetric.
1. $[\begin{array}{cc} 4 & 1 \\ - 3 & 2 \end{array}]$
  Solution dikhao
  Answer $[\begin{array}{cc} 4 & 1 \\ - 3 & 2 \end{array}] = [\begin{array}{cc} 4 & - 1 \\ - 1 & 2 \end{array}] + [\begin{array}{cc} 0 & 2 \\ - 2 & 0 \end{array}]$
  $\begin{array}{c} \begin{aligned} \frac{1}{2} (A + A^{'}) & = \frac{1}{2} [\begin{array}{cc} 8 & - 2 \\ - 2 & 4 \end{array}] \\ = [\begin{array}{cc} 4 & - 1 \\ - 1 & 2 \end{array}] \end{aligned} \\ \begin{aligned} \frac{1}{2} (A - A^{'}) & = \frac{1}{2} [\begin{array}{cc} 0 & 4 \\ - 4 & 0 \end{array}] \\ = [\begin{array}{cc} 0 & 2 \\ - 2 & 0 \end{array}] \end{aligned} \end{array}$
  Pehla part apne transpose ke barabar hai, doosra apne transpose ka negative hai, aur dono jud kar $A$ dete hain.
2. $[\begin{array}{ccc} 1 & 4 & - 2 \\ 0 & 3 & 5 \\ 6 & - 1 & 2 \end{array}]$
  Solution dikhao
  Answer $[\begin{array}{ccc} 1 & 4 & - 2 \\ 0 & 3 & 5 \\ 6 & - 1 & 2 \end{array}] = [\begin{array}{ccc} 1 & 2 & 2 \\ 2 & 3 & 2 \\ 2 & 2 & 2 \end{array}] + [\begin{array}{ccc} 0 & 2 & - 4 \\ - 2 & 0 & 3 \\ 4 & - 3 & 0 \end{array}]$
  $\begin{array}{c} \begin{aligned} \frac{1}{2} (A + A^{'}) & = \frac{1}{2} [\begin{array}{ccc} 2 & 4 & 4 \\ 4 & 6 & 4 \\ 4 & 4 & 4 \end{array}] \\ = [\begin{array}{ccc} 1 & 2 & 2 \\ 2 & 3 & 2 \\ 2 & 2 & 2 \end{array}] \end{aligned} \\ \begin{aligned} \frac{1}{2} (A - A^{'}) & = \frac{1}{2} [\begin{array}{ccc} 0 & 4 & - 8 \\ - 4 & 0 & 6 \\ 8 & - 6 & 0 \end{array}] \\ = [\begin{array}{ccc} 0 & 2 & - 4 \\ - 2 & 0 & 3 \\ 4 & - 3 & 0 \end{array}] \end{aligned} \end{array}$
  Pehla part apne transpose ke barabar hai, doosra apne transpose ka negative hai, aur dono jud kar $A$ dete hain.
3. $[\begin{array}{ccc} 2 & - 1 & 3 \\ 5 & 0 & - 4 \\ 1 & 2 & - 3 \end{array}]$
  Solution dikhao
  Answer $[\begin{array}{ccc} 2 & - 1 & 3 \\ 5 & 0 & - 4 \\ 1 & 2 & - 3 \end{array}] = [\begin{array}{ccc} 2 & 2 & 2 \\ 2 & 0 & - 1 \\ 2 & - 1 & - 3 \end{array}] + [\begin{array}{ccc} 0 & - 3 & 1 \\ 3 & 0 & - 3 \\ - 1 & 3 & 0 \end{array}]$
  $\begin{array}{c} \begin{aligned} \frac{1}{2} (A + A^{'}) & = \frac{1}{2} [\begin{array}{ccc} 4 & 4 & 4 \\ 4 & 0 & - 2 \\ 4 & - 2 & - 6 \end{array}] \\ = [\begin{array}{ccc} 2 & 2 & 2 \\ 2 & 0 & - 1 \\ 2 & - 1 & - 3 \end{array}] \end{aligned} \\ \begin{aligned} \frac{1}{2} (A - A^{'}) & = \frac{1}{2} [\begin{array}{ccc} 0 & - 6 & 2 \\ 6 & 0 & - 6 \\ - 2 & 6 & 0 \end{array}] \\ = [\begin{array}{ccc} 0 & - 3 & 1 \\ 3 & 0 & - 3 \\ - 1 & 3 & 0 \end{array}] \end{aligned} \end{array}$
  Pehla part apne transpose ke barabar hai, doosra apne transpose ka negative hai, aur dono jud kar $A$ dete hain.
4. $[\begin{array}{cc} 3 & - 2 \\ 6 & 1 \end{array}]$
  Solution dikhao
  Answer $[\begin{array}{cc} 3 & - 2 \\ 6 & 1 \end{array}] = [\begin{array}{cc} 3 & 2 \\ 2 & 1 \end{array}] + [\begin{array}{cc} 0 & - 4 \\ 4 & 0 \end{array}]$
  $\begin{array}{c} \begin{aligned} \frac{1}{2} (A + A^{'}) & = \frac{1}{2} [\begin{array}{cc} 6 & 4 \\ 4 & 2 \end{array}] \\ = [\begin{array}{cc} 3 & 2 \\ 2 & 1 \end{array}] \end{aligned} \\ \begin{aligned} \frac{1}{2} (A - A^{'}) & = \frac{1}{2} [\begin{array}{cc} 0 & - 8 \\ 8 & 0 \end{array}] \\ = [\begin{array}{cc} 0 & - 4 \\ 4 & 0 \end{array}] \end{aligned} \end{array}$
  Pehla part apne transpose ke barabar hai, doosra apne transpose ka negative hai, aur dono jud kar $A$ dete hain.
Quick answerMediumNCERT Exercise 3.3, Q11
Agar same order ke $A$ symmetric hai aur $B$ skew symmetric, to $A B - B A$ hai
symmetricskew symmetriczero matrixidentity matrix
Answer dekho
Answer (A) symmetric
$(A B - B A)^{'} = B^{'} A^{'} - A^{'} B^{'} = (- B) A - A (- B) = A B - B A$ , isliye yeh apne transpose ke barabar hai: symmetric.
Quick answerMediumNCERT Exercise 3.3, Q12
Agar $A = [\begin{array}{cc} \cos α & - \sin α \\ \sin α & \cos α \end{array}]$ aur $A + A^{'} = \sqrt{2} I$ , to $α$ ki ek value hai
$\frac{π}{6}$ $\frac{π}{3}$ $\frac{π}{2}$ $\frac{π}{4}$
Answer dekho
Answer (D) $\frac{π}{4}$
$A + A^{'} = [\begin{array}{cc} 2 \cos α & 0 \\ 0 & 2 \cos α \end{array}] = \sqrt{2} I$ se $\cos α = \frac{\sqrt{2}}{2}$ milta hai, isliye $α = \frac{π}{4}$ .

3.7 Invertible Matrices

Har non-zero number $a$ ka ek reciprocal $a^{- 1}$ hota hai jiske liye $a \cdot a^{- 1} = a^{- 1} \cdot a = 1$ . Matrix version mein 1 ki jagah identity $I$ aata hai, aur hum aisa matrix dhoondhte hain jo $A$ ko undo kare.

Definition

Invertible matrix

Order $m$ ka square matrix $A$ invertible hai jab same order ka koi square matrix $B$ $A B = B A = I$ de. Aisa $B$ $A$ ka inverse hai, jise $A^{- 1}$ likhte hain.

Example ${[\begin{array}{cc} 3 & 5 \\ 1 & 2 \end{array}]}^{- 1} = [\begin{array}{cc} 2 & - 5 \\ - 1 & 3 \end{array}]$

ExampleNCERT Definition 6 ke baad ka example · p. 68

Inverses ki ek jodi check karna

Dikhao ki $A = [\begin{array}{cc} 3 & 5 \\ 1 & 2 \end{array}]$ aur $B = [\begin{array}{cc} 2 & - 5 \\ - 1 & 3 \end{array}]$ ek doosre ke inverses hain.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Fig. 3.12 idea ko geometry se dikhata hai: ek matrix square ko shear karke parallelogram banata hai, aur uska inverse use seedha wapas shear kar deta hai.

2/3 $A$ har point ko uski height jitna right khiskata hai: $P \to (3, 2)$ .

Fig. 3.12

A^{- 1}

A

ko undo karta hai: pehle

A

aur phir

A^{- 1}

lagane se har point wahin rehta hai jahan se shuru hua tha.NCERT Definition 6 ka idea · p. 68

Ek grid jismein origin se 2 by 2 square bana hai, orientation dikhane ke liye ek chhota rangeen corner hai, aur uske upar wale edge par marked point P (1, 2) par hai. Step 1 square aur P dikhata hai. Step 2 A lagata hai, rows 1, 1 aur 0, 1 wala shear: har point apni height jitna right jaata hai, isliye square right ki taraf jhuka parallelogram ban jaata hai aur P (3, 2) par chala jaata hai. Step 3 A inverse lagata hai, rows 1, minus 1 aur 0, 1 ke saath, jo har point ko uski height jitna wapas left le jaata hai, aur square aur P (1, 2) par wapas aa jaate hain.

Definition se seedhe do baatein nikalti hain. Pehli, sirf square matrices ke inverses ho sakte hain: agar $A$ $m \times n$ hai jahan $m \neq n$ , to $A B$ aur $B A$ ke orders alag hain, isliye dono ek hi identity matrix ke barabar nahi ho sakte. Doosri, rishta dono taraf ka hai: equations $A B = B A = I$ utna hi yeh bhi kehti hain ki $A$ $B$ ko undo karta hai, isliye $B = A^{- 1}$ ke saath $A = B^{- 1}$ bhi hai.

Square aur non-zero hona kaafi nahi hai. Example 15 ke matrix $A = [\begin{array}{cc} 1 & 2 \\ 2 & 4 \end{array}]$ ka koi inverse nahi hai: wahan $A B = O$ hai jabki $B \neq O$ , aur ek inverse $C$ se $B = I B = (C A) B = C (A B) = C O = O$ milta, jo galat hai. Chapter 4 invertibility ka test (non-zero determinant) aur inverse ka formula deta hai.

Theorem 3Inverse ki uniqueness

Square matrix ka zyaada se zyaada ek inverse hota hai.

Theorem 4Product ka inverse

Same order ke do invertible square matrices ka product invertible hai, aur uska inverse inverses ka ulte order mein product hai: $(A B)^{- 1} = B^{- 1} A^{- 1}$ .

Practice · 3 questions

Practice 3.4: answers ke saath

Section 3.7 ki practice (Q1 NCERT Exercise 3.4 se match karta hai; Q2 aur Q3 extra hain).

Quick answerEasyNCERT Exercise 3.4, Q1
Order 2 ke square matrices $A$ aur $B$ ke liye kaun si condition guarantee karti hai ki $B = A^{- 1}$ ?
$A B = O$ $A B = B A$ $A + B = I$ $A B = B A = I$
Answer dekho
Answer (D) $A B = B A = I$
Definition ke liye dono products $I$ ke barabar hone chahiye. Sirf $A B = B A$ to tab bhi sahi hai, jaise jab $B = A$ ho.
Short answerEasy
Check karo ki $[\begin{array}{cc} 2 & 1 \\ 5 & 3 \end{array}]$ aur $[\begin{array}{cc} 3 & - 1 \\ - 5 & 2 \end{array}]$ ek doosre ke inverses hain ya nahi.
Solution dikhao
AnswerHaan: dono products $I$ ke barabar hain.
$[\begin{array}{cc} 2 & 1 \\ 5 & 3 \end{array}] [\begin{array}{cc} 3 & - 1 \\ - 5 & 2 \end{array}] = [\begin{array}{cc} 6 - 5 & - 2 + 2 \\ 15 - 15 & - 5 + 6 \end{array}] = I$ , aur doosre order mein product bhi $I$ hai.
Short answerMedium
Maan lo $A = [\begin{array}{cc} 1 & 2 \\ 0 & 1 \end{array}]$ aur $B = [\begin{array}{cc} 1 & 0 \\ 3 & 1 \end{array}]$ , jahan $A^{- 1} = [\begin{array}{cc} 1 & - 2 \\ 0 & 1 \end{array}]$ aur $B^{- 1} = [\begin{array}{cc} 1 & 0 \\ - 3 & 1 \end{array}]$ . $B^{- 1} A^{- 1}$ nikaalo aur check karo ki yeh $A B$ ka inverse hai. Kya $A^{- 1} B^{- 1}$ bhi $A B$ ka inverse hai?
Solution dikhao
Answer $B^{- 1} A^{- 1} = [\begin{array}{cc} 1 & - 2 \\ - 3 & 7 \end{array}]$ $A B = [\begin{array}{cc} 7 & 2 \\ 3 & 1 \end{array}]$ ka inverse hai; $A^{- 1} B^{- 1} = [\begin{array}{cc} 7 & - 2 \\ - 3 & 1 \end{array}]$ nahi hai.
$\begin{array}{c} A B = [\begin{array}{cc} 7 & 2 \\ 3 & 1 \end{array}] \\ B^{- 1} A^{- 1} = [\begin{array}{cc} 1 & - 2 \\ - 3 & 7 \end{array}] \end{array}$
$(A B) (B^{- 1} A^{- 1}) = [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}]$ aur $(B^{- 1} A^{- 1}) (A B) = [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}]$ , jaisa Theorem 4 predict karta hai.
Lekin $A^{- 1} B^{- 1} = [\begin{array}{cc} 7 & - 2 \\ - 3 & 1 \end{array}]$ , jo $B^{- 1} A^{- 1}$ se alag hai, isliye uniqueness (Theorem 3) se yeh $A B$ ka inverse nahi hai.

Miscellaneous examples

Aakhri teen examples chapter ke kai ideas ko jodte hain: mathematical induction ke saath matrix ki powers, products ke saath transposes, aur unknown matrix wali matrix equations.

Example 23NCERT Example 23 · p. 70

Rotation ki powers

Real number $θ$ ke liye maan lo $R (θ) = [\begin{array}{cc} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{array}]$ . Prove karo ki har natural number $n$ ke liye $R (θ)^{n} = R (n θ)$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Geometry mein, $R (θ)$ plane ke har point ko origin ke around angle $θ$ se anticlockwise ghumata hai, isliye ise $n$ baar lagane se $n θ$ ka turn hota hai (Fig. 3.13). $θ = 30 °$ ke liye baarah turns poora circle banate hain: $R (30 °)^{12} = R (360 °) = I$ .

4/4 $R (θ)^{3} = R (3 θ)$ : $90 °$ ka turn.

Fig. 3.13

R (θ)

se har multiplication angle mein

θ

jodta hai, isliye

R (θ)^{n} = R (n θ)

.NCERT Example 23 ka idea · p. 70

Origin par centre wala ek circle, axes ke saath. Step 1 positive x-axis ke saath ek arrow dikhata hai. Step 2 use R of theta se 30 degrees anticlockwise ghuma hua dikhata hai, start ki halki copy ke saath. Step 3 use 60 degrees ghuma hua dikhata hai, jis par R of theta squared barabar R of 2 theta likha hai. Step 4 use 90 degrees par dikhata hai, seedha upar point karta hua, jis par R of theta cubed barabar R of 3 theta likha hai. Chhote arcs har 30 degree turn mark karte hain.

Example 24NCERT Example 24 · p. 70

Symmetric matrices ka product kab symmetric hota hai?

Maan lo $A$ aur $B$ same order ke symmetric matrices hain. Prove karo ki $A B$ symmetric hai if and only if $A B = B A$ . $A = [\begin{array}{cc} 1 & 2 \\ 2 & 0 \end{array}]$ ke saath samjhao, pehle $B = [\begin{array}{cc} 0 & 1 \\ 1 & 3 \end{array}]$ ke saath aur phir $C = A + I$ ke saath.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 25NCERT Example 25 · p. 71

Unknown matrix nikaalna

Woh matrix $X$ nikaalo jiske liye $[\begin{array}{cc} 2 & 1 \\ 5 & 3 \end{array}] X - [\begin{array}{cc} 1 & 0 \\ 2 & 1 \end{array}] [\begin{array}{cc} 3 & 1 \\ 1 & 2 \end{array}] = O$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

2 mein se 1

R (θ)^{n} = R (n θ)

use karke batao,

R (30 °)^{6}

(1, 1)

entry kya hai?

Answer dekho

Answer -1

R (30 °)^{6} = R (180 °)

, jiski

(1, 1)

entry

\cos 180 ° = - 1

hai. (

30 °

ke chhe turns aadha turn banate hain.)

Recap

Summary aur key points

Matrix numbers ya functions ko ek rectangle mein rakhta hai; $m$ rows aur $n$ columns ke saath iska order $m \times n$ hai aur ismein $m n$ entries $a_{i j}$ hain (Section 3.2.1).
Row, column, square, diagonal, scalar, identity aur zero matrices ke naam unke shape ya pattern se hain; identity ⊂ scalar ⊂ diagonal ⊂ square (Section 3.3).
Do matrices equal hain jab unka order same ho aur corresponding entries barabar hon (Section 3.3.1).
Sums aur differences sirf same order ke matrices ke liye exist karte hain aur entry by entry chalte hain; $k A$ har entry ko $k$ se multiply karta hai (Section 3.4.1, Section 3.4.2).
Addition commutative aur associative hai, identity O aur inverse $- A$ ke saath; scalars sums par distribute hote hain (Section 3.4.3, Section 3.4.4).
$A B$ tab exist karta hai jab $A$ ke columns $B$ ki rows ke barabar hon: $(m \times n) (n \times p)$ se $m \times p$ milta hai, jahan $c_{i k} = \sum_{j} a_{i j} b_{j k}$ (Section 3.4.5).
Aam taur par $A B \neq B A$ , aur $A \neq O$ , $B \neq O$ ke saath bhi $A B = O$ ho sakta hai (Order matter karta hai: AB aur BA, Non-zero matrices se zero product).
Multiplication associative hai aur addition par distributive hai, aur $I A = A I = A$ (Section 3.4.6).
Transpose $A^{'} = [a_{j i}]$ ke liye $(A^{'})^{'} = A$ , $(k A)^{'} = k A^{'}$ , $(A + B)^{'} = A^{'} + B^{'}$ aur $(A B)^{'} = B^{'} A^{'}$ (Section 3.5.1).
$A$ symmetric hai agar $A^{'} = A$ aur skew symmetric agar $A^{'} = - A$ , aur tab uska diagonal zero hota hai (Section 3.6).
Har square matrix uniquely $\frac{1}{2} (A + A^{'}) + \frac{1}{2} (A - A^{'})$ mein split hota hai, symmetric plus skew symmetric (Theorem 2).
$B = A^{- 1}$ jab $A B = B A = I$ ; sirf square matrices ke inverses ho sakte hain, inverse unique hota hai, aur $(A B)^{- 1} = B^{- 1} A^{- 1}$ (Section 3.7).
Square matrix ki powers induction se nikaal sakte hain, jaise $R (θ)^{n} = R (n θ)$ mein (Example 23).

Flash cards · 10

Yaad rakhne wale rules

1 / 10

Miscellaneous exercises

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise 3.1, Q1	Practice 3.1 · Q1
NCERT Exercise 3.1, Q2	Practice 3.1 · Q2
NCERT Exercise 3.1, Q3	Practice 3.1 · Q3
NCERT Exercise 3.1, Q4	Practice 3.1 · Q4
NCERT Exercise 3.1, Q5	Practice 3.1 · Q5
NCERT Exercise 3.1, Q6	Practice 3.1 · Q6
NCERT Exercise 3.1, Q7	Practice 3.1 · Q7
NCERT Exercise 3.1, Q8	Practice 3.1 · Q8
NCERT Exercise 3.1, Q9	Practice 3.1 · Q9
NCERT Exercise 3.1, Q10	Practice 3.1 · Q10
NCERT Exercise 3.2, Q1	Practice 3.2 · Q1
NCERT Exercise 3.2, Q2	Practice 3.2 · Q2
NCERT Exercise 3.2, Q3	Practice 3.2 · Q3
NCERT Exercise 3.2, Q4	Practice 3.2 · Q4
NCERT Exercise 3.2, Q5	Practice 3.2 · Q5
NCERT Exercise 3.2, Q6	Practice 3.2 · Q6
NCERT Exercise 3.2, Q7	Practice 3.2 · Q7
NCERT Exercise 3.2, Q8	Practice 3.2 · Q8
NCERT Exercise 3.2, Q9	Practice 3.2 · Q9
NCERT Exercise 3.2, Q10	Practice 3.2 · Q10
NCERT Exercise 3.2, Q11	Practice 3.2 · Q11
NCERT Exercise 3.2, Q12	Practice 3.2 · Q12
NCERT Exercise 3.2, Q13	Practice 3.2 · Q13
NCERT Exercise 3.2, Q14	Practice 3.2 · Q14
NCERT Exercise 3.2, Q15	Practice 3.2 · Q15
NCERT Exercise 3.2, Q16	Practice 3.2 · Q16
NCERT Exercise 3.2, Q17	Practice 3.2 · Q17
NCERT Exercise 3.2, Q18	Practice 3.2 · Q18
NCERT Exercise 3.2, Q19	Practice 3.2 · Q19
NCERT Exercise 3.2, Q20	Practice 3.2 · Q20
NCERT Exercise 3.2, Q21	Practice 3.2 · Q21
NCERT Exercise 3.2, Q22	Practice 3.2 · Q22
NCERT Exercise 3.3, Q1	Practice 3.3 · Q1
NCERT Exercise 3.3, Q2	Practice 3.3 · Q2
NCERT Exercise 3.3, Q3	Practice 3.3 · Q3
NCERT Exercise 3.3, Q4	Practice 3.3 · Q4
NCERT Exercise 3.3, Q5	Practice 3.3 · Q5
NCERT Exercise 3.3, Q6	Practice 3.3 · Q6
NCERT Exercise 3.3, Q7	Practice 3.3 · Q7
NCERT Exercise 3.3, Q8	Practice 3.3 · Q8
NCERT Exercise 3.3, Q9	Practice 3.3 · Q9
NCERT Exercise 3.3, Q10	Practice 3.3 · Q10
NCERT Exercise 3.3, Q11	Practice 3.3 · Q11
NCERT Exercise 3.3, Q12	Practice 3.3 · Q12
NCERT Exercise 3.4, Q1	Practice 3.4 · Q1
NCERT Misc. Exercise 3, Q1	Miscellaneous exercises · Q1
NCERT Misc. Exercise 3, Q2	Miscellaneous exercises · Q2
NCERT Misc. Exercise 3, Q3	Miscellaneous exercises · Q3
NCERT Misc. Exercise 3, Q4	Miscellaneous exercises · Q4
NCERT Misc. Exercise 3, Q5	Miscellaneous exercises · Q5
NCERT Misc. Exercise 3, Q6	Miscellaneous exercises · Q6
NCERT Misc. Exercise 3, Q7	Miscellaneous exercises · Q7
NCERT Misc. Exercise 3, Q8	Miscellaneous exercises · Q8
NCERT Misc. Exercise 3, Q9	Miscellaneous exercises · Q9
NCERT Misc. Exercise 3, Q10	Miscellaneous exercises · Q10
NCERT Misc. Exercise 3, Q11	Miscellaneous exercises · Q11

Quick check: MCQs

3.1 Introduction2 questions
3.2.1 Matrix ka order4 questions
3.3.1 Matrices ki equality4 questions
3.4.4 Matrix ke scalar multiplication ki properties4 questions
3.4.6 Matrices ke multiplication ki properties4 questions
3.5.1 Matrices ke transpose ki properties4 questions
3.6 Symmetric aur Skew Symmetric Matrices4 questions
3.7 Invertible Matrices3 questions
Miscellaneous examples2 questions

Glossary · 14

Key terms (English + Hindi)

diagonal matrixविकर्ण आव्यूहvikarn aavyooh: Square matrix jiski main diagonal ke bahar ki saari entries 0 hain.Chapter mein pehli baar yahan aaya →

elementअवयवavayav: Matrix mein ek number ya function; $a_{i j}$ row $i$ aur column $j$ mein baithta hai.Chapter mein pehli baar yahan aaya →
equal matricesसमान आव्यूहsamaan aavyooh: Same order ke matrices jinki saari corresponding entries barabar hain.

identity matrixतत्समक आव्यूहtatsamak aavyooh: Square matrix $I$ jiske diagonal par 1 aur baaki jagah 0 hai; $A I = I A = A$ .Chapter mein pehli baar yahan aaya →
invertible matrixव्युत्क्रमणीय आव्यूहvyutkramaneey aavyooh: Square matrix $A$ jiske liye koi $B$ ho ki $A B = B A = I$ ; tab $B = A^{- 1}$ .

matrixआव्यूहaavyooh: Numbers ya functions ka ek rectangle, rows aur columns mein rakha hua aur ek object ki tarah handle kiya jaata hai.Chapter mein pehli baar yahan aaya →
matrix productआव्यूहों का गुणनफलaavyoohon ka gunanphal: Order $m \times n$ ke $A$ aur order $n \times p$ ke $B$ ke liye woh $m \times p$ matrix jiski $(i, k)$ entry $A$ ki row $i$ times $B$ ka column $k$ hai.

orderकोटिkoti: Matrix ka size $m \times n$ : $m$ rows aur $n$ columns.Chapter mein pehli baar yahan aaya →

scalar matrixअदिश आव्यूहadish aavyooh: Diagonal matrix jiski saari diagonal entries ek hi number $k$ ke barabar hain.Chapter mein pehli baar yahan aaya →
skew symmetric matrixविषम सममित आव्यूहvisham sammit aavyooh: Square matrix jismein $A^{'} = - A$ ; iski diagonal entries 0 hain.
square matrixवर्ग आव्यूहvarg aavyooh: Jitni rows utne columns wala matrix.Chapter mein pehli baar yahan aaya →
symmetric matrixसममित आव्यूहsammit aavyooh: Apne transpose ke barabar square matrix: $A^{'} = A$ .

transposeपरिवर्तparivart: Matrix $A^{'}$ (ya $A^{T}$ ) jiski rows $A$ ke columns hain.

zero matrixशून्य आव्यूहshoonya aavyooh: Matrix, jise O likhte hain, jiski har entry 0 hai.Chapter mein pehli baar yahan aaya →

Matrices: Class 12 Maths Chapter 3 Hinglish mein

Is chapter mein aap kya seekhenge

3.1 Introduction

3.2 Matrix

3.2.1 Matrix ka order

3.3 Matrices ke Types

3.3.1 Matrices ki equality

Practice 3.1: answers ke saath

3.4 Matrices par Operations

3.4.1 Matrices ka addition

3.4.2 Matrix ka scalar se multiplication

3.4.3 Matrix addition ki properties

3.4.4 Matrix ke scalar multiplication ki properties

3.4.5 Matrices ka multiplication

Order matter karta hai: AB aur BA

Non-zero matrices se zero product

3.4.6 Matrices ke multiplication ki properties

Practice 3.2: answers ke saath

3.5 Matrix ka Transpose

3.5.1 Matrices ke transpose ki properties

3.6 Symmetric aur Skew Symmetric Matrices

Practice 3.3: answers ke saath

3.7 Invertible Matrices

Practice 3.4: answers ke saath

Miscellaneous examples

Summary aur key points

Miscellaneous exercises

Practice questions answers ke saath (NCERT “Miscellaneous exercises”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye