Determinants: Class 12 Maths Chapter 4 Hinglish mein

Ek number jo areas, inverses aur solutions tay karta hai

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

Published 12 Oct 2026 · Updated 12 Oct 2026

90 min
1 simulation
65 practice questions
24 MCQs

Is chapter mein aap kya seekhenge

Order 1, 2 aur 3 ke determinants evaluate karo, aur 3 × 3 determinant ko us row ya column ke along expand karo jisse kaam sabse chhota ho
Yeh rule use karo ki n × n matrix ko k se multiply karne par uska determinant k to the power n se multiply ho jaata hai, aur aisi equations solve karo jinka unknown determinant ke andar baitha ho
Triangle ke vertices se determinant ki madad se uska area nikaalo, teen points ko collinearity ke liye test karo aur do points se guzarne wali line ki equation likho
Order 2 ya 3 ki square matrix ke minors, cofactors aur adjoint nikaalo
Samjhao ki square matrix ko uske adjoint se multiply karne par uska determinant times identity kyun milta hai, aur isse tay karo ki matrix invertible hai ya nahi aur uska inverse nikaalo
Linear equations ke system ko AX = B ke roop mein likho, uski consistency test karo aur use matrix method se solve karo

NCERT Mathematics Class 12, Chapter 4, pp. 76–103 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

4.1 Introduction

Graph paper par do seedhi lines banao. Woh ya to exactly ek point par milti hain, ya parallel hone ki wajah se kabhi nahi milti, ya ek line bilkul doosri ke upar padi hoti hai. $x$ aur $y$ mein linear equations ka ek pair algebra mein yahi sawaal poochta hai: iska ek solution hai, koi nahi, ya infinitely many?

Pair $a_{1} x + b_{1} y = c_{1}$ aur $a_{2} x + b_{2} y = c_{2}$ ke liye, coefficients se bana ek single number solve karna shuru karne se pehle hi iska jawab de deta hai. Pehli equation ko $b_{2}$ se aur doosri ko $b_{1}$ se multiply karo, phir $y$ hataane ke liye subtract karo: tumhe milta hai $(a_{1} b_{2} - a_{2} b_{1}) x = c_{1} b_{2} - c_{2} b_{1}$ . Agar $a_{1} b_{2} - a_{2} b_{1} \neq 0$ , to tum divide kar sakte ho, isliye $x$ fix ho jaata hai aur phir $y$ bhi: exactly ek solution. Agar yeh $0$ hai, to dono lines ka slope same hai, isliye woh parallel hain ya coincide karti hain (Fig. 4.1).

1/3 $x - y = - 1$ , $2 x + y = 7$ se $(2, 3)$ par milti hai, aur $a_{1} b_{2} - a_{2} b_{1} = (2) (- 1) - (1) (1) = - 3 \neq 0$ .

Fig. 4.1Number

a_{1} b_{2} - a_{2} b_{1}

sirf tab non-zero hota hai jab lines cross karti hain.

Ek coordinate grid jis par solid line 2x + y = 7 hai. Pehle step mein ek dashed doosri line x − y = −1 use point (2, 3) par cross karti hai, aur panel par likha hai one solution, saath mein a1b2 − a2b1 = −3. Doosre step mein dashed line 4x + 2y = 6 hai, jo pehli ke parallel hai, aur panel par likha hai no solution, number 0 ke barabar. Teesre step mein dashed line 4x + 2y = 14 hai, jo bilkul solid line ke upar padi hai, aur panel par likha hai infinitely many solutions, phir se number 0 ke saath.

Kyunki $a_{1} b_{2} - a_{2} b_{1}$ sirf coefficient matrix $[\begin{array}{cc} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array}]$ par depend karta hai, hum ise us matrix se jod dete hain aur ise matrix ka determinant (सारणिक) kehte hain. Hum $3 \times 3$ se badi nahi, aisi real square matrices tak hi rahenge. Pehle tum determinants jaldi evaluate karna seekhoge; phir woh apna kaam dikhate hain: corners se triangle ka area, matrix ke inverse ka formula, aur ek test jo try karne se pehle hi bata deta hai ki equations ka system solve ho sakta hai ya nahi.

Quick check

2 mein se 1

Equations ke kis pair ke liye

a_{1} b_{2} - a_{2} b_{1} = 0

hai?

x + 2 y = 5

aur

2 x - y = 0

3 x + y = 1

aur

x + 3 y = 1

x + 2 y = 5

aur

2 x + 4 y = 7

x - y = 2

aur

x + y = 2

Answer dekho

Answer (C) $x + 2 y = 5$ aur $2 x + 4 y = 7$

Yahan

a_{1} b_{2} - a_{2} b_{1} = 1 \cdot 4 - 2 \cdot 2 = 0

: doosri line ka slope pehli jaisa hi hai. Baaki pairs

- 5

8

aur

2

dete hain, isliye woh lines cross karti hain.

4.2 Determinant

Ek square matrix andar daalo, aur ek number bahar aata hai: woh output uska determinant hai. Yaani determinant square matrices se numbers tak ek function hai. Is book ki real matrices ke liye output real hota hai; complex entries wali matrices ke determinants complex ho sakte hain. Teen notations common hain: $∣ A ∣$ , $\det A$ aur $Δ$ .

Definition

Determinant

Ek rule jo har square matrix $A$ ko ek single number deta hai, jise $∣ A ∣$ , $\det A$ ya $Δ$ likhte hain. Determinant mein entries vertical bars ke beech hoti hain, $∣ \begin{array}{cc} a & b \\ c & d \end{array} ∣$ ; matrix mein brackets ke beech.

Example $A = [\begin{array}{cc} 7 & 2 \\ 3 & 1 \end{array}]$ numbers ki ek table hai, jabki $∣ A ∣ = ∣ \begin{array}{cc} 7 & 2 \\ 3 & 1 \end{array} ∣ = 1$ ek number hai.

4.2.1 Order one ki matrix ka determinant

$1 \times 1$ matrix $A = [a]$ ke liye determinant khud entry hi hai: $∣ A ∣ = a$ . Isliye $\det [- 7] = - 7$ . Yeh chhota sa case hi dikha deta hai ki bars ka matlab absolute value nahi hai.

4.2.2 Order two ki matrix ka determinant

$2 \times 2$ matrix ke liye, main diagonal (top-left se bottom-right) ki entries multiply karo, doosre diagonal ki entries multiply karo, aur pehle product mein se doosra product subtract karo (Fig. 4.2).

Order two

\begin{aligned} \det A & = ∣ \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} ∣ \\ = a_{11} a_{22} - a_{21} a_{12} \end{aligned}

3/3Subtract karo: $∣ \begin{array}{cc} 5 & 3 \\ 2 & 4 \end{array} ∣ = 20 - 6 = 14$ .

Fig. 4.2Main diagonal product minus doosre diagonal ka product.NCERT §4.2.2 · p. 77

Determinant jiski rows 5, 3 aur 2, 4 hain. Pehle step mein main diagonal, 5 aur 4, circle kiya gaya hai aur uske along ek arrow chalta hai; panel par 5 × 4 = 20 dikhta hai. Doosre step mein doosra diagonal, 3 aur 2, dashed arrow ke saath circle kiya gaya hai aur panel mein 3 × 2 = 6 judta hai. Teesre step mein panel par result 20 − 6 = 14 dikhta hai.

Dono rows swap karne se dono products swap ho jaate hain, isliye determinant ka sign palat jaata hai. Isliye kaunsa diagonal subtract karte ho, yeh matter karta hai.

Example 1NCERT Example 1 · p. 77

Ek numerical determinant

$∣ \begin{array}{cc} 6 & - 2 \\ 5 & 3 \end{array} ∣$ evaluate karo.

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

Example 2NCERT Example 2 · p. 77

Ek determinant jo x ko ignore karta hai

Dikhao ki $∣ \begin{array}{cc} x + 2 & x - 1 \\ x + 3 & x \end{array} ∣$ ki value har real number $x$ ke liye same hai.

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

Quick check

2 mein se 1

∣ \begin{array}{cc} 4 & - 3 \\ 2 & 5 \end{array} ∣

evaluate karo.

Answer dekho

Answer 26

4 \times 5 - (- 3) \times 2 = 20 + 6 = 26

. Lalchane wala

14

tab aata hai jab tum bhool jaate ho ki

- 3 \times 2

negative hai, isliye use subtract karne se

6

judta hai.

4.2.3 Order 3 × 3 ki matrix ka determinant

$3 \times 3$ determinant $2 \times 2$ wale determinants se banta hai. Ek row chuno, maan lo pehli. Uski har entry ko us $2 \times 2$ determinant se multiply karo jo us entry ki row aur column delete karne ke baad bachta hai. Phir teeno products ko signs $+$ , $-$ , $+$ ke saath jodo. Ise us row ke along determinant ka expansion kehte hain.

DerivationPehli row ke along expand karna

Maan lo $A = [a_{i j}]$ ek $3 \times 3$ matrix hai. Entry $a_{1 j}$ ko sign $(- 1)^{1 + j}$ milta hai aur woh $2 \times 2$ determinant jo row 1 aur column $j$ delete karne ke baad bachta hai:

\begin{aligned} ∣ A ∣ = & a_{11} ∣ \begin{array}{cc} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array} ∣ \\ - a_{12} ∣ \begin{array}{cc} a_{21} & a_{23} \\ a_{31} & a_{33} \end{array} ∣ \\ + a_{13} ∣ \begin{array}{cc} a_{21} & a_{22} \\ a_{31} & a_{32} \end{array} ∣ \end{aligned}

Har chhota determinant kholne par chhe products milte hain:

Six-term form

\begin{aligned} ∣ A ∣ = & a_{11} a_{22} a_{33} - a_{11} a_{23} a_{32} \\ - a_{12} a_{21} a_{33} + a_{12} a_{23} a_{31} \\ + a_{13} a_{21} a_{32} - a_{13} a_{22} a_{31} \end{aligned}

Ab doosri row try karo, jiske signs $-$ , $+$ , $-$ hain: $- a_{21} (a_{12} a_{33} - a_{13} a_{32}) + a_{22} (a_{11} a_{33} - a_{13} a_{31}) - a_{23} (a_{11} a_{32} - a_{12} a_{31})$ . Multiply karke terms match karo: jaise $- a_{23} a_{11} a_{32}$ six-term form ka doosra product hai, aur $+ a_{23} a_{12} a_{31}$ chautha. Chhe ke chhe sign ke saath match karte hain. Pehle column ke along expand karna, $a_{11} (\dots) - a_{21} (\dots) + a_{31} (\dots)$ , phir se wahi chhe products deta hai.

Iski ek wajah hai. Chhe products mein se har ek har row se exactly ek entry aur har column se exactly ek entry leta hai. Koi row ya column special nahi hai, isliye chhe ke chhe expansions (teen rows, teen columns) same value dete hain.

4/4 $Δ = - 4 - 8 + 63 = 51$ .

Fig. 4.3Chuni hui row ki har entry ko us determinant se multiply kiya jaata hai jo uski row aur column hataane ke baad bachta hai.

Determinant jiski rows 2, 1, 3; 4, −1, 0; 1, 5, 2 hain, aur columns ke upar signs plus, minus, plus hain. Pehle step mein entry 2 circle ki gayi hai, row 1 aur column 1 kaate gaye hain, aur bache chaar cells shaded hain; panel minor (−1)(2) − (0)(5) = −2 aur term +2 × (−2) = −4 compute karta hai. Doosre step mein column 2 ki entry 1 ke liye yahi hota hai: minor 8, term −1 × 8 = −8. Teesre step mein column 3 ki entry 3 ke liye: minor 21, term +3 × 21 = 63. Last step mein panel −4 − 8 + 63 jod kar 51 deta hai.

Tumhe har baar $(- 1)^{i + j}$ compute karne ki zaroorat nahi hai. Signs ek checkerboard banate hain jo top-left corner mein $+$ se shuru hota hai, isliye har row aur har column mein signs alternate hote hain (Fig. 4.4).

3/3 $C_{3}$ ke along expand karne mein signs $+, -, +$ lagte hain.

Fig. 4.4Row

i

aur column

j

wali jagah ka sign

(- 1)^{i + j}

hai.

Signs ka ek 3 by 3 grid: pehli row mein plus, minus, plus; doosri mein minus, plus, minus; teesri mein plus, minus, plus, aur plus wale squares shaded hain. Rows par R1 se R3 aur columns par C1 se C3 likha hai. Bagal mein: place (i, j) ka sign minus one to the power i + j hai, i + j even ho to plus aur odd ho to minus. Doosre step mein row 2 outline hai aur panel par likha hai: R2 ke along minus plus minus. Teesre step mein column 3 outline hai: C3 ke along plus minus plus.

Example 3NCERT Example 3 · p. 80

Zeros ko kaam karne do

$Δ = ∣ \begin{array}{ccc} 2 & - 1 & 5 \\ 0 & 3 & 0 \\ 4 & 1 & - 2 \end{array} ∣$ evaluate karo.

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

Example 4NCERT Example 4 · p. 80

Trigonometric entries wala ek determinant

Dikhao ki $Δ = ∣ \begin{array}{ccc} 1 & \sin θ & \cos θ \\ - \sin θ & 1 & 0 \\ - \cos θ & 0 & 1 \end{array} ∣$ , $θ$ par depend nahi karta.

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

Example 5NCERT Example 5 · p. 81

Determinant ke andar baithe unknown ke liye solve karna

Woh saare real $x$ nikaalo jinke liye $∣ \begin{array}{cc} 2 & x \\ x & 8 \end{array} ∣ = ∣ \begin{array}{cc} 3 & 1 \\ 5 & 4 \end{array} ∣$ .

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

PropertyPoori matrix ko scale karna

$n \times n$ matrix $A$ ki har entry ko ek number $k$ se multiply karo, to determinant $k^{n}$ se multiply ho jaata hai: $∣ k A ∣ = k^{n} ∣ A ∣$ .

$A$ ko $k$ se multiply karne par har entry $k$ se multiply hoti hai. six-term form dekho: har product mein teeno rows mein se har ek ki exactly ek entry hai, isliye har product, aur isliye $∣ A ∣$ bhi, $k \cdot k \cdot k = k^{3}$ le leta hai. Order 2 ke liye har product $a_{11} a_{22}$ ya $a_{21} a_{12}$ mein do factors hain aur woh $k^{2}$ leta hai; order 1 ke liye, $∣ [k a] ∣ = k a$ .

Check karo: $A = [\begin{array}{cc} 1 & 4 \\ 2 & 3 \end{array}]$ ka $∣ A ∣ = 3 - 8 = - 5$ hai, aur $3 A = [\begin{array}{cc} 3 & 12 \\ 6 & 9 \end{array}]$ ka $∣ 3 A ∣ = 27 - 72 = - 45 = 3^{2} \times (- 5)$ hai.

Ek aam galti hai $∣ k A ∣ = k ∣ A ∣$ likhna; yeh sirf $1 \times 1$ matrices ke liye sahi hai. Ek picture dikhati hai ki power kyun aati hai. $2 \times 2$ matrix ke liye, $∣ A ∣$ us parallelogram ka signed area hai jiski sides $A$ ke do columns hain, arrows ki tarah bane hue. Ek column double karne se area double hota hai; dono double karne se woh $2^{2}$ se multiply hota hai (Fig. 4.5). $3 \times 3$ matrices ke liye yahi idea volumes aur $k^{3}$ ke saath kaam karta hai.

3/3Dono columns double karne se woh $2^{2}$ se multiply hote hain: $∣ 2 A ∣ = 20 = 2^{2} ∣ A ∣$ .

Fig. 4.5Har column ko 2 se scale karne par area, aur determinant, 2 × 2 = 4 se scale hota hai.

Ek grid jis par origin se do arrows hain, (3, 1) aur (1, 2), jo area 5 ka ek shaded parallelogram banate hain, jo rows 3, 1 aur 1, 2 wale determinant ke barabar hai. Doosre step mein pehla arrow double hokar (6, 2) ho jaata hai; parallelogram, jo original ki dashed outline ke upar bana hai, ka area 10 hai. Teesre step mein dono arrows double hokar (6, 2) aur (2, 4) ho jaate hain, aur area 20 hai, jo 4 × 5 hai.

Practice · 8 questions

Practice 4.1: answers ke saath

Short answerEasyNCERT Exercise 4.1, Q1
$∣ \begin{array}{cc} - 3 & 7 \\ 2 & - 4 \end{array} ∣$ evaluate karo.
Solution dikhao
$(- 3) (- 4) - 7 \times 2 = 12 - 14 = - 2$ .
2
Short answerMediumNCERT Exercise 4.1, Q2
Har determinant evaluate karo.
1. $∣ \begin{array}{cc} \sec θ & \tan θ \\ \tan θ & \sec θ \end{array} ∣$ ( $\cos θ \neq 0$ ke liye)
  Solution dikhao
  Answer $1$
  $\sec^{2} θ - \tan^{2} θ = 1$ , jo $1 + \tan^{2} θ = \sec^{2} θ$ se aata hai.
2. $∣ \begin{array}{cc} x^{2} + 1 & x \\ x^{3} & x^{2} + 1 \end{array} ∣$
  Solution dikhao
  Answer $2 x^{2} + 1$
  $(x^{2} + 1)^{2} - x \cdot x^{3} = x^{4} + 2 x^{2} + 1 - x^{4} = 2 x^{2} + 1$ .
Short answerEasyNCERT Exercise 4.1, Q3
Maan lo $A = [\begin{array}{cc} 3 & - 1 \\ 2 & 5 \end{array}]$ . Dikhao ki $∣ 3 A ∣ = 9 ∣ A ∣$ .
Solution dikhao
$∣ A ∣ = 15 - (- 2) = 17$ . $3 A = [\begin{array}{cc} 9 & - 3 \\ 6 & 15 \end{array}]$ , isliye $∣ 3 A ∣ = 135 - (- 18) = 153$ .
$153 = 9 \times 17$ , isliye $∣ 3 A ∣ = 9 ∣ A ∣ = 3^{2} ∣ A ∣$ , jaisa scaling rule order 2 ke liye predict karta hai.
Short answerEasyNCERT Exercise 4.1, Q4
Maan lo $A = [\begin{array}{ccc} 2 & 0 & 0 \\ 1 & - 1 & 0 \\ 4 & 3 & 5 \end{array}]$ . Dikhao ki $∣ 2 A ∣ = 8 ∣ A ∣$ .
Solution dikhao
$R_{1}$ ke along: $∣ A ∣ = 2 ∣ \begin{array}{cc} - 1 & 0 \\ 3 & 5 \end{array} ∣ = 2 (- 5) = - 10$ .
$2 A = [\begin{array}{ccc} 4 & 0 & 0 \\ 2 & - 2 & 0 \\ 8 & 6 & 10 \end{array}]$ , isliye $∣ 2 A ∣ = 4 ∣ \begin{array}{cc} - 2 & 0 \\ 6 & 10 \end{array} ∣ = 4 (- 20) = - 80$ .
$- 80 = 8 \times (- 10)$ , isliye $∣ 2 A ∣ = 2^{3} ∣ A ∣$ .
5
Short answerMediumNCERT Exercise 4.1, Q5
Har determinant evaluate karo, koi convenient row ya column chun kar.
1. $∣ \begin{array}{ccc} 4 & 0 & - 1 \\ 2 & 0 & 3 \\ - 1 & 5 & 2 \end{array} ∣$
  Solution dikhao
  Column 2 mein do zeros hain. Uski akeli non-zero entry, 5, place $(3, 2)$ par hai jiska sign $-$ hai: $- 5 ∣ \begin{array}{cc} 4 & - 1 \\ 2 & 3 \end{array} ∣ = - 5 (12 + 2) = - 70$ .
2. $∣ \begin{array}{ccc} 2 & - 3 & 1 \\ 4 & 1 & - 2 \\ 3 & 2 & 5 \end{array} ∣$
  Solution dikhao
  $R_{1}$ ke along: $2 (5 + 4) + 3 (20 + 6) + 1 (8 - 3) = 18 + 78 + 5 = 101$ .
3. $∣ \begin{array}{ccc} 0 & 2 & - 5 \\ - 2 & 1 & 4 \\ 5 & - 4 & 0 \end{array} ∣$
  Solution dikhao
  $R_{1}$ ke along (pehli entry 0 hai): $- 2 ∣ \begin{array}{cc} - 2 & 4 \\ 5 & 0 \end{array} ∣ + (- 5) ∣ \begin{array}{cc} - 2 & 1 \\ 5 & - 4 \end{array} ∣ = - 2 (- 20) - 5 (8 - 5) = 40 - 15 = 25$ .
4. $∣ \begin{array}{ccc} 1 & - 2 & 3 \\ 0 & 4 & - 1 \\ 2 & - 3 & 0 \end{array} ∣$
  Solution dikhao
  $C_{1}$ ke along: $1 ∣ \begin{array}{cc} 4 & - 1 \\ - 3 & 0 \end{array} ∣ - 0 + 2 ∣ \begin{array}{cc} - 2 & 3 \\ 4 & - 1 \end{array} ∣ = 1 (0 - 3) + 2 (2 - 12) = - 3 - 20 = - 23$ .
Short answerMediumNCERT Exercise 4.1, Q6
Maan lo $A = [\begin{array}{ccc} 2 & - 1 & 3 \\ 1 & 4 & - 2 \\ 4 & 7 & - 1 \end{array}]$ . $∣ A ∣$ nikaalo, aur row 3 ko rows 1 aur 2 se compare karo.
Solution dikhao
Answer $∣ A ∣ = 0$ ; row 3, row 1 + 2 × row 2 ke barabar hai.
$R_{1}$ ke along: $2 (- 4 + 14) + 1 (- 1 + 8) + 3 (7 - 16) = 20 + 7 - 27 = 0$ .
Row 3 hai $(2, - 1, 3) + 2 (1, 4, - 2) = (4, 7, - 1)$ , yaani baaki rows ka ek combination. Jab bhi koi row is tarah doosri rows se bani ho, determinant 0 hota hai.
7
Short answerMediumNCERT Exercise 4.1, Q7
$x$ ki values nikaalo.
1. $∣ \begin{array}{cc} x & 3 \\ 4 & x \end{array} ∣ = ∣ \begin{array}{cc} 3 & - 2 \\ 2 & 3 \end{array} ∣$
  Solution dikhao
  Answer $x = \pm 5$
  $x^{2} - 12 = 9 + 4 = 13$ , isliye $x^{2} = 25$ aur $x = 5$ ya $x = - 5$ .
2. $∣ \begin{array}{cc} 3 x & 2 \\ x & 4 \end{array} ∣ = ∣ \begin{array}{cc} 9 & 3 \\ 2 & 4 \end{array} ∣$
  Solution dikhao
  Answer $x = 3$
  $12 x - 2 x = 36 - 6$ , isliye $10 x = 30$ aur $x = 3$ .
Quick answerEasyNCERT Exercise 4.1, Q8
Agar $∣ \begin{array}{cc} x & 2 \\ 4 & x \end{array} ∣ = ∣ \begin{array}{cc} 3 & 1 \\ 1 & 3 \end{array} ∣$ , to $x$ barabar hai
sirf $4$ $\pm 2 \sqrt{2}$ $\pm 4$ $16$
Answer dekho
Answer (C) $\pm 4$
$x^{2} - 8 = 9 - 1 = 8$ , isliye $x^{2} = 16$ aur $x = \pm 4$ . Dono signs kaam karte hain; $\pm 2 \sqrt{2}$ mein $- 8$ ko doosri taraf le jaana bhool gaye.

4.3 Triangle ka Area

Ek surveyor jo zameen ke plot ke corners jaanta hai, ya ek phone app jo unhe GPS se padhta hai, plot ko triangles mein baant kar area nikaalta hai. Har triangle ka area seedha uske corners ke coordinates se aata hai, aur formula ek determinant hai.

Pehle formula dekho; yeh kahan se aata hai, woh baad mein. Maan lo corners $A (x_{1}, y_{1})$ , $B (x_{2}, y_{2})$ aur $C (x_{3}, y_{3})$ hain.

Triangle ka area

\begin{array}{c} Δ = \frac{1}{2} ∣ \begin{array}{ccc} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array} ∣ \\ area of △ A B C = ∣ Δ ∣ \end{array}

Pehle column ke along expand karne se determinant $x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})$ ban jaata hai. Yahi Class 10 ke coordinate-geometry area formula ka bracket hai, isliye determinant use store karne ka bas ek saaf-suthra tareeka hai.

DerivationDeterminant area kyun deta hai

Triangle ko $x$ -axis ke upar aise rakho ki $x_{1} < x_{2} < x_{3}$ ho aur $B$ side $A C$ ke upar ho, aur axis par perpendiculars $A L$ , $B M$ , $C N$ girao (Fig. 4.6). Triangle woh hai jo do trapezia $A L M B$ aur $B M N C$ mein se trapezium $A L N C$ hataane ke baad bachta hai. Parallel sides $p$ , $q$ aur width $w$ wale trapezium ka area $\frac{1}{2} (p + q) w$ hota hai, isliye triangle ka area hai

\begin{aligned} \frac{1}{2} (y_{1} + y_{2}) (x_{2} - x_{1}) \\ + & \frac{1}{2} (y_{2} + y_{3}) (x_{3} - x_{2}) \\ - & \frac{1}{2} (y_{1} + y_{3}) (x_{3} - x_{1}) \end{aligned}

Multiply karne par terms $y_{1} x_{1}$ , $y_{2} x_{2}$ aur $y_{3} x_{3}$ cancel ho jaate hain, aur jo bachta hai woh $\frac{1}{2} [x_{1} (y_{3} - y_{2}) + x_{2} (y_{1} - y_{3}) + x_{3} (y_{2} - y_{1})] = - Δ$ hai. Is picture mein $A \to B \to C$ jaana clockwise ghoomta hai, aur clockwise order ke liye $Δ$ negative hota hai, isliye area $- Δ = ∣ Δ ∣$ hai. Vertices ko anticlockwise likhne se $+ Δ$ milta hai, aur triangle ki baaki placements bhi isi tarah kaam karti hain.

3/3AC ke neeche wala trapezium hataane par exactly triangle bachta hai.

Fig. 4.6ABC ka area = ALMB + BMNC − ALNC.

x-axis ke upar triangle ABC, jismein A left mein, B sabse upar aur C right mein hai; A, B aur C se dashed perpendiculars axis par points L, M aur N tak girte hain. Pehla step shaded triangle aur vertices ke coordinates dikhata hai. Doosre step mein do trapezia ALMB aur BMNC do colours mein shaded hain aur panel par likha hai ALMB + BMNC. Teesre step mein bada trapezium ALNC dashed border se outline hai, triangle phir se shaded hai, aur panel par likha hai minus ALNC barabar ABC ka area.

To determinant ka size ke saath ek sign bhi hota hai: jab vertices anticlockwise likhe jaate hain to woh positive hota hai aur clockwise likhe jaane par negative (Fig. 4.7). Area negative nahi ho sakta, isliye hum hamesha $∣ Δ ∣$ lete hain. Jab area diya ho aur ek vertex unknown ho, to dono signs possible hain: $Δ = + (area)$ aur $Δ = - (area)$ ko alag-alag solve karo.

1/3Order $(1, 1) \to (5, 2) \to (2, 4)$ anticlockwise ghoomta hai: determinant $+ 11$ hai, isliye $Δ = 5.5$ aur area $5.5$ square units hai.

Fig. 4.7Determinant ka sign vertices ka order record karta hai; uska size area deta hai.

Ek grid jis par points (1, 1), (5, 2) aur (2, 4) hain. Pehle step mein triangle shaded hai, vertices par number likhe hain 1st (1, 1), 2nd (5, 2), 3rd (2, 4), andar ek arrow anticlockwise ghoomta hai, aur panel par likha hai Δ = +11 aur area 5.5. Doosre step mein aakhri do vertices ka order swap hota hai, outline dashed hai, arrow clockwise ghoomta hai aur panel par likha hai Δ = −11, area wahi. Teesre step mein teesra point (1, 1) se (5, 2) wale segment par (3, 1.5) par aa jaata hai; triangle us segment mein simat jaata hai aur panel par likha hai Δ = 0, collinear.

Aakhri picture ek test deti hai. Teen points tabhi collinear hote hain jab $Δ = 0$ ho. Isse line ki equation bhi milti hai: ek point $P (x, y)$ , $(x_{1}, y_{1})$ aur $(x_{2}, y_{2})$ se guzarne wali line par tabhi hota hai jab teeno points collinear hon, yaani jab

Do points se guzarne wali line

∣ \begin{array}{ccc} x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \end{array} ∣ = 0

Example 6NCERT Example 6 · p. 82

Teen vertices se area

Vertices $P (- 2, 1)$ , $Q (3, 4)$ aur $R (5, - 2)$ wale triangle ka area nikaalo.

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

Example 7NCERT Example 7 · p. 83

Pehle ek line, phir diye gaye area par ek point

(a) Determinant use karke $A (2, 1)$ aur $B (4, 5)$ se guzarne wali line ki equation nikaalo. (b) $y$ -axis par woh saare points $D$ nikaalo jinke liye triangle $A B D$ ka area 6 square units ho.

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

SimulationVertex hilao, area predict karo

Ek coordinate grid jo across −6 se 8 tak aur upar −5 se 7 tak hai, jis par teen draggable points A(−2, −1), B(4, 2) aur C(0, 4) ek triangle mein jude hain. Har side apni length dikhati hai aur har point apne coordinates, jo drag karte waqt whole numbers par snap ho jaate hain, taaki tum vertices padh kar kisi bhi position ke liye area determinant compute kar sako.

Fig. 4.8Vertices ko drag karo; sides aur coordinates update ho jaate hain. Har position ke liye determinant nikaal kar dekho.

Try karo

Starting coordinates se dikhao ki area 12 square units hai. Phir C ko (2, 5) par drag karo. Compute karne se pehle naya area predict karo, aur line AB ki madad se result samjhao.
C ko (2, 1) ya (0, 0) par drag karo. Triangle ko kya hota hai, aur ab determinant kya hai?
C ko neeche (0, −3) par drag karo, taaki woh AB ke doosri taraf ho jaaye. Vertices ko A, B, C order mein lekar Δ compute karo. Uske sign ko kya hua?

Practice · 5 questions

Practice 4.2: answers ke saath

1
Short answerEasyNCERT Exercise 4.2, Q1
Diye gaye vertices wale triangle ka area nikaalo.
1. $(0, 0)$ , $(5, 2)$ , $(1, 4)$
  square units
  Solution dikhao
  $\frac{1}{2} ∣ \begin{array}{ccc} 0 & 0 & 1 \\ 5 & 2 & 1 \\ 1 & 4 & 1 \end{array} ∣ = \frac{1}{2} (20 - 2) = 9$ ( $R_{1}$ ke along expand karne par sirf aakhri entry bachti hai). Area 9 square units.
2. $(- 3, 2)$ , $(4, 5)$ , $(1, - 2)$
  square units
  Solution dikhao
  Determinant: $- 3 (5 + 2) - 2 (4 - 1) + 1 (- 8 - 5) = - 21 - 6 - 13 = - 40$ , isliye $Δ = - 20$ aur area 20 square units hai.
3. $(1, - 2)$ , $(6, 3)$ , $(- 2, 4)$
  square units
  Solution dikhao
  Determinant: $1 (3 - 4) + 2 (6 + 2) + 1 (24 + 6) = - 1 + 16 + 30 = 45$ , isliye area $22.5$ square units hai.
ProofMediumNCERT Exercise 4.2, Q2
Dikhao ki points $P (2 p, 3 p - 1)$ , $Q (2 q, 3 q - 1)$ aur $R (2 r, 3 r - 1)$ saare real $p$ , $q$ , $r$ ke liye collinear hain.
Solution dikhao
$∣ \begin{array}{ccc} 2 p & 3 p - 1 & 1 \\ 2 q & 3 q - 1 & 1 \\ 2 r & 3 r - 1 & 1 \end{array} ∣$ ko $C_{3}$ ke along expand karne par (signs $+, -, +$ ):
$+ [2 q (3 r - 1) - 2 r (3 q - 1)] - [2 p (3 r - 1) - 2 r (3 p - 1)] + [2 p (3 q - 1) - 2 q (3 p - 1)]$ .
Brackets simplify hokar $2 r - 2 q$ , $2 r - 2 p$ aur $2 q - 2 p$ bante hain, isliye total $(2 r - 2 q) - (2 r - 2 p) + (2 q - 2 p) = 0$ hai. Isliye points collinear hain. (Yeh saare $3 x - 2 y = 2$ par hain.)
3
Short answerMediumNCERT Exercise 4.2, Q3
k ki possible values nikaalo.
1. Vertices $(1, k)$ , $(4, - 2)$ , $(- 2, 4)$ wale triangle ka area 6 square units hai.
  Solution dikhao
  Answer $k = - 1$ ya $k = 3$
  $∣ \begin{array}{ccc} 1 & k & 1 \\ 4 & - 2 & 1 \\ - 2 & 4 & 1 \end{array} ∣ = 1 (- 2 - 4) - k (4 + 2) + 1 (16 - 4) = 6 - 6 k$ , isliye $Δ = 3 - 3 k$ .
  $3 - 3 k = 6$ se $k = - 1$ milta hai; $3 - 3 k = - 6$ se $k = 3$ milta hai.
2. Vertices $(2, 1)$ , $(k, 3)$ , $(- 1, 5)$ wale triangle ka area 5 square units hai.
  Solution dikhao
  Answer $k = 3$ ya $k = - 2$
  $∣ \begin{array}{ccc} 2 & 1 & 1 \\ k & 3 & 1 \\ - 1 & 5 & 1 \end{array} ∣ = 2 (3 - 5) - 1 (k + 1) + 1 (5 k + 3) = 4 k - 2$ , isliye $Δ = 2 k - 1$ .
  $2 k - 1 = 5$ se $k = 3$ milta hai; $2 k - 1 = - 5$ se $k = - 2$ milta hai.
4
Short answerEasyNCERT Exercise 4.2, Q4
Determinants use karke do points se guzarne wali line ki equation nikaalo.
1. $(2, 3)$ aur $(6, 11)$
  Solution dikhao
  Answer $y = 2 x - 1$
  $∣ \begin{array}{ccc} x & y & 1 \\ 2 & 3 & 1 \\ 6 & 11 & 1 \end{array} ∣ = x (3 - 11) - y (2 - 6) + (22 - 18) = - 8 x + 4 y + 4 = 0$ , yaani $y = 2 x - 1$ .
2. $(- 1, 4)$ aur $(5, 1)$
  Solution dikhao
  Answer $x + 2 y = 7$
  $x (4 - 1) - y (- 1 - 5) + (- 1 - 20) = 3 x + 6 y - 21 = 0$ , yaani $x + 2 y = 7$ .
Quick answerMediumNCERT Exercise 4.2, Q5
Points $(2, 1)$ , $(5, 3)$ aur $(k, k)$ area 4 square units ka triangle banate hain. $k$ ki possible values hain
sirf $k = 7$ $k = 3$ ya $k = - 5$ $k = 7$ ya $k = - 9$ $k = - 7$ ya $k = 9$
Answer dekho
Answer (C) $k = 7$ ya $k = - 9$
Determinant $2 (3 - k) - 1 (5 - k) + (5 k - 3 k) = k + 1$ hai, isliye $Δ = \frac{k + 1}{2} = \pm 4$ , jisse $k = 7$ ya $k = - 9$ milta hai. $\frac{1}{2}$ bhoolne par $3$ ya $- 5$ aata hai; sirf plus sign lene par $- 9$ chhoot jaata hai.

4.4 Minors aur Cofactors

Har expansion mein aane wale chhote determinants baar-baar wapas aate hain: expansions mein, adjoint mein aur inverse mein. Inka naam rakhna faayde ka hai, aur naam rakhne se hum ek expansion ko ek chhoti si line mein likh sakte hain.

Definition

Minor

Entry $a_{i j}$ ka minor $M_{i j}$ woh determinant hai jo row $i$ aur column $j$ delete karne ke baad bachta hai. Isliye minor hamesha apne determinant se ek size chhota hota hai: jab determinant order $n \geq 2$ ka ho, to minor order $n - 1$ ka.

Example $3 \times 3$ determinant mein har minor ek $2 \times 2$ determinant hota hai.

3/3 $3 + 2 = 5$ odd hai, isliye $A_{32} = (- 1)^{5} M_{32} = - 22$ .

Fig. 4.9Minor: entry ki row aur column delete karo. Cofactor: sign

(- 1)^{i + j}

lagao.

Determinant jiski rows 3, 0, −2; 5, 1, 4; −1, 6, 2 hain. Pehle step mein entry 6 circle ki gayi hai aur panel batata hai ki woh row 3 aur column 2 mein hai. Doosre step mein row 3 aur column 2 kaate gaye hain, bache chaar cells 3, −2, 5, 4 shaded hain, aur panel (3)(4) − (−2)(5) = 22 compute karta hai, jo minor M32 hai. Teesre step mein panel note karta hai ki 3 + 2 = 5 odd hai, isliye sign minus hai aur cofactor A32, −22 hai.

Example 8NCERT Example 8 · p. 84

Ek minor

$Δ = ∣ \begin{array}{ccc} 3 & 0 & - 2 \\ 5 & 1 & 4 \\ - 1 & 6 & 2 \end{array} ∣$ mein entry 4 ka minor nikaalo.

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

Definition

Cofactor

$a_{i j}$ ka cofactor $A_{i j} = (- 1)^{i + j} M_{i j}$ hai: minor, jismein uski place ka checkerboard sign laga ho.

ExampleFig. 4.9 mein $M_{32} = 22$ aur $A_{32} = - 22$ .

Cofactor

A_{i j} = (- 1)^{i + j} M_{i j}

Example 9NCERT Example 9 · p. 84

2 × 2 determinant ke saare minors aur cofactors

$∣ \begin{array}{cc} 5 & - 3 \\ 2 & 7 \end{array} ∣$ ki saari entries ke minors aur cofactors nikaalo.

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

Example 10NCERT Example 10 · p. 85

Letters mein minors aur cofactors

$Δ = ∣ \begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array} ∣$ ke liye, $a_{12}$ aur $a_{32}$ ke minors aur cofactors nikaalo.

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

PropertyCofactors mein expansion

$3 \times 3$ matrix $A$ aur kisi bhi row $i$ ke liye, $∣ A ∣ = a_{i 1} A_{i 1} + a_{i 2} A_{i 2} + a_{i 3} A_{i 3}$ . Kisi bhi column $j$ ke liye, $∣ A ∣ = a_{1 j} A_{1 j} + a_{2 j} A_{2 j} + a_{3 j} A_{3 j}$ .

Agar tum ek row ki entries lo lekin ek alag row ke cofactors, to kya hota hai? Answer hamesha zero hota hai, aur yahi agle section mein inverse ki chaabi hai.

PropertyGalat line ke cofactors

Agar ek row ki entries ko ek alag row ke cofactors se multiply karke products jode jaayein, to sum $0$ hota hai; columns ke liye bhi yahi sach hai. Jaise, $a_{21} A_{11} + a_{22} A_{12} + a_{23} A_{13} = 0$ .

Cofactors $A_{11}$ , $A_{12}$ , $A_{13}$ sirf rows 2 aur 3 se bante hain. Isliye woh us matrix $B$ ke bhi row-1 cofactors hain jo $A$ mein row 2 ko row 1 mein copy karke bani hai. $∣ B ∣$ ko uski pehli row ke along expand karne se exactly $a_{21} A_{11} + a_{22} A_{12} + a_{23} A_{13}$ milta hai.

Ab usi $∣ B ∣$ ko uski teesri row ke along expand karo. Wahan har minor $B$ ki rows 1 aur 2 se aata hai, jo barabar hain, isliye har ek $∣ \begin{array}{cc} p & q \\ p & q \end{array} ∣ = p q - q p = 0$ jaisa hai. Isliye $∣ B ∣ = 0$ , aur sum $0$ hai.

3/3Row 2 ki entries times row-1 cofactors: $3 (- 10) + 0 (- 23) + 5 (6) = 0$ .

Fig. 4.10Ek row ki entries times doosri row ke cofactors: do barabar rows wala determinant, jo 0 hai.

Determinant jiski rows 1, 4, −2; 3, 0, 5; −1, 2, 6 hain, aur rows 2 aur 3 shaded hain. Pehla step note karta hai ki row-1 cofactors −10, −23 aur 6 sirf rows 2 aur 3 se bane hain. Doosre step mein row 1 ki jagah row 2 ki copy, 3, 0, 5, aa jaati hai, jo doosre colour mein shaded hai; cofactors nahi badalte. Teesre step mein panel 3(−10) + 0(−23) + 5(6) = 0 compute karta hai, kyunki naye determinant mein do barabar rows hain.

Example 11NCERT Example 11 · p. 86

Har cofactor, aur ek sum jo zero ho jaata hai

$Δ = ∣ \begin{array}{ccc} 1 & 4 & - 2 \\ 3 & 0 & 5 \\ - 1 & 2 & 6 \end{array} ∣$ ke saare minors aur cofactors nikaalo. Unse $Δ$ nikaalo, aur check karo ki $a_{11} A_{12} + a_{21} A_{22} + a_{31} A_{32} = 0$ .

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

Neeche wala lab yeh calculations kisi bhi $3 \times 3$ matrix ke liye karta hai jo tum type karo. Chhe lines mein se har ek ke along baari-baari expand karo, phir ek line ki entries ko doosri line ke cofactors ke saath mix karo, aur total dekho.

InteractiveCofactor lab

Ek interactive 3 by 3 matrix jiski nau entries tum type kar sakte ho, aur presets: ek example, bahut zeros wali matrix, ek singular matrix aur ek random matrix. Expand tab mein tum woh row ya column chunte ho jiski entries use hoti hain (R1 se C3) aur jiske cofactors use hote hain. Teen term cards har entry, struck-out grid se nikla uska minor, uska sign, uska cofactor aur product dikhate hain, aur card par tap karne se woh minor matrix mein shaded ho jaata hai. Total ek message ke saath dikhta hai: line ke apne cofactors use karne par yeh determinant ke barabar hota hai, aur kisi doosri line ke cofactors use karne par hamesha 0 hota hai. Adjoint tab matrix of cofactors, uska transpose adj A, product A times adj A, aur determinant zero na ho to inverse fractions mein dikhata hai.

Fig. 4.11Ek matrix type karo, expand karne ke liye row ya column chuno, aur kisi term par tap karke uska minor dekho. Doosra tab adjoint aur inverse banata hai.

Quick check

4 mein se 1

∣ \begin{array}{ccc} 2 & 7 & 1 \\ 0 & 3 & - 4 \\ 5 & 1 & 6 \end{array} ∣

mein cofactor

A_{21}

nikaalo.

Answer dekho

Answer -41

Row 2 aur column 1 delete karo:

M_{21} = ∣ \begin{array}{cc} 7 & 1 \\ 1 & 6 \end{array} ∣ = 42 - 1 = 41

. Place

(2, 1)

ka sign

-

hai, isliye

A_{21} = - 41

. Answer

41

minor hai, cofactor nahi.

Practice · 5 questions

Practice 4.3: answers ke saath

1
Short answerEasyNCERT Exercise 4.3, Q1
Har entry ke minors aur cofactors likho.
1. $∣ \begin{array}{cc} - 5 & 2 \\ 6 & 1 \end{array} ∣$
  Solution dikhao
  Answer $M_{11} = 1, M_{12} = 6, M_{21} = 2, M_{22} = - 5$ ; $A_{11} = 1, A_{12} = - 6, A_{21} = - 2, A_{22} = - 5$
  Row 1, column 1 delete karne par $1$ bachta hai; row 1, column 2 par $6$ ; row 2, column 1 par $2$ ; row 2, column 2 par $- 5$ . Signs ke saath: $A_{11} = 1$ , $A_{12} = - 6$ , $A_{21} = - 2$ , $A_{22} = - 5$ .
2. $∣ \begin{array}{cc} p & q \\ r & s \end{array} ∣$
  Solution dikhao
  Answer $M_{11} = s, M_{12} = r, M_{21} = q, M_{22} = p$ ; $A_{11} = s, A_{12} = - r, A_{21} = - q, A_{22} = p$
  Letters ke saath wahi pattern: minors $s, r, q, p$ aur cofactors $s, - r, - q, p$ . Check karo: $p A_{11} + q A_{12} = p s - q r$ , jo determinant hai.
2
Short answerMediumNCERT Exercise 4.3, Q2
Har entry ke minors aur cofactors likho.
1. $∣ \begin{array}{ccc} 2 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 3 \end{array} ∣$
  Solution dikhao
  Answer $M_{11} = A_{11} = - 3$ , $M_{22} = A_{22} = 6$ , $M_{33} = A_{33} = - 2$ ; baaki har minor aur cofactor 0 hai.
  $M_{11} = (- 1) (3) - 0 = - 3$ , $M_{22} = (2) (3) = 6$ , $M_{33} = (2) (- 1) = - 2$ . Har off-diagonal minor mein zeros ki ek poori row ya column hoti hai, isliye woh 0 hai. Diagonal places ka sign $+$ hai, isliye cofactors minors ke barabar hain.
2. $∣ \begin{array}{ccc} 2 & - 1 & 0 \\ 1 & 3 & 4 \\ 0 & 5 & - 2 \end{array} ∣$
  Solution dikhao
  AnswerMinors: $- 26, - 2, 5; 2, - 4, 10; - 4, 8, 7$ . Cofactors: $- 26, 2, 5; - 2, - 4, - 10; - 4, - 8, 7$ .
  Row 1: $M_{11} = - 6 - 20 = - 26$ , $M_{12} = - 2 - 0 = - 2$ , $M_{13} = 5 - 0 = 5$ .
  Row 2: $M_{21} = 2 - 0 = 2$ , $M_{22} = - 4 - 0 = - 4$ , $M_{23} = 10 - 0 = 10$ .
  Row 3: $M_{31} = - 4 - 0 = - 4$ , $M_{32} = 8 - 0 = 8$ , $M_{33} = 6 + 1 = 7$ .
  Cofactors places $(1, 2), (2, 1), (2, 3), (3, 2)$ par sign palat dete hain. Check karo: $2 (- 26) + (- 1) (2) + 0 (5) = - 54 = Δ$ .
Short answerMediumNCERT Exercise 4.3, Q3
Teesri row ki entries ke cofactors use karke $Δ = ∣ \begin{array}{ccc} 4 & 1 & - 3 \\ 2 & 5 & 1 \\ 0 & - 2 & 6 \end{array} ∣$ evaluate karo.
Solution dikhao
$A_{31} = + ∣ \begin{array}{cc} 1 & - 3 \\ 5 & 1 \end{array} ∣ = 16$ , $A_{32} = - ∣ \begin{array}{cc} 4 & - 3 \\ 2 & 1 \end{array} ∣ = - 10$ , $A_{33} = + ∣ \begin{array}{cc} 4 & 1 \\ 2 & 5 \end{array} ∣ = 18$ .
$Δ = 0 (16) + (- 2) (- 10) + 6 (18) = 20 + 108 = 128$ .
ProofHardNCERT Exercise 4.3, Q4
Pehle column ki entries ke cofactors use karke dikhao ki $∣ \begin{array}{ccc} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} ∣ = (a - b) (b - c) (c - a)$ .
Solution dikhao
$A_{11} = b c^{2} - b^{2} c$ , $A_{21} = - (a c^{2} - a^{2} c) = a^{2} c - a c^{2}$ , $A_{31} = a b^{2} - a^{2} b$ .
Isliye $Δ = b c^{2} - b^{2} c + a^{2} c - a c^{2} + a b^{2} - a^{2} b$ .
Ab right side expand karo: $(a - b) (b - c) = a b - a c - b^{2} + b c$ , aur $(c - a)$ se multiply karne par $- a^{2} b + a^{2} c + a b^{2} - a c^{2} - b^{2} c + b c^{2}$ milta hai, wahi chhe terms.
Quick answerEasyNCERT Exercise 4.3, Q5
$Δ = ∣ a_{i j} ∣$ ek $3 \times 3$ determinant hai aur $A_{i j}$ , $a_{i j}$ ka cofactor hai. Kaunsa expression $Δ$ ke barabar hai?
$a_{12} A_{11} + a_{22} A_{21} + a_{32} A_{31}$ $a_{21} A_{31} + a_{22} A_{32} + a_{23} A_{33}$ $a_{12} M_{12} + a_{22} M_{22} + a_{32} M_{32}$ $a_{12} A_{12} + a_{22} A_{22} + a_{32} A_{32}$
Answer dekho
Answer (D) $a_{12} A_{12} + a_{22} A_{22} + a_{32} A_{32}$
Option D column 2 ke along expansion hai. A aur B do alag lines ko mix karte hain (dono ka sum 0 hai), aur C cofactors ke signs chhod deta hai.

4.5 Matrix ka Adjoint aur Inverse

Ek number $a \neq 0$ ka inverse bas $\frac{1}{a}$ hai. Kya square matrix ke liye bhi aisa koi formula hai? Hai, aur woh exactly un matrices ko bhi pehchaan leta hai jinka koi inverse hota hi nahi. Iska main ingredient cofactors hain, jo ek nayi matrix mein pack kiye jaate hain. (Chapter 3 mein tumne inverses elementary operations se nikaale the.)

4.5.1 Matrix ka adjoint

Definition

Adjoint

Square matrix $A$ ke liye, adjoint $adj A$ us matrix ka transpose hai jiski entries cofactors $A_{i j}$ hain.

Adjoint

\begin{aligned} adj A & = {[\begin{array}{ccc} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{array}]}^{T} \\ = [\begin{array}{ccc} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{array}] \end{aligned}

Ise aise padho: $A$ ki row 1 ke cofactors $adj A$ ke column 1 mein neeche jaate hain, row 2 ke cofactors column 2 mein, aur isi tarah aage (Fig. 4.12).

3/3Main diagonal mein reflect karo: rows columns ban jaati hain. Result $adj A$ hai.

Fig. 4.12Adjoint matrix of cofactors ka transpose hai.

Left mein matrix A jiski rows 1, 0, 2; 3, 1, 0; 0, 2, 1 hain, aur right ki ek matrix tak ek arrow. Pehle step mein row 1 ki entry 0 circle ki gayi hai, uske minor ke cells shaded hain, aur right mein sirf uska cofactor −3 dikhta hai, is note ke saath ki yeh minus (3·1 − 0·0) ke barabar hai. Doosre step mein right wali matrix saare nau cofactors se bhar jaati hai: 1, −3, 6; 4, 1, −2; −2, 6, 1. Teesre step mein ek dashed line main diagonal mark karti hai aur right wali matrix apna transpose ban jaati hai, adj A, jiski rows 1, 4, −2; −3, 1, 6; 6, −2, 1 hain.

Example 12NCERT Example 12 · p. 88

2 × 2 matrix ka adjoint

$A = [\begin{array}{cc} 4 & - 1 \\ 3 & 2 \end{array}]$ ke liye $adj A$ nikaalo.

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

Theorem 1Adjoint se multiply karna

Kisi bhi square matrix ko uske adjoint se, kisi bhi side se, multiply karne par uska determinant times same size ki identity matrix $I$ milti hai: $A (adj A) = ∣ A ∣ I = (adj A) A$ .

$n = 3$ lo. $A (adj A)$ ki row $i$ , column $j$ wali entry hai $A$ ki row $i$ times $adj A$ ka column $j$ . $adj A$ ke column $j$ mein $A$ ki row $j$ ke cofactors hain, isliye yeh entry $a_{i 1} A_{j 1} + a_{i 2} A_{j 2} + a_{i 3} A_{j 3}$ hai.

Jab $i = j$ , to yeh row $i$ ke along $∣ A ∣$ ka expansion hai. Jab $i \neq j$ , to yeh ek row ko doosri row ke cofactors ke saath jodta hai, jo $0$ deta hai (Property). Isliye product ki har diagonal place par $∣ A ∣$ hai aur baaki jagah $0$ : yeh $∣ A ∣ I$ hai. Rows ki jagah columns use karne se isi tarah $(adj A) A = ∣ A ∣ I$ milta hai.

3/3 $A (adj A) = 13 I = ∣ A ∣ I$ .

Fig. 4.13Diagonal entries

∣ A ∣

ke expansions hain; off-diagonal entries do rows ko mix karti hain aur zero ho jaati hain.

Ek line mein teen matrices: A jiski rows 1, 0, 2; 3, 1, 0; 0, 2, 1 hain, phir adj A jiski rows 1, 4, −2; −3, 1, 6; 6, −2, 1 hain, phir ek equals sign aur product. A ki row 1 poore samay shaded hai. Pehle step mein adj A ka column 1 shaded hai aur product ki top-left entry 13 dikhti hai, working 1(1) + 0(−3) + 2(6) = 13 ke saath. Doosre step mein column 2 shaded hai aur agli entry 0 dikhti hai, 1(4) + 0(1) + 2(−2) = 0 se. Teesre step mein poora product dikhta hai, diagonal par 13 aur baaki jagah 0, jis par |A| I likha hai.

Singular matrices aur inverse

Definition

Singular matrix

Square matrix $A$ singular hoti hai agar $∣ A ∣ = 0$ .

Example $[\begin{array}{cc} 3 & 6 \\ 1 & 2 \end{array}]$ singular hai, kyunki $3 \cdot 2 - 6 \cdot 1 = 0$ .

Definition

Non-singular matrix

Square matrix $A$ non-singular hoti hai agar $∣ A ∣ \neq 0$ .

Example $[\begin{array}{cc} 2 & 5 \\ 1 & 4 \end{array}]$ non-singular hai: uska determinant $3$ hai.

Agle do facts higher mathematics mein prove kiye jaate hain; yahan hum unhe seedha use karte hain aur ek ko numbers par check karte hain.

Theorem 2Non-singular matrices ke products

Barabar size ki do non-singular square matrices ka product, kisi bhi order mein lo, phir se non-singular hota hai.

Theorem 3Product ka determinant

Same order ki square matrices $A$ aur $B$ ke liye, $∣ A B ∣ = ∣ A ∣ ∣ B ∣$ .

CorollaryAdjoint ka determinant

Non-singular $n \times n$ matrix $A$ ke liye, $∣ adj A ∣ = ∣ A ∣^{n - 1}$ ; $3 \times 3$ matrix ke liye yeh $∣ A ∣^{2}$ hai.

Theorem 4Matrix ka inverse kab hota hai

Square matrix $A$ ka inverse tabhi exist karta hai jab $∣ A ∣ \neq 0$ ho, aur tab yeh $A^{- 1} = \frac{1}{∣ A ∣} adj A$ se milta hai.

Inverse (jab |A| ≠ 0)

A^{- 1} = \frac{1}{∣ A ∣} adj A

Isse do quick results milte hain. $∣ A ∣ ∣ A^{- 1} ∣ = ∣ I ∣ = 1$ se humein $∣ A^{- 1} ∣ = 1 / ∣ A ∣$ milta hai. Aur inverse unique hota hai, isliye tum koi bhi method use karo, tumhe same $A^{- 1}$ tak pahunchna chahiye; singular matrix ka koi inverse hota hi nahi, jabki har non-singular matrix ka ek inverse hota hai.

Example 13NCERT Example 13 · p. 90

Theorem 1 verify karo, phir invert karo

$A = [\begin{array}{ccc} 1 & - 1 & 0 \\ 2 & 1 & 3 \\ 1 & 0 & 2 \end{array}]$ ke liye check karo ki $A (adj A) = ∣ A ∣ I$ aur $A^{- 1}$ nikaalo.

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

Example 14NCERT Example 14 · p. 91

Product ka inverse

$A = [\begin{array}{cc} 3 & 1 \\ 5 & 2 \end{array}]$ aur $B = [\begin{array}{cc} 2 & - 1 \\ 1 & 1 \end{array}]$ ke liye verify karo ki $(A B)^{- 1} = B^{- 1} A^{- 1}$ .

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

Example 15NCERT Example 15 · p. 92

Matrix equation se inverse

Dikhao ki $A = [\begin{array}{cc} 5 & 2 \\ 3 & 1 \end{array}]$ , $A^{2} - 6 A - I = O$ ko satisfy karti hai, aur isse $A^{- 1}$ nikaalo.

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

Quick check

4 mein se 1

A = [\begin{array}{cc} 6 & - 2 \\ 4 & 1 \end{array}]

ke liye,

adj A

hai

[\begin{array}{cc} 1 & 2 \\ - 4 & 6 \end{array}]

[\begin{array}{cc} 1 & - 4 \\ 2 & 6 \end{array}]

[\begin{array}{cc} 1 & - 2 \\ 4 & 6 \end{array}]

[\begin{array}{cc} - 1 & 2 \\ - 4 & - 6 \end{array}]

Answer dekho

Answer (A) $[\begin{array}{cc} 1 & 2 \\ - 4 & 6 \end{array}]$

Diagonal entries (6 aur 1) swap karo aur baaki ke signs badlo:

- (- 2) = 2

- (4) = - 4

. Option B transpose karne se pehle wali matrix of cofactors hai.

Practice · 18 questions

Practice 4.4: answers ke saath

Short answerEasyNCERT Exercise 4.4, Q1
$[\begin{array}{cc} 3 & - 4 \\ 5 & 2 \end{array}]$ ka adjoint nikaalo.
Solution dikhao
Answer $[\begin{array}{cc} 2 & 4 \\ - 5 & 3 \end{array}]$
Cofactors: $A_{11} = 2$ , $A_{12} = - 5$ , $A_{21} = 4$ , $A_{22} = 3$ . Transpose karne par $[\begin{array}{cc} 2 & 4 \\ - 5 & 3 \end{array}]$ milta hai.
Short answerMediumNCERT Exercise 4.4, Q2
$[\begin{array}{ccc} 2 & 0 & - 1 \\ 1 & 3 & 2 \\ - 1 & 4 & 1 \end{array}]$ ka adjoint nikaalo.
Solution dikhao
Answer $[\begin{array}{ccc} - 5 & - 4 & 3 \\ - 3 & 1 & - 5 \\ 7 & - 8 & 6 \end{array}]$
Rows ke hisaab se cofactors: $A_{11} = 3 - 8 = - 5$ , $A_{12} = - (1 + 2) = - 3$ , $A_{13} = 4 + 3 = 7$ ; $A_{21} = - (0 + 4) = - 4$ , $A_{22} = 2 - 1 = 1$ , $A_{23} = - (8 - 0) = - 8$ ; $A_{31} = 0 + 3 = 3$ , $A_{32} = - (4 + 1) = - 5$ , $A_{33} = 6 - 0 = 6$ .
Transpose: $adj A = [\begin{array}{ccc} - 5 & - 4 & 3 \\ - 3 & 1 & - 5 \\ 7 & - 8 & 6 \end{array}]$ . (Check ke liye, $A$ ki row 1 times column 1 deta hai $- 10 + 0 - 7 = - 17 = ∣ A ∣$ .)
Short answerEasyNCERT Exercise 4.4, Q3
Check karo ki $A = [\begin{array}{cc} 3 & - 6 \\ - 2 & 4 \end{array}]$ hone par $A (adj A)$ aur $(adj A) A$ dono $∣ A ∣ I$ ke barabar hain.
Solution dikhao
$∣ A ∣ = 12 - 12 = 0$ aur $adj A = [\begin{array}{cc} 4 & 6 \\ 2 & 3 \end{array}]$ .
$A (adj A) = [\begin{array}{cc} 12 - 12 & 18 - 18 \\ - 8 + 8 & - 12 + 12 \end{array}] = O$ aur $(adj A) A = [\begin{array}{cc} 12 - 12 & - 24 + 24 \\ 6 - 6 & - 12 + 12 \end{array}] = O$ . Dono $0 \cdot I = ∣ A ∣ I$ ke barabar hain.
Short answerMediumNCERT Exercise 4.4, Q4
Check karo ki $A = [\begin{array}{ccc} 2 & 1 & - 1 \\ 0 & 3 & 1 \\ 4 & - 2 & 1 \end{array}]$ hone par $A (adj A)$ aur $(adj A) A$ dono $∣ A ∣ I$ ke barabar hain.
Solution dikhao
$∣ A ∣ = 2 (3 + 2) - 1 (0 - 4) + (- 1) (0 - 12) = 10 + 4 + 12 = 26$ .
$adj A = [\begin{array}{ccc} 5 & 1 & 4 \\ 4 & 6 & - 2 \\ - 12 & 8 & 6 \end{array}]$ .
Kisi bhi order mein multiply karne par $[\begin{array}{ccc} 26 & 0 & 0 \\ 0 & 26 & 0 \\ 0 & 0 & 26 \end{array}] = 26 I$ milta hai. Jaise, $A$ ki row 1 times $adj A$ ka column 1 hai $10 + 4 + 12 = 26$ , aur row 1 times column 2 hai $2 + 6 - 8 = 0$ .
Short answerEasyNCERT Exercise 4.4, Q5
$[\begin{array}{cc} 3 & 5 \\ 1 & 2 \end{array}]$ ka inverse nikaalo, agar exist karta hai.
Solution dikhao
Answer $[\begin{array}{cc} 2 & - 5 \\ - 1 & 3 \end{array}]$
$∣ A ∣ = 6 - 5 = 1$ , isliye $A^{- 1} = adj A = [\begin{array}{cc} 2 & - 5 \\ - 1 & 3 \end{array}]$ .
Short answerEasyNCERT Exercise 4.4, Q6
$[\begin{array}{cc} - 2 & 3 \\ 4 & - 5 \end{array}]$ ka inverse nikaalo, agar exist karta hai.
Solution dikhao
Answer $[\begin{array}{cc} \frac{5}{2} & \frac{3}{2} \\ 2 & 1 \end{array}]$
$∣ A ∣ = 10 - 12 = - 2$ aur $adj A = [\begin{array}{cc} - 5 & - 3 \\ - 4 & - 2 \end{array}]$ , isliye $A^{- 1} = - \frac{1}{2} adj A = [\begin{array}{cc} \frac{5}{2} & \frac{3}{2} \\ 2 & 1 \end{array}]$ .
Short answerMediumNCERT Exercise 4.4, Q7
$[\begin{array}{ccc} 2 & 1 & 3 \\ 0 & 1 & - 2 \\ 0 & 0 & 4 \end{array}]$ ka inverse nikaalo, agar exist karta hai.
Solution dikhao
Answer $\frac{1}{8} [\begin{array}{ccc} 4 & - 4 & - 5 \\ 0 & 8 & 4 \\ 0 & 0 & 2 \end{array}]$
$∣ A ∣ = 2 \times 1 \times 4 = 8$ (column 1 ke along neeche expand karo).
Cofactors: $A_{11} = 4$ , $A_{12} = 0$ , $A_{13} = 0$ , $A_{21} = - 4$ , $A_{22} = 8$ , $A_{23} = 0$ , $A_{31} = - 5$ , $A_{32} = 4$ , $A_{33} = 2$ .
$A^{- 1} = \frac{1}{8} [\begin{array}{ccc} 4 & - 4 & - 5 \\ 0 & 8 & 4 \\ 0 & 0 & 2 \end{array}] = [\begin{array}{ccc} \frac{1}{2} & - \frac{1}{2} & - \frac{5}{8} \\ 0 & 1 & \frac{1}{2} \\ 0 & 0 & \frac{1}{4} \end{array}]$ . Inverse bhi upper triangular hai.
Short answerMediumNCERT Exercise 4.4, Q8
$[\begin{array}{ccc} 1 & 0 & 0 \\ 2 & - 1 & 0 \\ 4 & 3 & 2 \end{array}]$ ka inverse nikaalo, agar exist karta hai.
Solution dikhao
Answer $[\begin{array}{ccc} 1 & 0 & 0 \\ 2 & - 1 & 0 \\ - 5 & \frac{3}{2} & \frac{1}{2} \end{array}]$
$∣ A ∣ = 1 \times (- 1) \times 2 = - 2$ .
$adj A = [\begin{array}{ccc} - 2 & 0 & 0 \\ - 4 & 2 & 0 \\ 10 & - 3 & - 1 \end{array}]$ .
$A^{- 1} = - \frac{1}{2} adj A = [\begin{array}{ccc} 1 & 0 & 0 \\ 2 & - 1 & 0 \\ - 5 & \frac{3}{2} & \frac{1}{2} \end{array}]$ .
Short answerMediumNCERT Exercise 4.4, Q9
$[\begin{array}{ccc} 1 & 2 & - 1 \\ 3 & 0 & 2 \\ - 2 & 1 & 1 \end{array}]$ ka inverse nikaalo, agar exist karta hai.
Solution dikhao
Answer $\frac{1}{19} [\begin{array}{ccc} 2 & 3 & - 4 \\ 7 & 1 & 5 \\ - 3 & 5 & 6 \end{array}]$
$R_{2}$ ke along: $∣ A ∣ = - 3 (2 + 1) + 0 - 2 (1 + 4) = - 9 - 10 = - 19$ .
$adj A = [\begin{array}{ccc} - 2 & - 3 & 4 \\ - 7 & - 1 & - 5 \\ 3 & - 5 & - 6 \end{array}]$ .
$A^{- 1} = - \frac{1}{19} adj A = \frac{1}{19} [\begin{array}{ccc} 2 & 3 & - 4 \\ 7 & 1 & 5 \\ - 3 & 5 & 6 \end{array}]$ .
Short answerMediumNCERT Exercise 4.4, Q10
$[\begin{array}{ccc} 2 & - 1 & 1 \\ 1 & 3 & - 2 \\ 0 & 4 & 1 \end{array}]$ ka inverse nikaalo, agar exist karta hai.
Solution dikhao
Answer $\frac{1}{27} [\begin{array}{ccc} 11 & 5 & - 1 \\ - 1 & 2 & 5 \\ 4 & - 8 & 7 \end{array}]$
$C_{1}$ ke along: $∣ A ∣ = 2 (3 + 8) - 1 (- 1 - 4) + 0 = 22 + 5 = 27$ .
$adj A = [\begin{array}{ccc} 11 & 5 & - 1 \\ - 1 & 2 & 5 \\ 4 & - 8 & 7 \end{array}]$ , isliye $A^{- 1} = \frac{1}{27} adj A$ .
Short answerMediumNCERT Exercise 4.4, Q11
$[\begin{array}{ccc} \cos α & 0 & \sin α \\ 0 & 1 & 0 \\ - \sin α & 0 & \cos α \end{array}]$ ka inverse nikaalo.
Solution dikhao
Answer $[\begin{array}{ccc} \cos α & 0 & - \sin α \\ 0 & 1 & 0 \\ \sin α & 0 & \cos α \end{array}]$
$R_{2}$ ke along: $∣ A ∣ = 1 (\cos^{2} α + \sin^{2} α) = 1$ .
Cofactors se $adj A = [\begin{array}{ccc} \cos α & 0 & - \sin α \\ 0 & 1 & 0 \\ \sin α & 0 & \cos α \end{array}]$ milta hai, aur kyunki $∣ A ∣ = 1$ , yahi $A^{- 1}$ hai. Yeh $A$ ka transpose hai: rotation ka inverse ulti taraf wapas ghuma deta hai.
Short answerMediumNCERT Exercise 4.4, Q12
Maan lo $A = [\begin{array}{cc} 2 & 5 \\ 1 & 3 \end{array}]$ aur $B = [\begin{array}{cc} 4 & 3 \\ 3 & 2 \end{array}]$ . Verify karo ki $(A B)^{- 1} = B^{- 1} A^{- 1}$ .
Solution dikhao
$A B = [\begin{array}{cc} 23 & 16 \\ 13 & 9 \end{array}]$ , $∣ A B ∣ = 207 - 208 = - 1$ , isliye $(A B)^{- 1} = [\begin{array}{cc} - 9 & 16 \\ 13 & - 23 \end{array}]$ .
$∣ A ∣ = 1$ , $A^{- 1} = [\begin{array}{cc} 3 & - 5 \\ - 1 & 2 \end{array}]$ ; $∣ B ∣ = - 1$ , $B^{- 1} = [\begin{array}{cc} - 2 & 3 \\ 3 & - 4 \end{array}]$ .
$B^{- 1} A^{- 1} = [\begin{array}{cc} - 6 - 3 & 10 + 6 \\ 9 + 4 & - 15 - 8 \end{array}] = [\begin{array}{cc} - 9 & 16 \\ 13 & - 23 \end{array}] = (A B)^{- 1}$ .
Short answerMediumNCERT Exercise 4.4, Q13
Agar $A = [\begin{array}{cc} 2 & - 1 \\ 3 & 1 \end{array}]$ , to dikhao ki $A^{2} - 3 A + 5 I = O$ . Isse $A^{- 1}$ nikaalo.
Solution dikhao
Answer $A^{- 1} = \frac{1}{5} [\begin{array}{cc} 1 & 1 \\ - 3 & 2 \end{array}]$
$A^{2} = [\begin{array}{cc} 1 & - 3 \\ 9 & - 2 \end{array}]$ , isliye $A^{2} - 3 A + 5 I = [\begin{array}{cc} 1 - 6 + 5 & - 3 + 3 \\ 9 - 9 & - 2 - 3 + 5 \end{array}] = O$ .
$∣ A ∣ = 5 \neq 0$ . $A^{- 1}$ se multiply karne par: $A - 3 I + 5 A^{- 1} = O$ , isliye $A^{- 1} = \frac{1}{5} (3 I - A) = \frac{1}{5} [\begin{array}{cc} 1 & 1 \\ - 3 & 2 \end{array}]$ .
Short answerMediumNCERT Exercise 4.4, Q14
$a$ aur $b$ aise nikaalo ki $A^{2} + a A + b I = O$ ho, jahan $A = [\begin{array}{cc} 4 & 3 \\ 1 & 2 \end{array}]$ .
Solution dikhao
Answer $a = - 6$ , $b = 5$
$A^{2} = [\begin{array}{cc} 19 & 18 \\ 6 & 7 \end{array}]$ . $A^{2} + a A + b I$ ki top-right entry $18 + 3 a = 0$ hai, isliye $a = - 6$ .
Top-left entry $19 + 4 a + b = 19 - 24 + b = 0$ hai, isliye $b = 5$ . Bottom row check karo: $6 - 6 = 0$ aur $7 - 12 + 5 = 0$ .
Short answerHardNCERT Exercise 4.4, Q15
$A = [\begin{array}{ccc} 2 & 1 & 0 \\ 1 & - 1 & 1 \\ 0 & 1 & 2 \end{array}]$ ke liye dikhao ki $A^{3} - 3 A^{2} - 2 A + 8 I = O$ . Isse $A^{- 1}$ nikaalo.
Solution dikhao
Answer $A^{- 1} = \frac{1}{8} [\begin{array}{ccc} 3 & 2 & - 1 \\ 2 & - 4 & 2 \\ - 1 & 2 & 3 \end{array}]$
$A^{2} = [\begin{array}{ccc} 5 & 1 & 1 \\ 1 & 3 & 1 \\ 1 & 1 & 5 \end{array}]$ aur $A^{3} = [\begin{array}{ccc} 11 & 5 & 3 \\ 5 & - 1 & 5 \\ 3 & 5 & 11 \end{array}]$ . Entry by entry, $A^{3} - 3 A^{2} - 2 A + 8 I = O$ (jaise top-left entry $11 - 15 - 4 + 8 = 0$ hai).
$∣ A ∣ = 2 (- 3) - 1 (2) + 0 = - 8 \neq 0$ . $A^{- 1}$ se multiply karne par: $A^{2} - 3 A - 2 I + 8 A^{- 1} = O$ , isliye $A^{- 1} = - \frac{1}{8} (A^{2} - 3 A - 2 I) = - \frac{1}{8} [\begin{array}{ccc} - 3 & - 2 & 1 \\ - 2 & 4 & - 2 \\ 1 & - 2 & - 3 \end{array}] = \frac{1}{8} [\begin{array}{ccc} 3 & 2 & - 1 \\ 2 & - 4 & 2 \\ - 1 & 2 & 3 \end{array}]$ .
Short answerHardNCERT Exercise 4.4, Q16
$A = [\begin{array}{ccc} 3 & 1 & 1 \\ 1 & 3 & 1 \\ 1 & 1 & 3 \end{array}]$ ke liye verify karo ki $A^{3} - 9 A^{2} + 24 A - 20 I = O$ aur isse $A^{- 1}$ nikaalo.
Solution dikhao
Answer $A^{- 1} = \frac{1}{10} [\begin{array}{ccc} 4 & - 1 & - 1 \\ - 1 & 4 & - 1 \\ - 1 & - 1 & 4 \end{array}]$
$A^{2}$ ke diagonal par $11$ hai aur baaki jagah $7$ ; $A^{3}$ ke diagonal par $47$ hai aur baaki jagah $39$ . Diagonal check: $47 - 99 + 72 - 20 = 0$ ; off-diagonal check: $39 - 63 + 24 = 0$ .
$∣ A ∣ = 3 (8) - 1 (2) + 1 (- 2) = 20 \neq 0$ . $A^{- 1}$ se multiply karne par: $A^{2} - 9 A + 24 I - 20 A^{- 1} = O$ , isliye $A^{- 1} = \frac{1}{20} (A^{2} - 9 A + 24 I)$ .
$A^{2} - 9 A + 24 I$ ke diagonal par $11 - 27 + 24 = 8$ hai aur baaki jagah $7 - 9 = - 2$ , isliye $A^{- 1} = \frac{1}{20} [\begin{array}{ccc} 8 & - 2 & - 2 \\ - 2 & 8 & - 2 \\ - 2 & - 2 & 8 \end{array}] = \frac{1}{10} [\begin{array}{ccc} 4 & - 1 & - 1 \\ - 1 & 4 & - 1 \\ - 1 & - 1 & 4 \end{array}]$ .
Quick answerEasyNCERT Exercise 4.4, Q17
$A$ ek non-singular $3 \times 3$ matrix hai jiska $∣ A ∣ = 5$ hai. $∣ adj A ∣$ nikaalo.
Answer dekho
Answer 25
$∣ adj A ∣ = 5^{2} = 25$ .
Quick answerMediumNCERT Exercise 4.4, Q18
$A$ ek invertible $3 \times 3$ matrix hai jiska $∣ A ∣ = - 4$ hai. To $∣ 2 A^{- 1} ∣$ barabar hai
$- 2$ $- \frac{1}{2}$ $- 8$ $\frac{1}{2}$
Answer dekho
Answer (A) $- 2$
$∣ 2 A^{- 1} ∣ = 2^{3} ∣ A^{- 1} ∣ = 8 \times (- \frac{1}{4}) = - 2$ . Answer $- \frac{1}{2}$ mein cube bhool gaye.

4.6 Determinants aur Matrices ke Applications

Mixed packs ki price tay karta dukaandaar, kisi joint par forces balance karta engineer aur nutrients match karta dietitian, sab aakhir mein kai unknowns mein kai linear equations par pahunchte hain. Matrices se hum aise system ko ek single equation ki tarah likh sakte hain, aur phir inverse use ek hi baar mein solve kar deta hai. Pehle, jo ho sakta hai uske liye do words.

Definition

Consistent system

Equations ka system consistent hota hai agar uska kam se kam ek solution ho.

Definition

Inconsistent system

Equations ka system inconsistent hota hai agar uska koi solution na ho.

Consistent system ka exactly ek solution ho sakta hai ya infinitely many. Is chapter mein hum sirf exactly ek solution wale systems ko solve karte hain; baaki ke liye hum bas tay karte hain ki solutions exist karte hain ya nahi.

4.6.1 Matrix ke inverse se linear equations ke system ka solution

$a_{1} x + b_{1} y + c_{1} z = d_{1}$ , $a_{2} x + b_{2} y + c_{2} z = d_{2}$ , $a_{3} x + b_{3} y + c_{3} z = d_{3}$ ke coefficients, unknowns aur right-hand sides ko teen matrices mein ikattha karo. Phir matrix multiplication teeno equations ko ek bana deta hai:

Matrix form AX = B

\underset{A}{\underset{⏟}{[\begin{array}{ccc} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array}]}} \underset{X}{\underset{⏟}{[\begin{array}{c} x \\ y \\ z \end{array}]}} = \underset{B}{\underset{⏟}{[\begin{array}{c} d_{1} \\ d_{2} \\ d_{3} \end{array}]}}

DerivationCase I: coefficient matrix non-singular hai

Agar $∣ A ∣ \neq 0$ , to $A^{- 1}$ exist karta hai. $A X = B$ ki dono sides ko left se $A^{- 1}$ se multiply karo aur regroup karo (matrix multiplication associative hai):

\begin{aligned} A^{- 1} (A X) & = A^{- 1} B \\ (A^{- 1} A) X & = A^{- 1} B \\ I X & = A^{- 1} B \end{aligned}

Matrix method

X = A^{- 1} B

Inverses unique hote hain, isliye koi aur $X$ is equation ko satisfy nahi kar sakta. Yeh recipe, coefficient matrix ko invert karo aur multiply karo, matrix method hai.

Case II: $∣ A ∣ = 0$ . Ab koi inverse nahi hai, lekin Theorem 1 phir bhi madad karta hai. $A X = B$ ko left se $adj A$ se multiply karo: kyunki $(adj A) A = ∣ A ∣ I = O$ , koi bhi solution $X$ hone par $(adj A) B = O$ hona zaroori hai. Isliye agar $(adj A) B \neq O$ , to koi solution exist nahi kar sakta aur system inconsistent hai. Agar $(adj A) B = O$ , to test chup rehta hai: system ke infinitely many solutions ho sakte hain (tab woh consistent hai) ya koi nahi, aur yeh tum khud equations dekh kar tay karte ho (Fig. 4.14).

3/3 $(adj A) B$ compute karo. Agar yeh $O$ nahi hai, to koi solution nahi. Agar yeh $O$ hai, to equations check karo: infinitely many solutions ya koi nahi.

Fig. 4.14Pehle

∣ A ∣

dekho; sirf agar woh

0

ho, tabhi tumhe

(adj A) B

chahiye.

Ek flow chart. Upar ek box AX = B ek diamond tak le jaata hai jo poochta hai ki kya |A| = 0 hai. Doosre step mein no branch dikhti hai: yeh ek green box tak jaati hai jis par likha hai X = A inverse B, exactly one solution. Teesre step mein yes branch uske bagal mein judti hai: yeh ek box find (adj A)B tak jaati hai, jo do mein batata hai: ek orange box not O, no solution, aur doosra box equals O, test further.

Example

Do singular systems

Tay karo ki har system consistent hai ya nahi. (a) $x + y + z = 6$ , $x + 2 y + 3 z = 14$ , $2 x + 3 y + 4 z = 20$ . (b) Wahi left-hand sides, right-hand sides $6$ , $14$ , $21$ ke saath.

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

Example 16NCERT Example 16 · p. 95

Do unknowns mein do equations

$4 x + 3 y = 11$ , $3 x - 2 y = 4$ ko matrix method se solve karo.

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

Example 17NCERT Example 17 · p. 95

Teen unknowns mein teen equations

Matrix method se solve karo: $2 x + y - z = 0$ , $x - 3 y + 2 z = 11$ , $3 x + 2 y + z = 7$ .

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

Example 18NCERT Example 18 · p. 96

Mela mein ticket ki keematein

Gaon ke ek mele mein ek family 2 adult, 3 child aur 1 senior ticket ke liye ₹270 deti hai. Doosri family 1 adult, 2 child aur 2 senior tickets ke liye ₹200 deti hai, aur teesri 3 adult, 1 child aur 2 senior tickets ke liye ₹280 deti hai. Matrix method se har tarah ke ticket ki price nikaalo.

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

Quick check

4 mein se 1

A X = B

ke liye, maan lo

∣ A ∣ = 0

aur

(adj A) B \neq O

. To system ka

exactly ek solution haiinfinitely many solutions hainkoi solution nahi haisolution ho bhi sakta hai aur nahi bhi

Answer dekho

Answer (C) koi solution nahi hai

Koi bhi solution hone par

(adj A) B = ∣ A ∣ X = O

hona zaroori hai. Kyunki yeh

O

nahi hai, koi solution nahi hai. “Ho bhi sakta hai aur nahi bhi” wala faisla sirf tab hota hai jab

(adj A) B = O

Miscellaneous example

Example 19NCERT Example 19 · p. 98

Product se inverse padh lena

$P = [\begin{array}{ccc} 1 & 0 & - 1 \\ 1 & 2 & - 2 \\ 1 & - 1 & 0 \end{array}]$ aur $Q = [\begin{array}{ccc} - 2 & 1 & 2 \\ - 2 & 1 & 1 \\ - 3 & 1 & 2 \end{array}]$ ke liye product $P Q$ nikaalo. Use use karke $x - z = 1$ , $x + 2 y - 2 z = 1$ , $x - y = 2$ solve karo.

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

Practice · 13 questions

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

ProofMediumNCERT Miscellaneous Exercises, Q1
Dikhao ki $∣ \begin{array}{ccc} 1 & x & \sin θ \\ - x & 1 & \cos θ \\ - \sin θ & - \cos θ & 1 \end{array} ∣$ , $θ$ par depend nahi karta.
Solution dikhao
$R_{1}$ ke along: $1 (1 + \cos^{2} θ) - x (- x + \sin θ \cos θ) + \sin θ (x \cos θ + \sin θ)$ .
$= 1 + \cos^{2} θ + x^{2} - x \sin θ \cos θ + x \sin θ \cos θ + \sin^{2} θ = x^{2} + 2$ , jismein koi $θ$ nahi hai.
Short answerMediumNCERT Miscellaneous Exercises, Q2
$∣ \begin{array}{ccc} \cos θ & - \sin θ \cos φ & \sin θ \sin φ \\ \sin θ & \cos θ \cos φ & - \cos θ \sin φ \\ 0 & \sin φ & \cos φ \end{array} ∣$ evaluate karo.
Solution dikhao
$R_{3}$ ke along: $(3, 2)$ wala term $- \sin φ (- \cos^{2} θ \sin φ - \sin^{2} θ \sin φ) = \sin^{2} φ$ hai.
$(3, 3)$ wala term $\cos φ (\cos^{2} θ \cos φ + \sin^{2} θ \cos φ) = \cos^{2} φ$ hai. Total $\sin^{2} φ + \cos^{2} φ = 1$ hai.
Short answerMediumNCERT Miscellaneous Exercises, Q3
Diya hai $A = [\begin{array}{ccc} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 2 \end{array}]$ aur $B^{- 1} = [\begin{array}{ccc} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 0 & 1 & 1 \end{array}]$ , $(A B)^{- 1}$ nikaalo.
Solution dikhao
Answer $\frac{1}{4} [\begin{array}{ccc} 1 & - 2 & 3 \\ 5 & - 6 & 3 \\ 0 & 4 & 0 \end{array}]$
$∣ A ∣ = 1 (2) - 2 (0 - 1) + 0 = 4$ aur $A^{- 1} = \frac{1}{4} [\begin{array}{ccc} 2 & - 4 & 2 \\ 1 & 2 & - 1 \\ - 1 & 2 & 1 \end{array}]$ .
$(A B)^{- 1} = B^{- 1} A^{- 1} = \frac{1}{4} [\begin{array}{ccc} 2 - 1 & - 4 + 2 & 2 + 1 \\ 4 + 1 & - 8 + 2 & 4 - 1 \\ 1 - 1 & 2 + 2 & - 1 + 1 \end{array}] = \frac{1}{4} [\begin{array}{ccc} 1 & - 2 & 3 \\ 5 & - 6 & 3 \\ 0 & 4 & 0 \end{array}]$ .
ProofHardNCERT Miscellaneous Exercises, Q4
Maan lo $A = [\begin{array}{ccc} 1 & 0 & 2 \\ - 1 & 1 & 1 \\ 2 & 1 & 1 \end{array}]$ . Verify karo ki (a) $(adj A)^{- 1} = adj (A^{- 1})$ aur (b) $(A^{- 1})^{- 1} = A$ .
Solution dikhao
$∣ A ∣ = 1 (1 - 1) - 0 + 2 (- 1 - 2) = - 6$ aur $adj A = [\begin{array}{ccc} 0 & 2 & - 2 \\ 3 & - 3 & - 3 \\ - 3 & - 1 & 1 \end{array}]$ , isliye $A^{- 1} = - \frac{1}{6} adj A$ .
(a) $adj A = ∣ A ∣ A^{- 1}$ , isliye $(adj A)^{- 1} = \frac{1}{∣ A ∣} A = - \frac{1}{6} A$ . Saath hi $adj (A^{- 1}) = ∣ A^{- 1} ∣ (A^{- 1})^{- 1} = \frac{1}{∣ A ∣} A = - \frac{1}{6} A$ . Dono $[\begin{array}{ccc} - \frac{1}{6} & 0 & - \frac{1}{3} \\ \frac{1}{6} & - \frac{1}{6} & - \frac{1}{6} \\ - \frac{1}{3} & - \frac{1}{6} & - \frac{1}{6} \end{array}]$ ke barabar hain; ise $adj A$ se multiply karne par $I$ milta hai, jo ise confirm karta hai.
(b) $A^{- 1} A = A A^{- 1} = I$ batata hai ki $A$ , $A^{- 1}$ ka ek inverse hai; inverses unique hote hain, isliye $(A^{- 1})^{- 1} = A$ . Numbers mein, $A^{- 1} \cdot A = - \frac{1}{6} (adj A) A = - \frac{1}{6} (- 6 I) = I$ .
Short answerMediumNCERT Miscellaneous Exercises, Q5
$∣ \begin{array}{ccc} p & q & q \\ q & p & q \\ q & q & p \end{array} ∣$ evaluate karo aur answer ko factors ke product ki tarah likho.
Solution dikhao
Answer $(p - q)^{2} (p + 2 q)$
$R_{1}$ ke along: $p (p^{2} - q^{2}) - q (p q - q^{2}) + q (q^{2} - p q) = p^{3} - 3 p q^{2} + 2 q^{3}$ .
Yeh $p = q$ par zero ho jaata hai, aur divide karne par $p^{3} - 3 p q^{2} + 2 q^{3} = (p - q)^{2} (p + 2 q)$ milta hai (expand karke check karo).
Short answerMediumNCERT Miscellaneous Exercises, Q6
$∣ \begin{array}{ccc} x + 1 & x & x \\ x & x + 2 & x \\ x & x & x + 3 \end{array} ∣$ evaluate karo.
Solution dikhao
Answer $11 x + 6$
$R_{1}$ ke along: $(x + 1) [(x + 2) (x + 3) - x^{2}] - x [x (x + 3) - x^{2}] + x [x^{2} - x (x + 2)]$ .
$= (x + 1) (5 x + 6) - x (3 x) + x (- 2 x) = 5 x^{2} + 11 x + 6 - 3 x^{2} - 2 x^{2} = 11 x + 6$ .
Short answerHardNCERT Miscellaneous Exercises, Q7
$\frac{4}{x} + \frac{3}{y} + \frac{8}{z} = 5$ , $\frac{2}{x} - \frac{6}{y} + \frac{4}{z} = 0$ , $\frac{6}{x} + \frac{9}{y} - \frac{4}{z} = 5$ solve karo.
Solution dikhao
Answer $x = 2$ , $y = 3$ , $z = 4$
$u = \frac{1}{x}$ , $v = \frac{1}{y}$ , $w = \frac{1}{z}$ ke saath: $4 u + 3 v + 8 w = 5$ , $2 u - 6 v + 4 w = 0$ , $6 u + 9 v - 4 w = 5$ .
$∣ A ∣ = 4 (24 - 36) - 3 (- 8 - 24) + 8 (18 + 36) = - 48 + 96 + 432 = 480$ aur $adj A = [\begin{array}{ccc} - 12 & 84 & 60 \\ 32 & - 64 & 0 \\ 54 & - 18 & - 30 \end{array}]$ .
$[\begin{array}{c} u \\ v \\ w \end{array}] = \frac{1}{480} [\begin{array}{c} - 60 + 0 + 300 \\ 160 + 0 + 0 \\ 270 + 0 - 150 \end{array}] = [\begin{array}{c} \frac{1}{2} \\ \frac{1}{3} \\ \frac{1}{4} \end{array}]$ , isliye $x = 2$ , $y = 3$ , $z = 4$ .
Quick answerMediumNCERT Miscellaneous Exercises, Q8
Agar $x$ , $y$ , $z$ non-zero hain, to $A = [\begin{array}{ccc} 0 & 0 & x \\ 0 & y & 0 \\ z & 0 & 0 \end{array}]$ ka inverse hai
$[\begin{array}{ccc} 0 & 0 & x^{- 1} \\ 0 & y^{- 1} & 0 \\ z^{- 1} & 0 & 0 \end{array}]$ $\frac{1}{x y z} [\begin{array}{ccc} 0 & 0 & x \\ 0 & y & 0 \\ z & 0 & 0 \end{array}]$ $[\begin{array}{ccc} 0 & 0 & - z^{- 1} \\ 0 & y^{- 1} & 0 \\ - x^{- 1} & 0 & 0 \end{array}]$ $[\begin{array}{ccc} 0 & 0 & z^{- 1} \\ 0 & y^{- 1} & 0 \\ x^{- 1} & 0 & 0 \end{array}]$
Answer dekho
Answer (D) $[\begin{array}{ccc} 0 & 0 & z^{- 1} \\ 0 & y^{- 1} & 0 \\ x^{- 1} & 0 & 0 \end{array}]$
$A$ ki row 1, $(0, 0, x)$ , inverse ke column 1 se milkar 1 deni chahiye, isliye bottom-left entry $x^{- 1}$ hai; isi tarah top-right entry $z^{- 1}$ hai. Option A inhe swap kar deta hai.
Quick answerHardNCERT Miscellaneous Exercises, Q9
Maan lo $A = [\begin{array}{ccc} 2 & \sin θ & 0 \\ - \sin θ & 2 & \sin θ \\ 0 & - \sin θ & 2 \end{array}]$ . Jab $θ$ saare real numbers par badalta hai, to $\det A$ ki sabse badi value kya hai?
Answer dekho
Answer 12
$R_{1}$ ke along: $2 (4 + \sin^{2} θ) - \sin θ (- 2 \sin θ) = 8 + 4 \sin^{2} θ$ . Kyunki $0 \leq \sin^{2} θ \leq 1$ , $\det A$ $8$ ( $θ = 0$ par) se $12$ ( $θ = \frac{π}{2}$ par) tak har value leta hai; sabse badi value $12$ hai.
Quick answerMedium
$A$ aur $B$ $3 \times 3$ matrices hain jinke $∣ A ∣ = 2$ aur $∣ B ∣ = - 3$ hain. $∣ 2 A B ∣$ nikaalo.
Answer dekho
Answer -48
$∣ 2 A B ∣ = 2^{3} ∣ A ∣ ∣ B ∣ = 8 \times 2 \times (- 3) = - 48$ .
11
Quick answerEasy
Har quantity (non-singular $3 \times 3$ matrix $A$ ke liye) ko uski value se match karo.
Left mein koi item tap karo, phir right mein uska match.
Answer dekho
Answer
$∣ A^{- 1} ∣$ : $1 / ∣ A ∣$
$∣ adj A ∣$ : $∣ A ∣^{2}$
$∣ 5 A ∣$ : $125 ∣ A ∣$
$∣ A^{T} ∣$ : $∣ A ∣$
$∣ A ∣ ∣ A^{- 1} ∣ = 1$ ; $∣ adj A ∣ = ∣ A ∣^{n - 1}$ jahan $n = 3$ ; $∣ k A ∣ = k^{3} ∣ A ∣$ ; aur determinant ko rows ya columns kisi ke bhi along expand kar sakte hain, isliye transpose karne se woh nahi badalta.
Quick answerEasy
$k$ ki kis value ke liye points $(1, 1)$ , $(2, 3)$ aur $(4, k)$ collinear hain?
Answer dekho
Answer 7
$1 (3 - k) - 1 (2 - 4) + (2 k - 12) = k - 7 = 0$ , isliye $k = 7$ .
Quick answerEasy
Jis square matrix ka determinant zero ho, use matrix kehte hain.
Answer dekho
Answer singular
$∣ A ∣ = 0$ wali matrix singular hoti hai; uska koi inverse nahi hota.

Flash cards · 10

Ek nazar mein formulas

1 / 10

Recap

Summary aur key points

Har square matrix ka ek determinant hota hai: $∣ [a] ∣ = a$ aur $∣ \begin{array}{cc} a & b \\ c & d \end{array} ∣ = a d - b c$ (Section 4.2.2).
$3 \times 3$ determinant ko kisi bhi row ya column ke along signs $(- 1)^{i + j}$ ke saath expand kar sakte hain; chhe ke chhe expansions same answer dete hain, isliye sabse zyada zeros wali line chuno (Section 4.2.3).
Order $n$ ki square matrix ke liye, $∣ k A ∣ = k^{n} ∣ A ∣$ (Property).
Triangle ka area $∣ Δ ∣$ hai, jahan rows $(x_{i}, y_{i}, 1)$ se bane determinant ke liye $Δ = \frac{1}{2} \det$ ; iska sign batata hai ki vertices anticlockwise chalte hain ya clockwise (Section 4.3).
Teen points tabhi collinear hote hain jab $Δ = 0$ ho, aur isse do points se guzarne wali line ki equation bhi milti hai (line ki equation).
Minor $M_{i j}$ woh determinant hai jo row $i$ aur column $j$ delete karne ke baad bachta hai; cofactor $A_{i j} = (- 1)^{i + j} M_{i j}$ hai (Section 4.4).
$∣ A ∣$ kisi bhi row ya column ki entries times unke apne cofactors ke sum ke barabar hai; kisi doosri line ke cofactors ke saath sum $0$ hota hai (Property).
$adj A$ matrix of cofactors ka transpose hai; order 2 ke liye diagonal swap karo aur baaki ka sign badlo (Section 4.5.1).
$A (adj A)$ aur $(adj A) A$ dono $∣ A ∣ I$ ke barabar hain (Theorem 1).
$A$ singular hai agar $∣ A ∣ = 0$ , warna non-singular; $∣ A B ∣ = ∣ A ∣ ∣ B ∣$ , $∣ A^{- 1} ∣ = 1 / ∣ A ∣$ aur $∣ adj A ∣ = ∣ A ∣^{n - 1}$ (Singular matrices aur inverse).
$A$ ka inverse tabhi hota hai jab $∣ A ∣ \neq 0$ , aur tab $A^{- 1} = \frac{1}{∣ A ∣} adj A$ (Theorem 4).
Agar $∣ A ∣ \neq 0$ , to system $A X = B$ ka unique solution $X = A^{- 1} B$ hai (Section 4.6.1).
Agar $∣ A ∣ = 0$ : jab $(adj A) B \neq O$ to system inconsistent hai; jab $(adj A) B = O$ to iske infinitely many solutions hain ya koi nahi, aur tum seedha check karte ho (Fig. 4.14).

Timeline

Historical note: determinants kaise bane

c. 200 BCE–100 CE
China mein counting-rod tables
Nine Chapters on the Mathematical Art kai equations ke coefficients ko rods ke grid ki tarah rakhta hai aur columns ko jod kar unknowns hataata hai.
1683
Seki Takakazu
Yeh Japanese mathematician equations se ek unknown hataane ke liye determinant jaise expansions likhte hain.
1693
Leibniz
Ek letter mein woh coefficients par do indices lagate hain aur batate hain ki system kab solve ho sakta hai.
1750
Cramer's rule
Gabriel Cramer ek rule publish karte hain jo n unknowns mein n equations ko aise expressions ke ratios ke roop mein solve karta hai.
1771–1773
Vandermonde, Laplace aur Lagrange
Vandermonde determinants ko unke apne liye padhte hain, Laplace unhe minors se expand karte hain, aur Lagrange unhe areas aur volumes ke liye use karte hain.
1801
Gauss
Woh determinant word use karte hain, lekin quadratic forms se judi ek quantity ke liye.
1812
Binet aur Cauchy
Dono product rule $∣ A B ∣ = ∣ A ∣ ∣ B ∣$ prove karte hain; Cauchy word ko uska aaj ka matlab dete hain.
1841
Jacobi
Unke papers determinants ko poore mathematics mein ek standard tool bana dete hain.

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 4.1, Q1	Practice 4.1 · Q1
NCERT Exercise 4.1, Q2	Practice 4.1 · Q2
NCERT Exercise 4.1, Q3	Practice 4.1 · Q3
NCERT Exercise 4.1, Q4	Practice 4.1 · Q4
NCERT Exercise 4.1, Q5	Practice 4.1 · Q5
NCERT Exercise 4.1, Q6	Practice 4.1 · Q6
NCERT Exercise 4.1, Q7	Practice 4.1 · Q7
NCERT Exercise 4.1, Q8	Practice 4.1 · Q8
NCERT Exercise 4.2, Q1	Practice 4.2 · Q1
NCERT Exercise 4.2, Q2	Practice 4.2 · Q2
NCERT Exercise 4.2, Q3	Practice 4.2 · Q3
NCERT Exercise 4.2, Q4	Practice 4.2 · Q4
NCERT Exercise 4.2, Q5	Practice 4.2 · Q5
NCERT Exercise 4.3, Q1	Practice 4.3 · Q1
NCERT Exercise 4.3, Q2	Practice 4.3 · Q2
NCERT Exercise 4.3, Q3	Practice 4.3 · Q3
NCERT Exercise 4.3, Q4	Practice 4.3 · Q4
NCERT Exercise 4.3, Q5	Practice 4.3 · Q5
NCERT Exercise 4.4, Q1	Practice 4.4 · Q1
NCERT Exercise 4.4, Q2	Practice 4.4 · Q2
NCERT Exercise 4.4, Q3	Practice 4.4 · Q3
NCERT Exercise 4.4, Q4	Practice 4.4 · Q4
NCERT Exercise 4.4, Q5	Practice 4.4 · Q5
NCERT Exercise 4.4, Q6	Practice 4.4 · Q6
NCERT Exercise 4.4, Q7	Practice 4.4 · Q7
NCERT Exercise 4.4, Q8	Practice 4.4 · Q8
NCERT Exercise 4.4, Q9	Practice 4.4 · Q9
NCERT Exercise 4.4, Q10	Practice 4.4 · Q10
NCERT Exercise 4.4, Q11	Practice 4.4 · Q11
NCERT Exercise 4.4, Q12	Practice 4.4 · Q12
NCERT Exercise 4.4, Q13	Practice 4.4 · Q13
NCERT Exercise 4.4, Q14	Practice 4.4 · Q14
NCERT Exercise 4.4, Q15	Practice 4.4 · Q15
NCERT Exercise 4.4, Q16	Practice 4.4 · Q16
NCERT Exercise 4.4, Q17	Practice 4.4 · Q17
NCERT Exercise 4.4, Q18	Practice 4.4 · Q18
NCERT Exercise 4.5, Q1	Practice 4.5 · Q1
NCERT Exercise 4.5, Q2	Practice 4.5 · Q2
NCERT Exercise 4.5, Q3	Practice 4.5 · Q3
NCERT Exercise 4.5, Q4	Practice 4.5 · Q4
NCERT Exercise 4.5, Q5	Practice 4.5 · Q5
NCERT Exercise 4.5, Q6	Practice 4.5 · Q6
NCERT Exercise 4.5, Q7	Practice 4.5 · Q7
NCERT Exercise 4.5, Q8	Practice 4.5 · Q8
NCERT Exercise 4.5, Q9	Practice 4.5 · Q9
NCERT Exercise 4.5, Q10	Practice 4.5 · Q10
NCERT Exercise 4.5, Q11	Practice 4.5 · Q11
NCERT Exercise 4.5, Q12	Practice 4.5 · Q12
NCERT Exercise 4.5, Q13	Practice 4.5 · Q13
NCERT Exercise 4.5, Q14	Practice 4.5 · Q14
NCERT Exercise 4.5, Q15	Practice 4.5 · Q15
NCERT Exercise 4.5, Q16	Practice 4.5 · Q16
NCERT Miscellaneous Exercises, Q1	Miscellaneous exercises · Q1
NCERT Miscellaneous Exercises, Q2	Miscellaneous exercises · Q2
NCERT Miscellaneous Exercises, Q3	Miscellaneous exercises · Q3
NCERT Miscellaneous Exercises, Q4	Miscellaneous exercises · Q4
NCERT Miscellaneous Exercises, Q5	Miscellaneous exercises · Q5
NCERT Miscellaneous Exercises, Q6	Miscellaneous exercises · Q6
NCERT Miscellaneous Exercises, Q7	Miscellaneous exercises · Q7
NCERT Miscellaneous Exercises, Q8	Miscellaneous exercises · Q8
NCERT Miscellaneous Exercises, Q9	Miscellaneous exercises · Q9

Quick check: MCQs

4.1 Introduction2 questions
4.2.2 Order two ki matrix ka determinant2 questions
4.2.3 Order 3 × 3 ki matrix ka determinant4 questions
4.3 Triangle ka Area4 questions
4.4 Minors aur Cofactors4 questions
Singular matrices aur inverse4 questions
4.6.1 Matrix ke inverse se linear equations ke system ka solution4 questions

Glossary · 12

Key terms (English + Hindi)

adjointसहखंडजsahkhandaj: Square matrix $A$ ki matrix of cofactors ka transpose, jise $adj A$ likhte hain.

cofactorसहखंडsahkhand: $A_{i j} = (- 1)^{i + j} M_{i j}$ : $a_{i j}$ ka minor, uske checkerboard sign ke saath.
collinear pointsसंरेख बिंदुsanrekh bindu: Points jo ek hi seedhi line par hon.Chapter mein pehli baar yahan aaya →
consistent systemसंगत निकायsangat nikaay: Equations ka system jiska kam se kam ek solution ho.Chapter mein pehli baar yahan aaya →

determinantसारणिकsaarnik: Square matrix $A$ ko diya gaya number, jise $∣ A ∣$ ya $\det A$ likhte hain. $2 \times 2$ matrix ke liye yeh $a d - b c$ hai.Chapter mein pehli baar yahan aaya →

expansion of a determinantसारणिक का प्रसारsaarnik ka prasaar: Determinant ko ek row ya column ki entries ke sum ki tarah likhna, jismein har entry apne cofactor se multiply hoti hai.Chapter mein pehli baar yahan aaya →

inconsistent systemअसंगत निकायasangat nikaay: Equations ka system jiska koi solution na ho.Chapter mein pehli baar yahan aaya →
inverse of a matrixव्युत्क्रमvyutkram: Woh matrix $A^{- 1}$ jiske liye $A A^{- 1} = A^{- 1} A = I$ . Jab $∣ A ∣ \neq 0$ , to yeh $\frac{1}{∣ A ∣} adj A$ ke barabar hai.Chapter mein pehli baar yahan aaya →

matrix methodआव्यूह विधिaavyooh vidhi: $A X = B$ ke roop mein likhe system ko $X = A^{- 1} B$ compute karke solve karna, jo $∣ A ∣ \neq 0$ hone par kaam karta hai.Chapter mein pehli baar yahan aaya →
minorउपसारणिकupsaarnik: $a_{i j}$ ka minor $M_{i j}$ woh determinant hai jo row $i$ aur column $j$ delete karne ke baad bachta hai.

non-singular matrixव्युत्क्रमणीय आव्यूहvyutkramaniya aavyooh: Woh square matrix jiska determinant 0 na ho. Iska hamesha inverse hota hai.Chapter mein pehli baar yahan aaya →

singular matrixअव्युत्क्रमणीय आव्यूहavyutkramaniya aavyooh: Woh square matrix jiska determinant 0 ho. Iska koi inverse nahi hota.Chapter mein pehli baar yahan aaya →

Determinants: Class 12 Maths Chapter 4 Hinglish mein

Is chapter mein aap kya seekhenge

4.1 Introduction

4.2 Determinant

4.2.1 Order one ki matrix ka determinant

4.2.2 Order two ki matrix ka determinant

4.2.3 Order 3 × 3 ki matrix ka determinant

Practice 4.1: answers ke saath

4.3 Triangle ka Area

Practice 4.2: answers ke saath

4.4 Minors aur Cofactors

Practice 4.3: answers ke saath

4.5 Matrix ka Adjoint aur Inverse

4.5.1 Matrix ka adjoint

Singular matrices aur inverse

Practice 4.4: answers ke saath

4.6 Determinants aur Matrices ke Applications

4.6.1 Matrix ke inverse se linear equations ke system ka solution

Practice 4.5: answers ke saath

Miscellaneous example

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

Summary aur key points

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye