##
If A = |2 3 4 1 -1 0 0 1 2|, find A
^{
− 1
}
. Hence, solve the system of equations

## x − y = 3

## 2x + 3y + 4z = 17

## y + 2z = 7

CBSE Class 12 Sample Paper for 2020 Boards

Paper Summary

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10 Important

Question 11 Important

Question 12

Question 13

Question 14 (OR 1st Question)

Question 14 (OR 2nd Question) Important

Question 15 (OR 1st Question)

Question 15 (OR 2nd Question) Important

Question 16

Question 17 Important

Question 18 (OR 1st Question)

Question 18 (OR 2nd Question)

Question 19

Question 20

Question 21 (OR 1st Question) Important

Question 21 (OR 2nd Question) Important

Question 22

Question 23

Question 24 (OR 1st Question)

Question 24 (OR 2nd Question)

Question 25

Question 26 Important

Question 27

Question 28 (OR 1st Question) Important

Question 28 (OR 2nd Question)

Question 29

Question 30 Important

Question 31 (OR 1st Question) Important

Question 31 (OR 2nd Question)

Question 32 Important

Question 33 (OR 1st Question) Important

Question 33 (OR 2nd Question) Important You are here

Question 34

Question 35 (OR 1st Question) Important

Question 35 (OR 2nd Question) Important

Question 36 Important

Class 12

Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at Oct. 24, 2019 by Teachoo

Question 33 (OR 2nd Question) If A = |2 3 4 1 -1 0 0 1 2|, find A− 1. Hence, solve the system of equations x − y = 3 2x + 3y + 4z = 17 y + 2z = 7 The equations can be written as 2x + 3y + 4z = 17 x − y = 3 y + 2z = 7 So, the equation is in the form of [2 3 4 1 -1 0 0 1 2] [ x y z] [17 3 7] i.e. AX = B X = A–1 B Here, A = [ 2 3 4 1 -1 0 0 1 2], x = [x y z] & B = [17 3 7] Finding A–1 We know that A-1 = 1/(|A|) adj (A) Calculating |A| |A|= [ 2 3 4 1 -1 0 0 1 2] = 2(−2 – 0) − 3 (2 – 0) + 4 (1 – 0) = –4 – 6 + 4 = −6 Since |A|≠ 0 ∴ The system of equation is consistent & has a unique solution Now finding adj (A) adj A = [ A11 A12 A13 A21 A22 A23 A21 A22 A23 A31 A32 A33] [ A11 A21 A31 A12 A22 A32 A13 A23 A33 ] A = [ 2 3 4 1 -1 0 0 1 2] 𝐴11 = –2 𝐴12 = −2 𝐴13 = 1 𝐴21 = –2 𝐴22 = 4 𝐴23 = –2 𝐴31 = 4 𝐴32 = 4 𝐴33 = –5 𝐴11 = −2 + 0 = –2 𝐴12 = −[2−0] = −2 𝐴13 = 1 – 0 = 1 𝐴21 = –[6−4] = –2 𝐴22 = 4 – 0 = 4 𝐴23 = –[2−0] = –2 𝐴31 = 0−(−4)= 4 𝐴32 = –[0−4] = 4 𝐴33 = −2−3 = –5 Thus adj A = [ -2 -2 4 -2 4 4 1 -2 -5 & |A| = –6 Now, A-1 = 1/(|A|) adj A A-1 = 1/-6 [-2 -2 4 -2 4 4 1 -2 -5] = 1/6 [2 2 -4 2 -4 -4 -1 2 5] Now, X = A–1B [X Y Z] = 1/6 [ 2 2 -4 2 -4 -4 -1 2 5] [ 17 3 7] [ XY Z] = 1/6 [2 (17) + 2(3) - 4(7) 2 (17) - 4 (3) - 4(7) (-1) (17) + 2(3) + 5 (7) [X Y Z] = 1/17 [2 (17) + 2 (3) - 4 (7) 2(17) - 4 (3) - 4(7) (-1) (17) + 2(3) + 5(7) [X Y Z] = 1/6 [34 + 6 - 28 34 - 12 - 28 -17 + 6 + 35] [X Y Z] = 1/6 [ 12 -6 24] [X Y Z] = [2 -1 4] ∴ x = 2, y = –1 and z = 4