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CBSE Class 12 Math 2018 Solved Paper

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Question : 25 of 29
Marks: +1, -0
If A = (2−3532−411−2), find A−1. Use it to solve the system of equations
2x - 3y + 5z = 11
3x + 2y - 4z = - 5
x + y - 2z = - 3
OR
Using elementary row transformations, find the inverse of the matrix
A = (123257−2−4−5)
Solution:  
A = (2−3532−411−2)
AA−1 = I
(2−3532−411−2) A−1 = (100010001)
R1 ↔ R3
(11−232−42−35) A−1 = (100010001)
R2 → R2−3R1 . R3 → R3−2R1
(11−20−120−59) A−1 = (00101−310−2)
R2 → - R2
(11−20−120−59) A−1 = (0010−1310−2)
R1 → R1−R2 , R3 → R3+5R2
(10001−200−1) A−1 = (01−20−131−513)
R3 → - R3
(10001−200−1) A−1 = (01−20−13−15−13)
R2 → R2+2R3
(100010001) A−1 = (01−2−29−23−15−13)
A−1 = (01−2−29−23−15−13)
Consider,
AX = B where B = [11−53] and X = [xyz]
⇒ A−1 AX = A−1 B
⇒ X = A−1 B
⇒ X = (01−2−29−23−15−13) [11−53]
⇒ [11−53] = [123]
⇒ x = 1 , y = 2 , z = 3
OR
Given A = (123257−2−4−5)
Consider,
|123257−2−4−5| = 1 (- 25 + 28) - 2 (- 10 + 14) + 3 (- 8 + 10)
= 3 - 8 + 6
= 1 ≠ 0
A−1 exist.
A A−1 = I
(123257−2−4−5) A−1 = (100010001)
R2 → R2−2R1,R3 → R3+2R1
(123011001) A−1 = (100−210201)
R1 → R1−2R2
(100010001) A−1 = (5−20−210201)
R1 → R1−R3,R2 → R2−R3
(100010001) A−1 = (3−2−1−41−120−1)
A−1 = (3−2−1−41−1201)
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