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CBSE Class 12 Math 2018 Solved Paper

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Question : 25
Total: 29
If A = (
2−35
32−4
11−2
), 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 = (
123
257
−2−4−5
)
Solution:  
A = (
2−35
32−4
11−2
)
AA−1 = I
(
2−35
32−4
11−2
) A−1 = (
100
010
001
)
R1 ↔ R3
(
11−2
32−4
2−35
) A−1 = (
100
010
001
)
R2 → R2−3R1 . R3 → R3−2R1
(
11−2
0−12
0−59
) A−1 = (
001
01−3
10−2
)
R2 → - R2
(
11−2
0−12
0−59
) A−1 = (
001
0−13
10−2
)
R1 → R1−R2 , R3 → R3+5R2
(
100
01−2
00−1
) A−1 = (
01−2
0−13
1−513
)
R3 → - R3
(
100
01−2
00−1
) A−1 = (
01−2
0−13
−15−13
)
R2 → R2+2R3
(
100
010
001
) A−1 = (
01−2
−29−23
−15−13
)
A−1 = (
01−2
−29−23
−15−13
)
Consider,
AX = B where B = [
11
−5
3
] and X = [
x
y
z
]
⇒ A−1 AX = A−1 B
⇒ X = A−1 B
⇒ X = (
01−2
−29−23
−15−13
) [
11
−5
3
]
⇒ [
11
−5
3
] = [
1
2
3
]
⇒ x = 1 , y = 2 , z = 3
OR
Given A = (
123
257
−2−4−5
)
Consider,
|
123
257
−2−4−5
| = 1 (- 25 + 28) - 2 (- 10 + 14) + 3 (- 8 + 10)
= 3 - 8 + 6
= 1 ≠ 0
A−1 exist.
A A−1 = I
(
123
257
−2−4−5
) A−1 = (
100
010
001
)
R2 → R2−2R1,R3 → R3+2R1
(
123
011
001
) A−1 = (
100
−210
201
)
R1 → R1−2R2
(
100
010
001
) A−1 = (
5−20
−210
201
)
R1 → R1−R3,R2 → R2−R3
(
100
010
001
) A−1 = (
3−2−1
−41−1
20−1
)
A−1 = (
3−2−1
−41−1
201
)
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