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CBSE Class 12 Math 2012 Solved Paper

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Question : 23
Total: 29
Using matrices solve the following system of linear equations:
x - y + 2z = 7
3x + 4y - 5z = - 5
2x - y + 3z = 12
OR
Using elementary operations, find the inverse of the following matrix:
(
−112
123
311
)
Solution:  
The given system of equation can be written in the form of AX = B, where
A = [
1−12
34−5
2−13
] , X = [
X
Y
Z
] and B = [
7
−5
12
]
Now,
|A| = 1 (12 - 5) + 1 (9 + 10) + 2 (- 3 - 8) = 7 + 19 - 22 = 4 ≠ 0
Thus, A is non-singular. Therefore, its inverse exists.
Now, A11 = 7 , A12 = - 19 , A13 = - 11
A21 = 1 , A22 = - 1 , A23 = - 1
A31 = - 3 , A32 = 11 , A33 = 7
∴ A−1 =
1
|A|
 (adj A) =
1
4
[
71−3
−19−111
−11−17
]
OR
Consider the given matrix.
Let A = [
−112
123
311
]
We know that, A = In A
Perform sequence of elementary row operations on A on the left hand side and the term In on the right hand side till we obtain the result
In = BA
Thus, B = A−1
Here, I3 = [
100
010
001
]
Thus,we have,
[
−112
123
311
] = [
100
010
001
] A
R1 ↔ R2
[
123
−112
311
] = [
010
100
001
] A
R2 → R2+R1
R3 → R3−3R1
[
123
035
0−5−8
] = [
010
110
0−31
] A
R1 → R1+R2
[
158
035
0−5−8
] = [
120
110
0−31
] A
R1 → R1+R3
[
100
035
0−5−8
] = [
1−11
110
0−31
] A
R2 →
R2
3
[
100
01
5
3
0−5−8
] = [
1−11
1
3
1
3
0
0−31
] A
R32 → R2+5R2
[
100
01
5
3
00
1
3
] = [
1−11
1
3
1
3
0
5
3
−
4
3
1
] A
[
100
01
5
3
001
] = [
1−11
1
3
1
3
0
5−43
] A
R2 → R2−
5
3
R3
[
100
010
001
] = [
1−11
−87−5
5−43
] A
Thus the inverse of the matrix A is given by
[
1−11
−87−5
5−43
]
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