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CBSE Class 12 Math 2012 Solved Paper

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Question : 23 of 29
Marks: +1, -0
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:
(−112123311)\begin{pmatrix} -1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{pmatrix}
Solution:  
The given system of equation can be written in the form of AX = B, where
A = [1−1234−52−13]\begin{bmatrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{bmatrix} , X = [XYZ]\begin{bmatrix} X \\ Y \\ Z \end{bmatrix} and B = [7−512]\begin{bmatrix} 7 \\ -5 \\ 12 \end{bmatrix}
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, A11A_{11} = 7 , A12A_{12} = - 19 , A13A_{13} = - 11
A21A_{21} = 1 , A22A_{22} = - 1 , A23A_{23} = - 1
A31A_{31} = - 3 , A32A_{32} = 11 , A33A_{33} = 7
∴ A−1A^{-1} = 1∣A∣\frac{1}{|A|} (adj A) = 14[71−3−19−111−11−17]\frac{1}{4} \begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix}
OR
Consider the given matrix.
Let A = [−112123311]\begin{bmatrix} -1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix}
We know that, A = InI_n A
Perform sequence of elementary row operations on A on the left hand side and the term InI_n on the right hand side till we obtain the result
InI_n = BA
Thus, B = A−1A^{-1}
Here, I3I_3 = [100010001]\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
Thus,we have,
[−112123311]\begin{bmatrix} -1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix} = [100010001]\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} A
R1R_1 ↔ R2R_2
[123−112311]\begin{bmatrix} 1 & 2 & 3 \\ -1 & 1 & 2 \\ 3 & 1 & 1 \end{bmatrix} = [010100001]\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} A
R2R_2 → R2+R1R_2 + R_1
R3R_3 → R3−3R1R_3 - 3R_1
[1230350−5−8]\begin{bmatrix} 1 & 2 & 3 \\ 0 & 3 & 5 \\ 0 & -5 & -8 \end{bmatrix} = [0101100−31]\begin{bmatrix} 0 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & -3 & 1 \end{bmatrix} A
R1R_1 → R1+R2R_1 + R_2
[1580350−5−8]\begin{bmatrix} 1 & 5 & 8 \\ 0 & 3 & 5 \\ 0 & -5 & -8 \end{bmatrix} = [1201100−31]\begin{bmatrix} 1 & 2 & 0 \\ 1 & 1 & 0 \\ 0 & -3 & 1 \end{bmatrix} A
R1R_1 → R1+R3R_1 + R_3
[1000350−5−8]\begin{bmatrix} 1 & 0 & 0 \\ 0 & 3 & 5 \\ 0 & -5 & -8 \end{bmatrix} = [1−111100−31]\begin{bmatrix} 1 & -1 & 1 \\ 1 & 1 & 0 \\ 0 & -3 & 1 \end{bmatrix} A
R2R_2 → R23\frac{R_2}{3}
[10001530−5−8]\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & \frac{5}{3} \\ 0 & -5 & -8 \end{bmatrix} = [1−11131300−31]\begin{bmatrix} 1 & -1 & 1 \\ \frac{1}{3} & \frac{1}{3} & 0 \\ 0 & -3 & 1 \end{bmatrix} A
R32R_{32} → R2+5R2R_2 + 5R_2
[10001530013]\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & \frac{5}{3} \\ 0 & 0 & \frac{1}{3} \end{bmatrix} = [1−111313053−431]\begin{bmatrix} 1 & -1 & 1 \\ \frac{1}{3} & \frac{1}{3} & 0 \\ \frac{5}{3} & -\frac{4}{3} & 1 \end{bmatrix} A
[1000153001]\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & \frac{5}{3} \\ 0 & 0 & 1 \end{bmatrix} = [1−11131305−43]\begin{bmatrix} 1 & -1 & 1 \\ \frac{1}{3} & \frac{1}{3} & 0 \\ 5 & -4 & 3 \end{bmatrix} A
R2R_2 → R2−53R3R_2 - \frac{5}{3} R_3
[100010001]\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = [1−11−87−55−43]\begin{bmatrix} 1 & -1 & 1 \\ -8 & 7 & -5 \\ 5 & -4 & 3 \end{bmatrix} A
Thus the inverse of the matrix A is given by
[1−11−87−55−43]\begin{bmatrix} 1 & -1 & 1 \\ -8 & 7 & -5 \\ 5 & -4 & 3 \end{bmatrix}
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