Class 12 Determinants · Very Long Answer5 Marksmatrix_method_linear_systems

Class 12 Maths · Determinants · Very Long Answer #103

Solve by the matrix method: (3)x + (-2)y + (3)z = 8; (2)x + (1)y + (-1)z = 1; (4)x + (-3)y + (2)z = 4.

✅ Step-by-Step Solution

Final Answer: x=1, y=2, z=3

Write MX=B, where M=\(\begin{bmatrix}3 & -2 & 3\\2 & 1 & -1\\4 & -3 & 2\end{bmatrix}\), X=\(\begin{bmatrix}x\\y\\z\end{bmatrix}\), and B=\(\begin{bmatrix}8\\1\\4\end{bmatrix}\).
The determinant is -17≠0. The signed-minor cofactor matrix is \(\begin{bmatrix}-1 & -8 & -10\\-5 & -6 & 1\\-1 & 9 & 7\end{bmatrix}\).
Transpose that cofactor matrix: adj(M)=\(\begin{bmatrix}-1 & -5 & -1\\-8 & -6 & 9\\-10 & 1 & 7\end{bmatrix}\), hence M⁻¹=adj(M)/(-17)=\(\begin{bmatrix}1/17 & 5/17 & 1/17\\8/17 & 6/17 & -9/17\\10/17 & -1/17 & -7/17\end{bmatrix}\).
Multiply the inverse by the right side: X=M⁻¹B=\(\begin{bmatrix}1\\2\\3\end{bmatrix}\). The resulting entries satisfy MX=B by direct multiplication.
Therefore x=1, y=2, z=3.

[1 mark] Write MX=B, where M=\(\begin{bmatrix}3 & -2 & 3\\2 & 1 & -1\\4 & -3 & 2\end{bmatrix}\), X=\(\begin{bmatrix}x\\y\\z\end{bmatrix}\), and B=\(\begin{bmatrix}8\\1\\4\end{bmatrix}\).
[1 mark] The determinant is -17≠0. The signed-minor cofactor matrix is \(\begin{bmatrix}-1 & -8 & -10\\-5 & -6 & 1\\-1 & 9 & 7\end{bmatrix}\).
[1 mark] Transpose that cofactor matrix: adj(M)=\(\begin{bmatrix}-1 & -5 & -1\\-8 & -6 & 9\\-10 & 1 & 7\end{bmatrix}\), hence M⁻¹=adj(M)/(-17)=\(\begin{bmatrix}1/17 & 5/17 & 1/17\\8/17 & 6/17 & -9/17\\10/17 & -1/17 & -7/17\end{bmatrix}\).
[1 mark] Multiply the inverse by the right side: X=M⁻¹B=\(\begin{bmatrix}1\\2\\3\end{bmatrix}\). The resulting entries satisfy MX=B by direct multiplication.
[1 mark] Therefore x=1, y=2, z=3.

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