Class 12 Determinants · Very Long Answer5 Marksmatrix_method_linear_systems

Class 12 Maths · Determinants · Very Long Answer #107

Solve by the matrix method: (1)x + (1)y + (-1)z = 3; (2)x + (3)y + (1)z = 10; (3)x + (-1)y + (-7)z = 1.

✅ Step-by-Step Solution

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

Write MX=B, where M=\(\begin{bmatrix}1 & 1 & -1\\2 & 3 & 1\\3 & -1 & -7\end{bmatrix}\), X=\(\begin{bmatrix}x\\y\\z\end{bmatrix}\), and B=\(\begin{bmatrix}3\\10\\1\end{bmatrix}\).
The determinant is 8≠0. The signed-minor cofactor matrix is \(\begin{bmatrix}-20 & 17 & -11\\8 & -4 & 4\\4 & -3 & 1\end{bmatrix}\).
Transpose that cofactor matrix: adj(M)=\(\begin{bmatrix}-20 & 8 & 4\\17 & -4 & -3\\-11 & 4 & 1\end{bmatrix}\), hence M⁻¹=adj(M)/(8)=\(\begin{bmatrix}-5/2 & 1 & 1/2\\17/8 & -1/2 & -3/8\\-11/8 & 1/2 & 1/8\end{bmatrix}\).
Multiply the inverse by the right side: X=M⁻¹B=\(\begin{bmatrix}3\\1\\1\end{bmatrix}\). The resulting entries satisfy MX=B by direct multiplication.
Therefore x=3, y=1, z=1.

[1 mark] Write MX=B, where M=\(\begin{bmatrix}1 & 1 & -1\\2 & 3 & 1\\3 & -1 & -7\end{bmatrix}\), X=\(\begin{bmatrix}x\\y\\z\end{bmatrix}\), and B=\(\begin{bmatrix}3\\10\\1\end{bmatrix}\).
[1 mark] The determinant is 8≠0. The signed-minor cofactor matrix is \(\begin{bmatrix}-20 & 17 & -11\\8 & -4 & 4\\4 & -3 & 1\end{bmatrix}\).
[1 mark] Transpose that cofactor matrix: adj(M)=\(\begin{bmatrix}-20 & 8 & 4\\17 & -4 & -3\\-11 & 4 & 1\end{bmatrix}\), hence M⁻¹=adj(M)/(8)=\(\begin{bmatrix}-5/2 & 1 & 1/2\\17/8 & -1/2 & -3/8\\-11/8 & 1/2 & 1/8\end{bmatrix}\).
[1 mark] Multiply the inverse by the right side: X=M⁻¹B=\(\begin{bmatrix}3\\1\\1\end{bmatrix}\). The resulting entries satisfy MX=B by direct multiplication.
[1 mark] Therefore x=3, y=1, z=1.

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