Class 12 Determinants · Very Long Answer5 Marksmatrix_method_linear_systems

Class 12 Maths · Determinants · Very Long Answer #105

Solve by the matrix method: (1)x + (-1)y + (1)z = 4; (2)x + (1)y + (-3)z = 0; (1)x + (1)y + (1)z = 2.

✅ Step-by-Step Solution

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

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

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

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