Class 12 Determinants · Very Long Answer5 Marksmatrix_method_linear_systems

Class 12 Maths · Determinants · Very Long Answer #71

Solve by the matrix method: (1)x + (2)y + (3)z = 6; (2)x + (-1)y + (1)z = 2; (3)x + (2)y + (-2)z = 3.

✅ Step-by-Step Solution

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

Write MX=B, where M=\(\begin{bmatrix}1 & 2 & 3\\2 & -1 & 1\\3 & 2 & -2\end{bmatrix}\), X=\(\begin{bmatrix}x\\y\\z\end{bmatrix}\), and B=\(\begin{bmatrix}6\\2\\3\end{bmatrix}\).
The determinant is 35≠0. The signed-minor cofactor matrix is \(\begin{bmatrix}0 & 7 & 7\\10 & -11 & 4\\5 & 5 & -5\end{bmatrix}\).
Transpose that cofactor matrix: adj(M)=\(\begin{bmatrix}0 & 10 & 5\\7 & -11 & 5\\7 & 4 & -5\end{bmatrix}\), hence M⁻¹=adj(M)/(35)=\(\begin{bmatrix}0 & 2/7 & 1/7\\1/5 & -11/35 & 1/7\\1/5 & 4/35 & -1/7\end{bmatrix}\).
Multiply the inverse by the right side: X=M⁻¹B=\(\begin{bmatrix}1\\1\\1\end{bmatrix}\). The resulting entries satisfy MX=B by direct multiplication.
Therefore x=1, y=1, z=1.

[1 mark] Write MX=B, where M=\(\begin{bmatrix}1 & 2 & 3\\2 & -1 & 1\\3 & 2 & -2\end{bmatrix}\), X=\(\begin{bmatrix}x\\y\\z\end{bmatrix}\), and B=\(\begin{bmatrix}6\\2\\3\end{bmatrix}\).
[1 mark] The determinant is 35≠0. The signed-minor cofactor matrix is \(\begin{bmatrix}0 & 7 & 7\\10 & -11 & 4\\5 & 5 & -5\end{bmatrix}\).
[1 mark] Transpose that cofactor matrix: adj(M)=\(\begin{bmatrix}0 & 10 & 5\\7 & -11 & 5\\7 & 4 & -5\end{bmatrix}\), hence M⁻¹=adj(M)/(35)=\(\begin{bmatrix}0 & 2/7 & 1/7\\1/5 & -11/35 & 1/7\\1/5 & 4/35 & -1/7\end{bmatrix}\).
[1 mark] Multiply the inverse by the right side: X=M⁻¹B=\(\begin{bmatrix}1\\1\\1\end{bmatrix}\). The resulting entries satisfy MX=B by direct multiplication.
[1 mark] Therefore x=1, y=1, z=1.

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