Class 12 Determinants · Very Long Answer5 Marksmatrix_method_linear_systems

Class 12 Maths · Determinants · Very Long Answer #104

Solve by the matrix method: (1)x + (2)y + (1)z = 7; (1)x + (3)z = 11; (2)x + (-3)y = 1.

✅ Step-by-Step Solution

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

Write MX=B, where M=\(\begin{bmatrix}1 & 2 & 1\\1 & 0 & 3\\2 & -3 & 0\end{bmatrix}\), X=\(\begin{bmatrix}x\\y\\z\end{bmatrix}\), and B=\(\begin{bmatrix}7\\11\\1\end{bmatrix}\).
The determinant is 18≠0. The signed-minor cofactor matrix is \(\begin{bmatrix}9 & 6 & -3\\-3 & -2 & 7\\6 & -2 & -2\end{bmatrix}\).
Transpose that cofactor matrix: adj(M)=\(\begin{bmatrix}9 & -3 & 6\\6 & -2 & -2\\-3 & 7 & -2\end{bmatrix}\), hence M⁻¹=adj(M)/(18)=\(\begin{bmatrix}1/2 & -1/6 & 1/3\\1/3 & -1/9 & -1/9\\-1/6 & 7/18 & -1/9\end{bmatrix}\).
Multiply the inverse by the right side: X=M⁻¹B=\(\begin{bmatrix}2\\1\\3\end{bmatrix}\). The resulting entries satisfy MX=B by direct multiplication.
Therefore x=2, y=1, z=3.

[1 mark] Write MX=B, where M=\(\begin{bmatrix}1 & 2 & 1\\1 & 0 & 3\\2 & -3 & 0\end{bmatrix}\), X=\(\begin{bmatrix}x\\y\\z\end{bmatrix}\), and B=\(\begin{bmatrix}7\\11\\1\end{bmatrix}\).
[1 mark] The determinant is 18≠0. The signed-minor cofactor matrix is \(\begin{bmatrix}9 & 6 & -3\\-3 & -2 & 7\\6 & -2 & -2\end{bmatrix}\).
[1 mark] Transpose that cofactor matrix: adj(M)=\(\begin{bmatrix}9 & -3 & 6\\6 & -2 & -2\\-3 & 7 & -2\end{bmatrix}\), hence M⁻¹=adj(M)/(18)=\(\begin{bmatrix}1/2 & -1/6 & 1/3\\1/3 & -1/9 & -1/9\\-1/6 & 7/18 & -1/9\end{bmatrix}\).
[1 mark] Multiply the inverse by the right side: X=M⁻¹B=\(\begin{bmatrix}2\\1\\3\end{bmatrix}\). The resulting entries satisfy MX=B by direct multiplication.
[1 mark] Therefore x=2, y=1, z=3.

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