Class 12 Determinants · Very Long Answer5 Marksmatrix_method_linear_systems

Class 12 Maths · Determinants · Very Long Answer #100

Solve by the matrix method: (2)x + (-3)y + (5)z = 11; (3)x + (2)y + (-4)z = -5; (1)x + (1)y + (-2)z = -3.

✅ Step-by-Step Solution

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

Write MX=B, where M=\(\begin{bmatrix}2 & -3 & 5\\3 & 2 & -4\\1 & 1 & -2\end{bmatrix}\), X=\(\begin{bmatrix}x\\y\\z\end{bmatrix}\), and B=\(\begin{bmatrix}11\\-5\\-3\end{bmatrix}\).
The determinant is -1≠0. The signed-minor cofactor matrix is \(\begin{bmatrix}0 & 2 & 1\\-1 & -9 & -5\\2 & 23 & 13\end{bmatrix}\).
Transpose that cofactor matrix: adj(M)=\(\begin{bmatrix}0 & -1 & 2\\2 & -9 & 23\\1 & -5 & 13\end{bmatrix}\), hence M⁻¹=adj(M)/(-1)=\(\begin{bmatrix}0 & 1 & -2\\-2 & 9 & -23\\-1 & 5 & -13\end{bmatrix}\).
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=\(\begin{bmatrix}2 & -3 & 5\\3 & 2 & -4\\1 & 1 & -2\end{bmatrix}\), X=\(\begin{bmatrix}x\\y\\z\end{bmatrix}\), and B=\(\begin{bmatrix}11\\-5\\-3\end{bmatrix}\).
[1 mark] The determinant is -1≠0. The signed-minor cofactor matrix is \(\begin{bmatrix}0 & 2 & 1\\-1 & -9 & -5\\2 & 23 & 13\end{bmatrix}\).
[1 mark] Transpose that cofactor matrix: adj(M)=\(\begin{bmatrix}0 & -1 & 2\\2 & -9 & 23\\1 & -5 & 13\end{bmatrix}\), hence M⁻¹=adj(M)/(-1)=\(\begin{bmatrix}0 & 1 & -2\\-2 & 9 & -23\\-1 & 5 & -13\end{bmatrix}\).
[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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