Class 12 Determinants · Very Long Answer5 Marksmatrix_method_linear_systems

Class 12 Maths · Determinants · Very Long Answer #106

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

✅ Step-by-Step Solution

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

Write MX=B, where M=\(\begin{bmatrix}2 & 3 & 3\\1 & -2 & 1\\3 & -1 & -2\end{bmatrix}\), X=\(\begin{bmatrix}x\\y\\z\end{bmatrix}\), and B=\(\begin{bmatrix}5\\-4\\3\end{bmatrix}\).
The determinant is 40≠0. The signed-minor cofactor matrix is \(\begin{bmatrix}5 & 5 & 5\\3 & -13 & 11\\9 & 1 & -7\end{bmatrix}\).
Transpose that cofactor matrix: adj(M)=\(\begin{bmatrix}5 & 3 & 9\\5 & -13 & 1\\5 & 11 & -7\end{bmatrix}\), hence M⁻¹=adj(M)/(40)=\(\begin{bmatrix}1/8 & 3/40 & 9/40\\1/8 & -13/40 & 1/40\\1/8 & 11/40 & -7/40\end{bmatrix}\).
Multiply the inverse by the right side: X=M⁻¹B=\(\begin{bmatrix}1\\2\\-1\end{bmatrix}\). The resulting entries satisfy MX=B by direct multiplication.
Therefore x=1, y=2, z=-1.

[1 mark] Write MX=B, where M=\(\begin{bmatrix}2 & 3 & 3\\1 & -2 & 1\\3 & -1 & -2\end{bmatrix}\), X=\(\begin{bmatrix}x\\y\\z\end{bmatrix}\), and B=\(\begin{bmatrix}5\\-4\\3\end{bmatrix}\).
[1 mark] The determinant is 40≠0. The signed-minor cofactor matrix is \(\begin{bmatrix}5 & 5 & 5\\3 & -13 & 11\\9 & 1 & -7\end{bmatrix}\).
[1 mark] Transpose that cofactor matrix: adj(M)=\(\begin{bmatrix}5 & 3 & 9\\5 & -13 & 1\\5 & 11 & -7\end{bmatrix}\), hence M⁻¹=adj(M)/(40)=\(\begin{bmatrix}1/8 & 3/40 & 9/40\\1/8 & -13/40 & 1/40\\1/8 & 11/40 & -7/40\end{bmatrix}\).
[1 mark] Multiply the inverse by the right side: X=M⁻¹B=\(\begin{bmatrix}1\\2\\-1\end{bmatrix}\). The resulting entries satisfy MX=B by direct multiplication.
[1 mark] Therefore x=1, y=2, z=-1.

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