Class 12 Maths · Determinants · Very Long Answer #69

📋 CBSE Stepwise Marking Scheme

• [1 mark] The supplied product is \(\begin{bmatrix}8 & 0 & 0\\0 & 8 & 0\\0 & 0 & 8\end{bmatrix}\)=8I₃. Hence the system matrix M=\(\begin{bmatrix}1 & -1 & 1\\1 & -2 & -2\\2 & 1 & 3\end{bmatrix}\) has inverse M⁻¹=(1/8)\(\begin{bmatrix}-4 & 4 & 4\\-7 & 1 & 3\\5 & -3 & -1\end{bmatrix}\).
• [1 mark] The determinant is 8≠0. The signed-minor cofactor matrix is \(\begin{bmatrix}-4 & -7 & 5\\4 & 1 & -3\\4 & 3 & -1\end{bmatrix}\).
• [1 mark] Transpose that cofactor matrix: adj(M)=\(\begin{bmatrix}-4 & 4 & 4\\-7 & 1 & 3\\5 & -3 & -1\end{bmatrix}\), hence M⁻¹=adj(M)/(8)=\(\begin{bmatrix}-1/2 & 1/2 & 1/2\\-7/8 & 1/8 & 3/8\\5/8 & -3/8 & -1/8\end{bmatrix}\).
• [1 mark] Multiply the inverse by the right side: X=M⁻¹B=\(\begin{bmatrix}3\\-2\\-1\end{bmatrix}\). The resulting entries satisfy MX=B by direct multiplication.
• [1 mark] Therefore x=3, y=-2, z=-1.

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