Class 12 Maths · Determinants · Very Long Answer #108

📋 CBSE Stepwise Marking Scheme

• [1 mark] First the supplied A has inverse \(\begin{bmatrix}-3 & -2 & -4\\2 & 1 & 2\\2 & 1 & 3\end{bmatrix}\). The system coefficient matrix is M=Aᵀ=\(\begin{bmatrix}1 & -2 & 0\\2 & -1 & -1\\0 & -2 & 1\end{bmatrix}\), so M⁻¹=(A⁻¹)ᵀ. Write MX=B with B=\(\begin{bmatrix}10\\8\\7\end{bmatrix}\).
• [1 mark] The determinant is 1≠0. The signed-minor cofactor matrix is \(\begin{bmatrix}-3 & -2 & -4\\2 & 1 & 2\\2 & 1 & 3\end{bmatrix}\).
• [1 mark] Transpose that cofactor matrix: adj(M)=\(\begin{bmatrix}-3 & 2 & 2\\-2 & 1 & 1\\-4 & 2 & 3\end{bmatrix}\), hence M⁻¹=adj(M)/(1)=\(\begin{bmatrix}-3 & 2 & 2\\-2 & 1 & 1\\-4 & 2 & 3\end{bmatrix}\).
• [1 mark] Multiply the inverse by the right side: X=M⁻¹B=\(\begin{bmatrix}0\\-5\\-3\end{bmatrix}\). The resulting entries satisfy MX=B by direct multiplication.
• [1 mark] Therefore x=0, y=-5, z=-3.

Master Chapter 4: Determinants