Class 12 Maths · Determinants · Very Long Answer #109

📋 CBSE Stepwise Marking Scheme

• [1 mark] The supplied product AB=\(\begin{bmatrix}6 & 0 & 0\\0 & 6 & 0\\0 & 0 & 6\end{bmatrix}\)=6I₃, so A⁻¹=B/6. The system is AX=[3,17,7]ᵀ with A=\(\begin{bmatrix}1 & -1 & 0\\2 & 3 & 4\\0 & 1 & 2\end{bmatrix}\).
• [1 mark] The determinant is 6≠0. The signed-minor cofactor matrix is \(\begin{bmatrix}2 & -4 & 2\\2 & 2 & -1\\-4 & -4 & 5\end{bmatrix}\).
• [1 mark] Transpose that cofactor matrix: adj(M)=\(\begin{bmatrix}2 & 2 & -4\\-4 & 2 & -4\\2 & -1 & 5\end{bmatrix}\), hence M⁻¹=adj(M)/(6)=\(\begin{bmatrix}1/3 & 1/3 & -2/3\\-2/3 & 1/3 & -2/3\\1/3 & -1/6 & 5/6\end{bmatrix}\).
• [1 mark] Multiply the inverse by the right side: X=M⁻¹B=\(\begin{bmatrix}2\\-1\\4\end{bmatrix}\). The resulting entries satisfy MX=B by direct multiplication.
• [1 mark] Therefore x=2, y=-1, z=4.

Master Chapter 4: Determinants