Class 12 Maths · Determinants · Very Long Answer #62

📋 CBSE Stepwise Marking Scheme

• [1 mark] First the supplied A has inverse \(\begin{bmatrix}7/31 & 2/31\\5/31 & -3/31\end{bmatrix}\). The system coefficient matrix is M=Aᵀ=\(\begin{bmatrix}3 & 5\\2 & -7\end{bmatrix}\), so M⁻¹=(A⁻¹)ᵀ. Write MX=B with B=\(\begin{bmatrix}11\\-3\end{bmatrix}\).
• [1 mark] The determinant is -31≠0. The signed-minor cofactor matrix is \(\begin{bmatrix}-7 & -2\\-5 & 3\end{bmatrix}\).
• [1 mark] Transpose that cofactor matrix: adj(M)=\(\begin{bmatrix}-7 & -5\\-2 & 3\end{bmatrix}\), hence M⁻¹=adj(M)/(-31)=\(\begin{bmatrix}7/31 & 5/31\\2/31 & -3/31\end{bmatrix}\).
• [1 mark] Multiply the inverse by the right side: X=M⁻¹B=\(\begin{bmatrix}2\\1\end{bmatrix}\). The resulting entries satisfy MX=B by direct multiplication.
• [1 mark] Therefore x=2, y=1.

Master Chapter 4: Determinants