Class 12 Maths · Determinants · Very Long Answer #64

📋 CBSE Stepwise Marking Scheme

• [1 mark] Expand the determinant: det(A)=1≠0, so A is invertible.
• [1 mark] Compute the signed minors Cᵢⱼ=(−1)ⁱ⁺ʲMᵢⱼ to obtain the cofactor matrix \(\begin{bmatrix}3 & 1 & 2\\2 & 1 & 2\\6 & 2 & 5\end{bmatrix}\).
• [1 mark] Transpose the cofactor matrix to get adj(A)=\(\begin{bmatrix}3 & 2 & 6\\1 & 1 & 2\\2 & 2 & 5\end{bmatrix}\). Then A⁻¹=adj(A)/(1)=\(\begin{bmatrix}3 & 2 & 6\\1 & 1 & 2\\2 & 2 & 5\end{bmatrix}\).
• [1 mark] Use the reverse-order rule (AB)⁻¹=B⁻¹A⁻¹. Multiplication gives \(\begin{bmatrix}10 & 7 & 21\\-49 & -34 & -103\\17 & 12 & 36\end{bmatrix}\).
• [1 mark] Therefore (AB)⁻¹=\(\begin{bmatrix}10 & 7 & 21\\-49 & -34 & -103\\17 & 12 & 36\end{bmatrix}\).

Master Chapter 4: Determinants