Class 12 Maths · Determinants · Very Long Answer #67

📋 CBSE Stepwise Marking Scheme

• [1 mark] The supplied product equals I₃ by row-column multiplication: \(\begin{bmatrix}1 & -1 & 2\\0 & 2 & -3\\3 & -2 & 4\end{bmatrix}\)\(\begin{bmatrix}-2 & 0 & 1\\9 & 2 & -3\\6 & 1 & -2\end{bmatrix}\)=I₃. The alternative coefficient matrix is the transpose of M=\(\begin{bmatrix}1 & -1 & 2\\0 & 2 & -3\\3 & -2 & 4\end{bmatrix}\), so its inverse is \(\begin{bmatrix}-2 & 9 & 6\\0 & 2 & 1\\1 & -3 & -2\end{bmatrix}\).
• [1 mark] The determinant is -1≠0. The signed-minor cofactor matrix is \(\begin{bmatrix}2 & 0 & -1\\-9 & -2 & 3\\-6 & -1 & 2\end{bmatrix}\).
• [1 mark] Transpose that cofactor matrix: adj(M)=\(\begin{bmatrix}2 & -9 & -6\\0 & -2 & -1\\-1 & 3 & 2\end{bmatrix}\), hence M⁻¹=adj(M)/(-1)=\(\begin{bmatrix}-2 & 9 & 6\\0 & 2 & 1\\1 & -3 & -2\end{bmatrix}\).
• [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