Class 12 Maths · Determinants · Very Long Answer #68

📋 CBSE Stepwise Marking Scheme

• [1 mark] Write MX=B, where M=\(\begin{bmatrix}1 & -1 & 2\\0 & 2 & -3\\3 & -2 & 4\end{bmatrix}\), X=\(\begin{bmatrix}x\\y\\z\end{bmatrix}\), and B=\(\begin{bmatrix}1\\1\\3\end{bmatrix}\).
• [1 mark] The determinant is -1≠0. The signed-minor cofactor matrix is \(\begin{bmatrix}2 & -9 & -6\\0 & -2 & -1\\-1 & 3 & 2\end{bmatrix}\).
• [1 mark] Transpose that cofactor matrix: adj(M)=\(\begin{bmatrix}2 & 0 & -1\\-9 & -2 & 3\\-6 & -1 & 2\end{bmatrix}\), hence M⁻¹=adj(M)/(-1)=\(\begin{bmatrix}-2 & 0 & 1\\9 & 2 & -3\\6 & 1 & -2\end{bmatrix}\).
• [1 mark] Multiply the inverse by the right side: X=M⁻¹B=\(\begin{bmatrix}1\\2\\1\end{bmatrix}\). The resulting entries satisfy MX=B by direct multiplication.
• [1 mark] Therefore x=1, y=2, z=1.

Master Chapter 4: Determinants