Class 12 Maths · Determinants · Very Long Answer #99

📋 CBSE Stepwise Marking Scheme

• [1 mark] Write MX=B, where M=\(\begin{bmatrix}1 & 1 & 1\\1 & 0 & 2\\3 & 1 & 1\end{bmatrix}\), X=\(\begin{bmatrix}x\\y\\z\end{bmatrix}\), and B=\(\begin{bmatrix}6\\7\\12\end{bmatrix}\).
• [1 mark] The determinant is 4≠0. The signed-minor cofactor matrix is \(\begin{bmatrix}-2 & 5 & 1\\0 & -2 & 2\\2 & -1 & -1\end{bmatrix}\).
• [1 mark] Transpose that cofactor matrix: adj(M)=\(\begin{bmatrix}-2 & 0 & 2\\5 & -2 & -1\\1 & 2 & -1\end{bmatrix}\), hence M⁻¹=adj(M)/(4)=\(\begin{bmatrix}-1/2 & 0 & 1/2\\5/4 & -1/2 & -1/4\\1/4 & 1/2 & -1/4\end{bmatrix}\).
• [1 mark] Multiply the inverse by the right side: X=M⁻¹B=\(\begin{bmatrix}3\\1\\2\end{bmatrix}\). The resulting entries satisfy MX=B by direct multiplication.
• [1 mark] Therefore x=3, y=1, z=2.

Master Chapter 4: Determinants