Class 12 Maths · Determinants · Very Long Answer #60

📋 CBSE Stepwise Marking Scheme

• [1 mark] Write MX=B, where M=\(\begin{bmatrix}5 & -1 & 4\\2 & 3 & -5\\5 & -2 & 6\end{bmatrix}\), X=\(\begin{bmatrix}x\\y\\z\end{bmatrix}\), and B=\(\begin{bmatrix}5\\2\\-1\end{bmatrix}\).
• [1 mark] The determinant is 1≠0. The signed-minor cofactor matrix is \(\begin{bmatrix}8 & -37 & -19\\-2 & 10 & 5\\-7 & 33 & 17\end{bmatrix}\).
• [1 mark] Transpose that cofactor matrix: adj(M)=\(\begin{bmatrix}8 & -2 & -7\\-37 & 10 & 33\\-19 & 5 & 17\end{bmatrix}\), hence M⁻¹=adj(M)/(1)=\(\begin{bmatrix}8 & -2 & -7\\-37 & 10 & 33\\-19 & 5 & 17\end{bmatrix}\).
• [1 mark] Multiply the inverse by the right side: X=M⁻¹B=\(\begin{bmatrix}43\\-198\\-102\end{bmatrix}\). The resulting entries satisfy MX=B by direct multiplication.
• [1 mark] Therefore x=43, y=-198, z=-102.

Master Chapter 4: Determinants