MHTCET 22 Apr 2025 Shift 2 Paper

Concept:The defining property of the inverse matrix is $AA^{-1} = I$.This property generates linear equations for the unknown entries $a$, $b$, and $c$.Explanation:Substitute the given matrices into $AA^{-1} = I$.$$\frac{1}{11}\begin{bmatrix} -1 & 7 & -24 \\ 2 & a & 4 \\ 2 & -3 & 15 \end{bmatrix}\begin{bmatrix} 3 & 3 & 4 \\ 2 & -3 & 4 \\ b & -1 & c \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$Perform the matrix multiplication.$$\frac{1}{11}\begin{bmatrix} 11-24b & 0 & 24-24c \\ 6+2a+4b & 2-3a & 8+4a+4c \\ 15b & 0 & -4+15c \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$Now equate the corresponding entries.From the $(1,1)$ entry: $\frac{11-24b}{11} = 1 \Rightarrow b = 0$.From the $(2,2)$ entry: $\frac{2-3a}{11} = 1 \Rightarrow 2-3a = 11 \Rightarrow a = -3$.From the $(3,3)$ entry: $\frac{-4+15c}{11} = 1 \Rightarrow -4+15c = 11 \Rightarrow c = 1$.These values also satisfy the remaining entries, so the solution is consistent.Answer:$a = -3$, $b = 0$, $c = 1$.Hence the values in order are $-3, 0, 1$, which matches option C.