BITSAT 2024 Shift 1 Paper

Given the matrix $A=\begin{pmatrix} 3 & -2 & 4 \\ 1 & 2 & -1 \\ 0 & 1 & 1 \end{pmatrix}$ and $A^{-1}=\;\frac{1}{k} \operatorname{adj}(A)$, we want to determine the value of $k$.We know from the properties of matrices that:$A^{-1}=\;\frac{\operatorname{adj}(A)}{|A|}$By comparing this equation with the given relationship $A^{-1}=\;\frac{1}{k} \operatorname{adj}(A)$, we can identify that:$k=|A|$Therefore, we need to calculate the determinant $|A|$ of the matrix $A$ :$|A|=\begin{vmatrix} 3 & -2 & 4 \\ 1 & 2 & -1 \\ 0 & 1 & 1 \end{vmatrix}$To find $|A|$, expand across the first row:$|A|=3\begin{vmatrix} 2 & -1 \\ 1 & 1 \end{vmatrix}-(-2)\begin{vmatrix} 1 & -1 \\ 0 & 1 \end{vmatrix}+4\begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix}$Calculate each of the $2 \times 2$ determinants:$\;\begin{vmatrix} 2 & -1 \\ 1 & 1 \end{vmatrix}=(2 \times 1)-(-1 \times 1)=2+1=3$$\;\begin{vmatrix} 1 & -1 \\ 0 & 1 \end{vmatrix}=(1 \times 1)-(0 \times -1)=1$$\;\begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix}=(1 \times 1)-(0 \times 2)=1$Substitute these values back into the determinant calculation: $|A|=3 \times 3+2 \times 1+4 \times 1=9+2+4=15$Thus, the value of $k$ is $|A|=15$.