Science:Math Exam Resources/Courses/MATH152/April 2022/Question A13/Solution 1

We start by rearranging the equation for ${\textstyle X}$, knowing that the matrix ${\textstyle A}$ is invertible (it has non-zero determinant). This gives ${\textstyle X=(C-B^{T})A^{-1}}$. Now,

$${\begin{aligned}\det(X)&=\det((C-B^{T})A^{-1})\\&=\det(C-B^{T})\det(A^{-1})\\&={\frac {\det(C-B^{T})}{det(A)}}.\end{aligned}}$$

Now, compute ${\textstyle C-B^{T}}$

$$C-B^{T}=\left[{\begin{array}{cc}2&3\\0&3\end{array}}\right].$$

We can compute ${\textstyle \det(C-B^{T})=6}$ and ${\textstyle \det(A)=2}$, and put this together with the above expression for ${\textstyle \det(X)}$ to get ${\textstyle \det(X)=6/2=3}$.