Answer: (i) \(\dfrac12\begin{pmatrix}3&2\\5&4\end{pmatrix}\). (ii) \(x=\dfrac{29}{4}\), \(y=\dfrac{53}{4}\).
(i) Let
\(A=\begin{pmatrix}4&-2\\-5&3\end{pmatrix}.\)
The determinant is
\(4\cdot3-(-2)(-5)=12-10=2.\)
Therefore
\(A^{-1}=\frac12 \begin{pmatrix}3&2\\5&4\end{pmatrix}.\)
(ii) Rewrite the equations as
\(8x-4y=5, \qquad -10x+6y=7.\)
Divide both equations by \(2\):
\(4x-2y=\frac52, \qquad -5x+3y=\frac72.\)
So
\(\begin{pmatrix}4&-2\\-5&3\end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}\frac52\\\frac72\end{pmatrix}.\)
Premultiply by the inverse matrix:
\(\begin{pmatrix}x\\y\end{pmatrix} = \frac12 \begin{pmatrix}3&2\\5&4\end{pmatrix} \begin{pmatrix}\frac52\\\frac72\end{pmatrix}.\)
This gives
\(x=\frac12\left(3\cdot\frac52+2\cdot\frac72\right)=\frac{29}{4},\)
and
\(y=\frac12\left(5\cdot\frac52+4\cdot\frac72\right)=\frac{53}{4}.\)
