Answer:
[tex]\large\boxed{\left[\begin{array}{ccc}6&5\\5&4\end{array}\right]X=\left[\begin{array}{ccc}3\\4\end{array}\right] }[/tex]
Step-by-step explanation:
[tex]\text{Substitute:}\\\\\left[\begin{array}{ccc}-6&5\\5&4\end{array}\right] \cdot\left[\begin{array}{ccc}8\\-9\end{array}\right] =\left[\begin{array}{ccc}(-6)(8)+(5)(-9)\\(5)(8)+(4)(-9)\end{array}\right] =\left[\begin{array}{ccc}-93\\4\end{array}\right] \neq\left[\begin{array}{ccc}3\\4\end{array}\right][/tex]
[tex]\left[\begin{array}{ccc}6&5\\5&4\end{array}\right] \cdot\left[\begin{array}{ccc}8\\-9\end{array}\right] =\left[\begin{array}{ccc}(6)(8)+(5)(-9)\\(5)(8)+(4)(-9)\end{array}\right] =\left[\begin{array}{ccc}3\\4\end{array}\right]\qquad\bold{CORRECT\ :)}[/tex]
