Question

plzz answer urgent........​

Answer 1

Answer:

                $\left[\begin{matrix}1&-4\\3&-2\end{matrix}\right]$

Step-by-step explanation:

We need to get the inverse of  $\left[\begin{matrix}5&-7\\-2&3\end{matrix}\right]$  and multiply both sides of the equation by that.

First, the determinant is...

$\Delta=\left|\begin{matrix}5&-7\\-2&3\end{matrix}\right|=5\times3-(-7)\times(-2)=15-14=1$

Then, the inverse is...

$\displaystyle\left[\begin{matrix}5&-7\\-2&3\end{matrix}\right]^{-1} = \frac{1}{\Delta}\left[\begin{matrix}3&7\\2&5\end{matrix}\right] = \left[\begin{matrix}3&7\\2&5\end{matrix}\right]$

Using this, we solve the equation for X...

$\displaystyle\ \ \ \left[\begin{matrix}5&-7\\-2&3\end{matrix}\right]\cdot X=\left[\begin{matrix}-16&-6\\7&2\end{matrix}\right]\\\\\Rightarrow\left[\begin{matrix}5&-7\\-2&3\end{matrix}\right]^{-1}\left[\begin{matrix}5&-7\\-2&3\end{matrix}\right]\cdot X=\left[\begin{matrix}5&-7\\-2&3\end{matrix}\right]^{-1}\left[\begin{matrix}-16&-6\\7&2\end{matrix}\right]$

$\displaystyle\Rightarrow X=\left[\begin{matrix}3&7\\2&5\end{matrix}\right]\left[\begin{matrix}-16&-6\\7&2\end{matrix}\right]=\left[\begin{matrix}1&-4\\3&-2\end{matrix}\right]$

Hope that helps!