Find the value of $a,b,c$ and $d$ from the equation:
$\left[\begin{array}{cc}a-b& 2a+c\\ 2a-b& 3c+d\end{array}\right]=\left[\begin{array}{cc}-1& 5\\ 0& 13\end{array}\right]$

$\left[\begin{array}{cc}a-b& 2a+c\\ 2a-b& 3c+d\end{array}\right]=\left[\begin{array}{cc}-1& 5\\ 0& 13\end{array}\right]$

As the two matrices are equal, their corresponding elements are also equal.
Comparing the corresponding elements, we get:

$a-b=-1$ .....(1)
$2a-b=0$ .....(2)
$2a+c=5$ .....(3)
$3c+d=13$ .....(4)

From (2), we have
$b=2a$

Then , from (1), we have:
$a-2a=-1$
$\Rightarrow a=1$
$\Rightarrow b=2$

Now, from (3), we have:
$2\times 1+c=5$
$\Rightarrow c=3$

From (4), we have:
$3\times 3+d=13$
$\Rightarrow 9+d=13\Rightarrow d=4$
$\therefore a=1,b=2,c=3,\text{and}d=4$