Answer
$$B=3A_1-4A_2-A_3.$$
Work Step by Step
Let $$
B=\left[\begin{array}{ccc}{2} & {3} \\ {-4} & {2}\end{array}\right], \quad A_1=\left[\begin{array}{ccc}{1} & {0} \\ {0} & {1}\end{array}\right], \quad A_2=\left[\begin{array}{ccc}{0} & {-1} \\ {1} & {0}\end{array}\right], \quad A_3=\left[\begin{array}{ccc}{1} & { 1} \\ {0} & {1}\end{array}\right]
$$
We have
$$B=aA_1+bA_2+cA_3=a\left[\begin{array}{ccc}{1} & {0} \\ {0} & {1}\end{array}\right]+b\left[\begin{array}{ccc}{0} & {-1} \\ {1} & {0}\end{array}\right]+c\left[\begin{array}{ccc}{1} & { 1} \\ {0} & {1}\end{array}\right]
$$
and hence we get the system $$a+c=2, \quad -b+c=3,\quad b=-4, \quad a+c=2.$$
The above system has the solution $a=3$, $b=-4$ and $c=-1$
so $$B=3A_1-4A_2-A_3.$$
