Answer
$B$ can not be expressed as a linear combination of $A_1$, $A_2$ and $A_3$.
Work Step by Step
Let $$
B=\left[\begin{array}{ccc}{3} & {1} &{1} \\ {0}& {1} & {0}\end{array}\right], \quad A_1=\left[\begin{array}{ccc}{1} & {0} &{-1}\\ {0} & {1}&{0}\end{array}\right], \quad A_2=\left[\begin{array}{ccc}{-1} & {2}&{0} \\ {0}&{1} & {0}\end{array}\right], \quad A_3=\left[\begin{array}{ccc}{1} & { 1} &{1} \\ {0} & {0}&{0}\end{array}\right]
$$
We have
$$B=aA_1+bA_2+cA_3=a\left[\begin{array}{ccc}{1} & {0} &{-1}\\ {0} & {1}&{0}\end{array}\right]+b\left[\begin{array}{ccc}{-1} & {2}&{0} \\ {0}&{1} & {0}\end{array}\right]+c\left[\begin{array}{ccc}{1} & { 1} &{1} \\ {0} & {0}&{0}\end{array}\right]
$$
and hence we get the system $$a-b+c=3, \quad 2b+c=1,\quad -a+c=1, \quad a+b=1.$$
One can see easily thst the above system is inconsistence and has no solutions. So, $B$ can not be expressed as a linear combination of $A_1$, $A_2$ and $A_3$.
