Solution:
We have $\boldsymbol{A}=\left(\begin{array}{ccc}\mathbf{4} & -\mathbf{1} & -\mathbf{4} \\ \mathbf{3} & \mathbf{0} & -\mathbf{4} \\ \mathbf{3} & -\mathbf{1} & -\mathbf{3}\end{array}\right)$.
To show: $A^{2}=A$.
Now if number of columns in left matrix is equals to the number of rows in right matrix multiplication of two matrices is possible.
Next discuss the order of the matrices which are given. The order of matrix $\boldsymbol{A}$ is $\mathbf{3} \times \mathbf{3}$.
Thus the multiplications we can proceed as the multiplication $A^{2}$ is possible.
$A^{2}=A A$
$\begin{aligned}
=&\left(\begin{array}{ccc}
4 & -1 & -4 \\
3 & 0 & -4 \\
3 & -1 & -3
\end{array}\right)\left(\begin{array}{ccc}
4 & -1 & -4 \\
3 & 0 & -4 \\
3 & -1 & -3
\end{array}\right) \\
=&\left(\begin{array}{ccc}
16-3-12 & -4+0+4 & -16+4+12 \\
12+0-12 & -3+0+4 & -12+0+12 \\
12-3-9 & -3+0+3 & -12+4+9
\end{array}\right) \\
=&\left(\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right) \\
&=\mathbf{I}
\end{aligned}$