If $A = \left[\begin{array}{ccc} -3 &1 &-3\cr 3 &4 &-3\cr 2 &-1 &1 \end{array}\right]$ and $B = \left[\begin{array}{ccc} -1 &2 &3\cr 1 &-3 &-2\cr 0 &3 &3 \end{array}\right]$, then

$4 A - 3 B=$ $\left[\Rule{0pt}{3.6em}{0pt}\right.$$\left]\Rule{0pt}{3.6em}{0pt}\right.$ and

$A^{T}=$ $\left[\Rule{0pt}{3.6em}{0pt}\right.$$\left]\Rule{0pt}{3.6em}{0pt}\right.$.