If $A=\left[\begin{array}{rrr}1 & 1 & -1 \\ 2 & 0 & 3 \\ 3 & -1 & 2\end{array}\right]$,$B=\left[\begin{array}{rr}1 & 3 \\ 0 & 2 \\ -1 & 4\end{array}\right]$ and $C=\left[\begin{array}{rrrr}1 & 2 & 3 & -4 \\ 2 & 0 & -2 & 1\end{array}\right]$,find $A(BC)$,$(AB)C$ and show that $(AB)C=A(BC)$.

(N/A) First,we calculate $AB$:
$AB = \left[\begin{array}{rrr}1 & 1 & -1 \\ 2 & 0 & 3 \\ 3 & -1 & 2\end{array}\right] \left[\begin{array}{rr}1 & 3 \\ 0 & 2 \\ -1 & 4\end{array}\right] = \left[\begin{array}{rr}1+0+1 & 3+2-4 \\ 2+0-3 & 6+0+12 \\ 3+0-2 & 9-2+8\end{array}\right] = \left[\begin{array}{rr}2 & 1 \\ -1 & 18 \\ 1 & 15\end{array}\right]$
Next,we calculate $(AB)C$:
$(AB)C = \left[\begin{array}{rr}2 & 1 \\ -1 & 18 \\ 1 & 15\end{array}\right] \left[\begin{array}{rrrr}1 & 2 & 3 & -4 \\ 2 & 0 & -2 & 1\end{array}\right] = \left[\begin{array}{rrrr}2+2 & 4+0 & 6-2 & -8+1 \\ -1+36 & -2+0 & -3-36 & 4+18 \\ 1+30 & 2+0 & 3-30 & -4+15\end{array}\right] = \left[\begin{array}{rrrr}4 & 4 & 4 & -7 \\ 35 & -2 & -39 & 22 \\ 31 & 2 & -27 & 11\end{array}\right]$
Now,we calculate $BC$:
$BC = \left[\begin{array}{rr}1 & 3 \\ 0 & 2 \\ -1 & 4\end{array}\right] \left[\begin{array}{rrrr}1 & 2 & 3 & -4 \\ 2 & 0 & -2 & 1\end{array}\right] = \left[\begin{array}{rrrr}1+6 & 2+0 & 3-6 & -4+3 \\ 0+4 & 0+0 & 0-4 & 0+2 \\ -1+8 & -2+0 & -3-8 & 4+4\end{array}\right] = \left[\begin{array}{rrrr}7 & 2 & -3 & -1 \\ 4 & 0 & -4 & 2 \\ 7 & -2 & -11 & 8\end{array}\right]$
Finally,we calculate $A(BC)$:
$A(BC) = \left[\begin{array}{rrr}1 & 1 & -1 \\ 2 & 0 & 3 \\ 3 & -1 & 2\end{array}\right] \left[\begin{array}{rrrr}7 & 2 & -3 & -1 \\ 4 & 0 & -4 & 2 \\ 7 & -2 & -11 & 8\end{array}\right] = \left[\begin{array}{rrrr}7+4-7 & 2+0+2 & -3-4+11 & -1+2-8 \\ 14+0+21 & 4+0-6 & -6+0-33 & -2+0+24 \\ 21-4+14 & 6+0-4 & -9+4-22 & -3-2+16\end{array}\right] = \left[\begin{array}{rrrr}4 & 4 & 4 & -7 \\ 35 & -2 & -39 & 22 \\ 31 & 2 & -27 & 11\end{array}\right]$
Since $(AB)C = A(BC)$,the associative property of matrix multiplication is verified.