Question
If $A =\left[\begin{array}{ll}1 & 4 \\ 1 & 0\end{array}\right], B =\left[\begin{array}{cc}2 & 1 \\ 3 & -1\end{array}\right]$ and $C =\left[\begin{array}{ll}2 & 3 \\ 0 & 5\end{array}\right]$ compute $( AB ) C$ $=(C B) A$ ?
✓
Answer
Given
$
\begin{aligned}
& A=\left[\begin{array}{ll}
1 & 4 \\
1 & 0
\end{array}\right] \\
& B=\left[\begin{array}{cc}
2 & 1 \\
3 & -1
\end{array}\right] \\
& C=\left[\begin{array}{ll}
2 & 3 \\
0 & 5
\end{array}\right] \\
& (A B) C=\left[\left[\begin{array}{ll}
1 & 4 \\
1 & 0
\end{array}\right] \times\left[\begin{array}{cc}
2 & 1 \\
3 & -1
\end{array}\right]\right]\left[\begin{array}{ll}
2 & 3 \\
0 & 5
\end{array}\right] \\
& =\left[\begin{array}{cc}
2+12 & 1-14 \\
2+0 & 1+0
\end{array}\right]\left[\begin{array}{ll}
2 & 3 \\
0 & 5
\end{array}\right] \\
& =\left[\begin{array}{cc}
14 & -13 \\
2 & 1
\end{array}\right]\left[\begin{array}{ll}
2 & 3 \\
0 & 5
\end{array}\right] \\
& =\left[\begin{array}{cc}
28+0 & 42-15 \\
4+0 & 6+5
\end{array}\right] \\
& =\left[\begin{array}{cc}
28 & 27 \\
4 & 11
\end{array}\right] \\
& \text { (CB) } A=\left[\left[\begin{array}{ll}
2 & 3 \\
0 & 5
\end{array}\right]\left[\begin{array}{cc}
2 & 1 \\
3 & -1
\end{array}\right]\right]\left[\begin{array}{ll}
1 & 4 \\
1 & 0
\end{array}\right] \\
& =\left[\begin{array}{cc}
4+9 & 2-3 \\
0+15 & 0-5
\end{array}\right]\left[\begin{array}{ll}
1 & 4 \\
1 & 0
\end{array}\right] \\
& =\left[\begin{array}{ll}
13 & 1 \\
15 & 5
\end{array}\right]\left[\begin{array}{ll}
1 & 4 \\
1 & 0
\end{array}\right] \\
& =\left[\begin{array}{cc}
13-1 & 52+0 \\
15-5 & 60+0
\end{array}\right] \\
& =\left[\begin{array}{ll}
12 & 52 \\
10 & 60
\end{array}\right] \\
&
\end{aligned}
$
It is clear from above that
$
(A B) C \neq (C B) A .
$
$
\begin{aligned}
& A=\left[\begin{array}{ll}
1 & 4 \\
1 & 0
\end{array}\right] \\
& B=\left[\begin{array}{cc}
2 & 1 \\
3 & -1
\end{array}\right] \\
& C=\left[\begin{array}{ll}
2 & 3 \\
0 & 5
\end{array}\right] \\
& (A B) C=\left[\left[\begin{array}{ll}
1 & 4 \\
1 & 0
\end{array}\right] \times\left[\begin{array}{cc}
2 & 1 \\
3 & -1
\end{array}\right]\right]\left[\begin{array}{ll}
2 & 3 \\
0 & 5
\end{array}\right] \\
& =\left[\begin{array}{cc}
2+12 & 1-14 \\
2+0 & 1+0
\end{array}\right]\left[\begin{array}{ll}
2 & 3 \\
0 & 5
\end{array}\right] \\
& =\left[\begin{array}{cc}
14 & -13 \\
2 & 1
\end{array}\right]\left[\begin{array}{ll}
2 & 3 \\
0 & 5
\end{array}\right] \\
& =\left[\begin{array}{cc}
28+0 & 42-15 \\
4+0 & 6+5
\end{array}\right] \\
& =\left[\begin{array}{cc}
28 & 27 \\
4 & 11
\end{array}\right] \\
& \text { (CB) } A=\left[\left[\begin{array}{ll}
2 & 3 \\
0 & 5
\end{array}\right]\left[\begin{array}{cc}
2 & 1 \\
3 & -1
\end{array}\right]\right]\left[\begin{array}{ll}
1 & 4 \\
1 & 0
\end{array}\right] \\
& =\left[\begin{array}{cc}
4+9 & 2-3 \\
0+15 & 0-5
\end{array}\right]\left[\begin{array}{ll}
1 & 4 \\
1 & 0
\end{array}\right] \\
& =\left[\begin{array}{ll}
13 & 1 \\
15 & 5
\end{array}\right]\left[\begin{array}{ll}
1 & 4 \\
1 & 0
\end{array}\right] \\
& =\left[\begin{array}{cc}
13-1 & 52+0 \\
15-5 & 60+0
\end{array}\right] \\
& =\left[\begin{array}{ll}
12 & 52 \\
10 & 60
\end{array}\right] \\
&
\end{aligned}
$
It is clear from above that
$
(A B) C \neq (C B) A .
$
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