Question
Let $A=\left[\begin{array}{ll}1 & 2 \\ 1 & 3\end{array}\right], B=\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right], C=\left[\begin{array}{ll}2 & 0 \\ 1 & 2\end{array}\right]$ Show that $(A-B) C=A C-B C$
✓
Answer
$\begin{aligned} & A=\left[\begin{array}{ll}1 & 2 \\ 1 & 3\end{array}\right], B=\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right], C=\left[\begin{array}{ll}2 & 0 \\ 1 & 2\end{array}\right] \\ & A-B=\left[\begin{array}{ll}1 & 2 \\ 1 & 3\end{array}\right]-\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right] \\ & =\left[\begin{array}{cc}-3 & 2 \\ 0 & -2\end{array}\right] \\ & (A-B) C=\left[\begin{array}{cc}-3 & 2 \\ 0 & -2\end{array}\right] \times\left[\begin{array}{ll}2 & 0 \\ 1 & 2\end{array}\right] \\ & =\left[\begin{array}{cc}-6+2 & 0+4 \\ 0-2 & 0-4\end{array}\right] \\ & =\left[\begin{array}{cc}-4 & 4 \\ -2 & -4\end{array}\right] \ldots(1) \\ & A C=\left[\begin{array}{ll}1 & 2 \\ 1 & 3\end{array}\right] \times\left[\begin{array}{ll}2 & 0 \\ 1 & 2\end{array}\right] \\ & =\left[\begin{array}{ll}2+2 & 0+4 \\ 2+3 & 0+6\end{array}\right] \\ & =\left[\begin{array}{cc}4 & 4 \\ 5 & 6\end{array}\right]\end{aligned}$
$
\begin{aligned}
& BC =\left[\begin{array}{ll}
4 & 0 \\
1 & 5
\end{array}\right] \times\left[\begin{array}{ll}
2 & 0 \\
1 & 2
\end{array}\right] \\
& =\left[\begin{array}{cc}
8+0 & 0+0 \\
2+5 & 0+10
\end{array}\right] \\
& =\left[\begin{array}{cc}
8 & 0 \\
7 & 10
\end{array}\right] \\
& A C-B C=\left[\begin{array}{ll}
4 & 4 \\
5 & 6
\end{array}\right]-\left[\begin{array}{cc}
8 & 0 \\
7 & 10
\end{array}\right] \\
& =\left[\begin{array}{cc}
-4 & 4 \\
-2 & -4
\end{array}\right] \ldots \text { (2) }
\end{aligned}
$
From (1) and (2) we get
$
(A-B) C=A C-B C
$
$
\begin{aligned}
& BC =\left[\begin{array}{ll}
4 & 0 \\
1 & 5
\end{array}\right] \times\left[\begin{array}{ll}
2 & 0 \\
1 & 2
\end{array}\right] \\
& =\left[\begin{array}{cc}
8+0 & 0+0 \\
2+5 & 0+10
\end{array}\right] \\
& =\left[\begin{array}{cc}
8 & 0 \\
7 & 10
\end{array}\right] \\
& A C-B C=\left[\begin{array}{ll}
4 & 4 \\
5 & 6
\end{array}\right]-\left[\begin{array}{cc}
8 & 0 \\
7 & 10
\end{array}\right] \\
& =\left[\begin{array}{cc}
-4 & 4 \\
-2 & -4
\end{array}\right] \ldots \text { (2) }
\end{aligned}
$
From (1) and (2) we get
$
(A-B) C=A C-B C
$
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