Given that A= [ cc 1 & 3 5 & -1 ], B= [ ccc 1 & -1 & 2 3 & 5 & 2 … - Vidyadip

Question

Given that $A=\left[\begin{array}{cc}1 & 3 \\ 5 & -1\end{array}\right], B=\left[\begin{array}{ccc}1 & -1 & 2 \\ 3 & 5 & 2\end{array}\right], C=\left[\begin{array}{ccc}1 & 3 & 2 \\ -4 & 1 & 3\end{array}\right]$ verify that $A(B+C)=A B+A C$

✓

Answer

$\begin{aligned} & \text { Given } A=\left[\begin{array}{cc}1 & 3 \\ 5 & -1\end{array}\right], B=\left[\begin{array}{ccc}1 & -1 & 2 \\ 3 & 5 & 2\end{array}\right] C=\left[\begin{array}{ccc}1 & 3 & 2 \\ -4 & 1 & 3\end{array}\right] \\ & B+C=\left[\begin{array}{ccc}1 & -1 & 2 \\ 3 & 5 & 2\end{array}\right]+\left[\begin{array}{ccc}1 & 3 & 2 \\ -4 & 1 & 3\end{array}\right] \\ & =\left[\begin{array}{ccc}2 & 2 & 4 \\ -1 & 6 & 5\end{array}\right] \\ & A(B+C)=\left[\begin{array}{cc}1 & 3 \\ 5 & -1\end{array}\right] \times\left[\begin{array}{ccc}2 & 2 & 4 \\ -1 & 6 & 5\end{array}\right] \\ & =\left[\begin{array}{ccc}2-3 & 2+18 & 4+15 \\ 10+1 & 10-6 & 20-5\end{array}\right] \\ & =\left[\begin{array}{ccc}-1 & 20 & 19 \\ 11 & 4 & 15\end{array}\right] \\ & A B=\left[\begin{array}{cc}1 & 3 \\ 5 & -1\end{array}\right] \times\left[\begin{array}{ccc}1 & -1 & 2 \\ 3 & 5 & 2\end{array}\right] \\ & =\left[\begin{array}{ccc}1+9 & -1+15 & 2+6 \\ 5-3 & -5-5 & 10-2\end{array}\right] \\ & =\left[\begin{array}{ccc}10 & 14 & 8 \\ 2 & -10 & 8\end{array}\right] \\ & \end{aligned}$
$
\begin{aligned}
& A C=\left[\begin{array}{cc}
1 & 3 \\
5 & -1
\end{array}\right] \times\left[\begin{array}{ccc}
1 & 3 & 2 \\
-4 & 1 & 3
\end{array}\right] \\
& =\left[\begin{array}{ccc}
1-12 & 3+3 & 2+9 \\
5+4 & 15-1 & 10-3
\end{array}\right] \\
& =\left[\begin{array}{ccc}
-11 & 6 & 11 \\
9 & 14 & 7
\end{array}\right] \\
& A B+A C=\left[\begin{array}{ccc}
10 & 14 & 8 \\
2 & -10 & 8
\end{array}\right]+\left[\begin{array}{ccc}
-11 & 6 & 11 \\
9 & 14 & 7
\end{array}\right] \\
& =\left[\begin{array}{ccc}
-1 & 20 & 19 \\
11 & 4 & 15
\end{array}\right] \ldots(2)
\end{aligned}
$
From (1) and (2) we get
$
A(B+C)=A B+A C
$

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