If verify that

Question:

 If verify that

A (B + C) = (AB + AC).

$\mathrm{A}=\left[\begin{array}{ll}2 & 1\end{array}\right], \quad \mathrm{B}=\left[\begin{array}{lll}5 & 3 & 4 \\ 8 & 7 & 6\end{array}\right]$ and $\mathrm{C}=\left[\begin{array}{ccc}-1 & 2 & 1 \\ 1 & 0 & 2\end{array}\right]$

Solution:

Given, $A=\left[\begin{array}{ll}2 & 1\end{array}\right], B=\left[\begin{array}{lll}5 & 3 & 4 \\ 8 & 7 & 6\end{array}\right]$ and $C=\left[\begin{array}{ccc}-1 & 2 & 1 \\ 1 & 0 & 2\end{array}\right]$

Now,

$A(B+C)=\left[\begin{array}{ll}2 & 1\end{array}\right]\left[\begin{array}{lll}5-1 & 3+2 & 4+1 \\ 8+1 & 7+0 & 6+2\end{array}\right]$

$=\left[\begin{array}{ll}2 & 1\end{array}\right]\left[\begin{array}{lll}4 & 5 & 5 \\ 9 & 7 & 8\end{array}\right]$

$=[8+910+710+8]$ 

$=\left[\begin{array}{lll}17 & 17 & 18\end{array}\right]$ ...(i)

And,

$A B=\left[\begin{array}{ll}2 & 1\end{array}\right]\left[\begin{array}{lll}5 & 3 & 4 \\ 8 & 7 & 6\end{array}\right]=[10+86+78+6]=\left[\begin{array}{lll}18 & 13 & 14\end{array}\right]$

And, 

$A C=\left[\begin{array}{ll}2 & 1\end{array}\right]\left[\begin{array}{ccc}-1 & 2 & 1 \\ 1 & 0 & 2\end{array}\right]=[-2+14+02+2]=\left[\begin{array}{lll}-1 & 4 & 4\end{array}\right]$

So, $A B+A C=\left[\begin{array}{llll}18 & 13 & 14\end{array}\right]+\left[\begin{array}{lll}-1 & 4 & 4\end{array}\right]$

$=\left[\begin{array}{lll}17 & 17 & 18\end{array}\right]$ $\ldots$ (ii)

From equations (i) and (ii),

$A(B+C)=(A B+A C)$

Hence, verified

 

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