Prove the following

Question:

 If

(i) (AB) C = A (BC)

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

Solution:

Given, $A=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right], B=\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]$ and $C=\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]$

(i) $A B=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]=\left[\begin{array}{cc}2+6 & 3-8 \\ -4+3 & -6-4\end{array}\right]=\left[\begin{array}{cc}8 & -5 \\ -1 & -10\end{array}\right]$

and $(A B) C=\left[\begin{array}{cc}8 & -5 \\ -1 & -10\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]=\left[\begin{array}{cc}8+5 & 0 \\ -1+10 & 0\end{array}\right]=\left[\begin{array}{cc}13 & 0 \\ 9 & 0\end{array}\right]$...(i)

Again, $(B C)=\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]=\left[\begin{array}{cc}2-3 & 0 \\ 3+4 & 0\end{array}\right]=\left[\begin{array}{cc}-1 & 0 \\ 7 & 0\end{array}\right]$

And $A(B C)=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}-1 & 0 \\ 7 & 0\end{array}\right]=\left[\begin{array}{cc}-1+14 & 0 \\ 2+7 & 0\end{array}\right]=\left[\begin{array}{cc}13 & 0 \\ 9 & 0\end{array}\right]$$\ldots(i i)$

From (i) and (ii), we get

Hence, $(\mathbf{A B}) \mathbf{C}=\mathbf{A}(\mathbf{B C})$

(ii) $B+C=\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]+\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]=\left[\begin{array}{cc}3 & 3 \\ 2 & -4\end{array}\right]$

Now,

$A \cdot(B+C)=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}3 & 3 \\ 2 & -4\end{array}\right]$

$=\left[\begin{array}{cc}3+4 & 3-8 \\ -6+2 & -6-4\end{array}\right]$

$=\left[\begin{array}{cc}7 & -5 \\ -4 & -10\end{array}\right]$ ...(iii)

$A B=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]=\left[\begin{array}{cc}2+6 & 3-8 \\ -4+3 & -6-4\end{array}\right]=\left[\begin{array}{cc}8 & -5 \\ -1 & -10\end{array}\right]$

and, $A C=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]=\left[\begin{array}{cc}1-2 & 0 \\ -2-1 & 0\end{array}\right]=\left[\begin{array}{cc}-1 & 0 \\ -3 & 0\end{array}\right]$

Thus, $A B+A C=\left[\begin{array}{cc}8 & -5 \\ -1 & -10\end{array}\right]+\left[\begin{array}{cc}-1 & 0 \\ -3 & 0\end{array}\right]=\left[\begin{array}{cc}7 & -5 \\ -4 & -10\end{array}\right]$ $\ldots$ (iv)

Hence from equation (iii) and (iv), we have

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

 

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