Let A= | ll 3 & 2 0 & 5 | and B= | ll 1 & 0 1 & 2 |, find (i) (A+B)(A-B) (ii) A^2-B^2. Is (i) equal to (ii) ?

l (i) A= | ll 3 & 2 0 & 5 |_ 2 2 , B= | ll 1 & 0 1 & 2 |_ 2 2 A+B= | ll 3+1 & 2+0 0+1 & 5+2 |= | ll 4 & 2 1 & 7 |_ 2 2 A-B= | ll 3-1 & 2-0 0-1 & 5-2 |= | cc 2 & 2 -1 & 3 |_ 2 2 .(A+B)(A-B)= | cc 4 & 2 1 & 7 | cc 2 & 2 -1 & 3 , = | ll 8-2 & 8+6 2-7 & 2+21 | = | cc 6 & 14 -5 & 23 | . .(1) (ii) A ^2 = | ll 3 & 2 0 & 5 | | ll 3 & 2 0 & 5 | = | cc 9+0 & 6+10 0+0 & 0+25 | = | cc 9 & 16 0 & 25 |_ 2 2 B^2= | ll 1 & 0 1 & 2 | | ll 1 & 0 1 & 2 | = | ll 1+0 & 0+0 1+2 & 0+4 | = | ll 1 & 0 3 & 4 |_ 2 2 |A^2-B^2= | ll 9 & 16 0 & 25 |- | ll 1 & 0 3 & 4 |= | cc 8 & 16 -3 & 21 |_ 2 2 .

Let A= | ll 3 & 2 0 & 5 | and B= | ll 1 & 0 1 & 2 |, find (i) (A+…

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Question

Let $A=\left|\begin{array}{ll}3 & 2 \\ 0 & 5\end{array}\right|$ and $B=\left|\begin{array}{ll}1 & 0 \\ 1 & 2\end{array}\right|$, find $(i) \ (A+B)(A-B) \ (ii) \ A^2-B^2$. Is $(i)$ equal to $(ii)$ ?

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Answer

$\begin{array}{l}\text { (i) } A=\left|\begin{array}{ll}3 & 2 \\ 0 & 5\end{array}\right|_{2 \times 2}, B=\left|\begin{array}{ll}1 & 0 \\ 1 & 2\end{array}\right|_{2 \times 2} \end{array} $
$ A+B=\left|\begin{array}{ll}3+1 & 2+0 \\ 0+1 & 5+2\end{array}\right|=\left|\begin{array}{ll}4 & 2 \\ 1 & 7\end{array}\right|_{2 \times 2} $
$ A-B=\left|\begin{array}{ll}3-1 & 2-0 \\ 0-1 & 5-2\end{array}\right|=\left|\begin{array}{cc}2 & 2 \\ -1 & 3\end{array}\right|_{2 \times 2} $
$ \left.(A+B)(A-B)=\left|\begin{array}{cc}4 & 2 \\ 1 & 7\end{array}\right| \begin{array}{cc}2 & 2 \\ -1 & 3\end{array} \right\rvert\, $
$ =\left|\begin{array}{ll}8-2 & 8+6 \\ 2-7 & 2+21\end{array}\right|$
$=\left|\begin{array}{cc}6 & 14 \\ -5 & 23\end{array}\right| \ldots \ldots . .(1)$
$(ii) A ^2$
$=\left|\begin{array}{ll}3 & 2 \\ 0 & 5\end{array}\right|\left|\begin{array}{ll}3 & 2 \\ 0 & 5\end{array}\right|$
$=\left|\begin{array}{cc}9+0 & 6+10 \\ 0+0 & 0+25\end{array}\right|$
$=\left|\begin{array}{cc}9 & 16 \\ 0 & 25\end{array}\right|_{2 \times 2}$
$B^2=\left|\begin{array}{ll}1 & 0 \\ 1 & 2\end{array}\right|\left|\begin{array}{ll}1 & 0 \\ 1 & 2\end{array}\right|$
$=\left|\begin{array}{ll}1+0 & 0+0 \\ 1+2 & 0+4\end{array}\right|$
$=\left|\begin{array}{ll}1 & 0 \\ 3 & 4\end{array}\right|_{2 \times 2}$
$ \left|A^2-B^2=\left|\begin{array}{ll} 9 & 16 \\ 0 & 25 \end{array}\right|-\left|\begin{array}{ll} 1 & 0 \\ 3 & 4 \end{array}\right|=\left|\begin{array}{cc} 8 & 16 \\ -3 & 21 \end{array}\right|_{2 \times 2} \ldots\right.$