If A= [ ll 3 & 1 2 & 1 ] and B= [ cc 1 & -2 5 & 3 ], then show that (A - B)^2 A2 - 2AB + B^2.

l A-B= [ ll 3 & 1 2 & 1 ]- [ cc 1 & -2 5 & 3 ] = [ ll 3-1 & 1+2 2-5 & 1-3 ]= [ cc 2 & 3 -3 & -2 ] (A - B)^2 = (A - B)(A - B) l ( A - B ) 2= [ cc 2 & 3 -3 & -2 ] [ cc 2 & 3 -3 & -2 ] = [ cc 4-9 & 6-6 -6+6 & -9+4 ] = [ cc -5 & 0 0 & -5 ] l and A^2= [ ll 3 & 1 2 & 1 ] [ ll 3 & 1 2 & 1 ] = [ ll 9+2 & 3+1 6+2 & 2+1 ]= [ cc 11 & 4 8 & 3 ] l and B^2= [ cc 1 & -2 5 & 3 ] [ cc 1 & -2 5 & 3 ] = [ cc 1-10 & -2-6 5+15 & -10+9 ] = [ cc -9 & -8 20 & -1 ] l and A B= [ ll 3 & 1 2 & 1 ] [ cc 1 & -2 5 & 3 ] = [ ll 3+5 & -6+3 2+5 & -4+3 ]= [ ll 8 & -3 7 & -1 ] l Now A^2-2 A B+B^2 = [ cc 11 & 4 8 & 3 ]-2 [ cc 8 & -3 7 & -1 ]+ [ cc -9 & -8 20 & -1 ] = [ cc 11 & 4 8 & 3 ]- [ ll 16 & -6 14 & -2 ]+ [ cc -9 & -8 20 & -1 ] = [ cc 11-6-9 & 4+6-8 8-14+20 & 3+2-1 ] = [ cc -14 & 2 14 & 4 ] Hence, from above calculations, we get (A - B)^2 A^2 - 2AB + B^2.

If A= [ ll 3 & 1 2 & 1 ] and B= [ cc 1 & -2 5 & 3 ], then show th…

Download App

Question

If $A=\left[\begin{array}{ll}3 & 1 \\ 2 & 1\end{array}\right]$ and $B=\left[\begin{array}{cc}1 & -2 \\ 5 & 3\end{array}\right],$ then show that $(A - B)^2 \neq A2 - 2AB + B^2.$

✓

Answer

$\begin{array}{l}A-B=\left[\begin{array}{ll}3 & 1 \\ 2 & 1\end{array}\right]-\left[\begin{array}{cc}1 & -2 \\ 5 & 3\end{array}\right] \\ =\left[\begin{array}{ll}3-1 & 1+2 \\ 2-5 & 1-3\end{array}\right]=\left[\begin{array}{cc}2 & 3 \\ -3 & -2\end{array}\right]\end{array}$
$(A - B)^2 = (A - B)(A - B)$
$\begin{array}{l}\Rightarrow( A - B ) 2=\left[\begin{array}{cc}2 & 3 \\ -3 & -2\end{array}\right]\left[\begin{array}{cc}2 & 3 \\ -3 & -2\end{array}\right] \\ =\left[\begin{array}{cc}4-9 & 6-6 \\ -6+6 & -9+4\end{array}\right] \\ =\left[\begin{array}{cc}-5 & 0 \\ 0 & -5\end{array}\right]\end{array}$
$\begin{array}{l}\text { and } A^2=\left[\begin{array}{ll}3 & 1 \\ 2 & 1\end{array}\right]\left[\begin{array}{ll}3 & 1 \\ 2 & 1\end{array}\right] \\ =\left[\begin{array}{ll}9+2 & 3+1 \\ 6+2 & 2+1\end{array}\right]=\left[\begin{array}{cc}11 & 4 \\ 8 & 3\end{array}\right]\end{array}$
$\begin{array}{l}\text { and } B^2=\left[\begin{array}{cc}1 & -2 \\ 5 & 3\end{array}\right]\left[\begin{array}{cc}1 & -2 \\ 5 & 3\end{array}\right] \\ =\left[\begin{array}{cc}1-10 & -2-6 \\ 5+15 & -10+9\end{array}\right] \\ =\left[\begin{array}{cc}-9 & -8 \\ 20 & -1\end{array}\right]\end{array}$
$\begin{array}{l}\text { and } A B=\left[\begin{array}{ll}3 & 1 \\ 2 & 1\end{array}\right]\left[\begin{array}{cc}1 & -2 \\ 5 & 3\end{array}\right] \\ =\left[\begin{array}{ll}3+5 & -6+3 \\ 2+5 & -4+3\end{array}\right]=\left[\begin{array}{ll}8 & -3 \\ 7 & -1\end{array}\right]\end{array}$
$\begin{array}{l}\text { Now } A^2-2 A B+B^2 \\ =\left[\begin{array}{cc}11 & 4 \\ 8 & 3\end{array}\right]-2\left[\begin{array}{cc}8 & -3 \\ 7 & -1\end{array}\right]+\left[\begin{array}{cc}-9 & -8 \\ 20 & -1\end{array}\right] \\ =\left[\begin{array}{cc}11 & 4 \\ 8 & 3\end{array}\right]-\left[\begin{array}{ll}16 & -6 \\ 14 & -2\end{array}\right]+\left[\begin{array}{cc}-9 & -8 \\ 20 & -1\end{array}\right] \\ =\left[\begin{array}{cc}11-6-9 & 4+6-8 \\ 8-14+20 & 3+2-1\end{array}\right] \\ =\left[\begin{array}{cc}-14 & 2 \\ 14 & 4\end{array}\right]\end{array}$
Hence$,$ from above calculations, we get
$(A - B)^2 \neq A^2 - 2AB + B^2.$