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Matrices — Mathematics STD 10 — Question

ICSE BoardEnglish MediumSTD 10MathematicsMatrices5 Marks
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 $(A - B)^2 \neq A2 - 2AB + B^2.$
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Answer

$A-B=\left[\begin{array}{ll}3 & 1 \\ 2 & 1\end{array}\right]-\left[\begin{array}{cc}1 & -2 \\ 5 & 3\end{array}\right]$
$=\left[\begin{array}{ll}3-1 & 1+2 \\ 2-5 & 1-3\end{array}\right]=\left[\begin{array}{cc}2 & 3 \\ -3 & -2\end{array}\right]$
$(A - B)^2 = (A - B)(A - B)$
$\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]$
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]$
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]$
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]$
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]$
Hence, from above calculations, we get
$(A - B)^2 \neq A^2 - 2AB + B^2.$

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