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Question Bank [2022] — Maths (commerce) STD 12 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 12 Commerce / ArtsMaths (commerce)Question Bank [2022]3 Marks
Question
If $A=\left[\begin{array}{cc}1 & 2 \\ -1 & -2\end{array}\right], B=\left[\begin{array}{cc}2 & a \\ -1 & b\end{array}\right]$ and $(A+B)^2=A^2+B^2$, find the values of $a$ and $b$.
✓

Answer

Given $(A+B)^2=A^2+B^2$
$\therefore A^2+A B+B A+B^2=A^2+B^2$
$\therefore A B+B A=0$
$\therefore A B=-B A$
$\therefore\left[\begin{array}{cc}1 & 2 \\-1 & 2\end{array}\right]\left[\begin{array}{cc}2 & a \\-1 & b\end{array}\right]=-\left[\begin{array}{cc}2 & a \\-1 & b\end{array}\right]\left[\begin{array}{cc}1 & 2 \\-1 & -2\end{array}\right] $
$\therefore\left[\begin{array}{cc}2-2 & a+2 b \\-2+2 & -a-2 b\end{array}\right]=-\left[\begin{array}{cc}2-a & 4-2 a \\-1-b & -2-2 b\end{array}\right] $
$\therefore\left[\begin{array}{cc}0 & a+2 b \\0 & -a-2 b\end{array}\right]=\left[\begin{array}{cc}-2+a & -4+2 a \\1+b & 2+2 b \end{array}\right]$
$\therefore$ By equality of matrices, we get
$-2+a=0 \text { and } 1+b=0$
$\therefore a=2$ and $b=-1$.

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