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Home  >>  CBSE XII  >>  Math  >>  Matrices
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If $A=\begin{bmatrix}1 & -1\\2 & 3\end{bmatrix}$ and $B=\begin{bmatrix}2 & 1\\1 & 0\end{bmatrix}$ prove that $(A+B)^2\neq A^2+2AB+B^2$

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$A^2=A.A=\begin{bmatrix}1 & -1\\2 & 3\end{bmatrix}\begin{bmatrix}1 & -1\\2 & 3\end{bmatrix}$
$\quad=\begin{bmatrix}1(1)-1(2) & 1(-1)+(-1)(3)\\2(1)+3(2) & 2(-1)+3(3)\end{bmatrix}$
$\quad=\begin{bmatrix}1-2 & -1-3\\2+6 & -2+9\end{bmatrix}$
$\quad=\begin{bmatrix}-1 & -4\\8 & 7\end{bmatrix}$
$AB=\begin{bmatrix}1 & -1\\2 & 3\end{bmatrix}\begin{bmatrix}2 & 1\\1& 0\end{bmatrix}$
$\quad=\begin{bmatrix}1(2)-1(1) & 1(1)-(1)(0)\\2(2)+3(1) & 2(1)+3(0)\end{bmatrix}$
$\quad=\begin{bmatrix}2-1 & 1-0\\4+3 & 2+0\end{bmatrix}$
$\quad=\begin{bmatrix}1 & 1\\7 & 2\end{bmatrix}$
$2AB=2\begin{bmatrix}1 & 1\\7 & 2\end{bmatrix}$
$\qquad=\begin{bmatrix}2 & 2\\14 & 4\end{bmatrix}$
$B^2=B.B=\begin{bmatrix}2 & 1\\1 & 0\end{bmatrix}\begin{bmatrix}2 & 1\\1 & 0\end{bmatrix}$
$\quad=\begin{bmatrix}2(2)+1(1) & 2(1)+(1)(0)\\1(2)+0(1) & 1(1)+0(0)\end{bmatrix}$
$\quad=\begin{bmatrix}4+1 & 2+0\\2+0 & 1+0\end{bmatrix}$
$\quad=\begin{bmatrix}5 & 2\\2 & 1\end{bmatrix}$
$A+B=\begin{bmatrix}1 & -1\\2 & 3\end{bmatrix}+\begin{bmatrix}2 & 1\\1 & 0\end{bmatrix}$
$\qquad=\begin{bmatrix}3 & 0\\3 & 3\end{bmatrix}$
$(A+B)^2=\begin{bmatrix}3 & 0\\3 & 3\end{bmatrix}\begin{bmatrix}3 & 0\\3 & 3\end{bmatrix}$
$\quad\qquad=\begin{bmatrix}3(3)+0(3) & 3(0)+0(3)\\3(3)+3(3) & 3(0)+3(3)\end{bmatrix}$
$\quad\qquad=\begin{bmatrix}9+0 & 0+0\\9+9 & 0+9\end{bmatrix}$
$\quad\qquad=\begin{bmatrix}9 & 0\\18 & 9\end{bmatrix}$---------(1)
$A^2+2AB+B^2=\begin{bmatrix}-1 & -4\\8 & 7\end{bmatrix}+\begin{bmatrix}2 & 2\\14 & 4\end{bmatrix}+\begin{bmatrix}5 & 2\\2 & 1\end{bmatrix}$
$\qquad\qquad\;\;\;\;\;\;\;=\begin{bmatrix}6 & 0\\24 & 12\end{bmatrix}$---------(2)
From (1) & (2) it is clear that $(A+B)^2\neq A^2+2AB+B^2$
answered Apr 4, 2013 by sreemathi.v
 

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