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Question
If `A=[[0,-x],[x,0]],[[0,1],[1,0]]` and `x^2=-1,` then show that `(A+B)^2=A^2+B^2`
Solution
Given:\[A = \begin{bmatrix}0 & - x \\ x & 0\end{bmatrix}, B = \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}\] and x2 = −1
To show: (A + B)2 = A2 + B2
LHS:
\[A + B = \begin{bmatrix}0 & - x \\ x & 0\end{bmatrix} + \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}\]
\[ = \begin{bmatrix}0 + 0 & - x + 1 \\ x + 1 & 0 + 0\end{bmatrix}\]
\[ = \begin{bmatrix}0 & - x + 1 \\ x + 1 & 0\end{bmatrix}\]
\[ \left( A + B \right)^2 = \begin{bmatrix}0 & - x + 1 \\ x + 1 & 0\end{bmatrix}\begin{bmatrix}0 & - x + 1 \\ x + 1 & 0\end{bmatrix}\]
\[ = \begin{bmatrix}0 + \left( 1 - x \right)\left( 1 + x \right) & 0 + 0 \\ 0 + 0 & \left( x + 1 \right)\left( 1 - x \right) + 0\end{bmatrix}\]
\[ = \begin{bmatrix}1 - x^2 & 0 \\ 0 & 1 - x^2\end{bmatrix} . . . (1)\]
R. H. S
\[A = \begin{bmatrix}0 & - x \\ x & 0\end{bmatrix}\]
\[ A^2 = \begin{bmatrix}0 & - x \\ x & 0\end{bmatrix}\begin{bmatrix}0 & - x \\ x & 0\end{bmatrix}\]
\[ = \begin{bmatrix}0 - x^2 & 0 + 0 \\ 0 + 0 & - x^2 + 0\end{bmatrix}\]
\[ = \begin{bmatrix}- x^2 & 0 \\ 0 & - x^2\end{bmatrix} . . . (2)\]
\[\]
\[ B = \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}\]
\[ B^2 = \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}\]
\[ = \begin{bmatrix}0 + 1 & 0 + 0 \\ 0 + 0 & 1 + 0\end{bmatrix}\]
\[ = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} . . . (3)\]
\[\]
\[Adding (2) and (3), we get\]
\[ A^2 + B^2 = \begin{bmatrix}- x^2 & 0 \\ 0 & - x^2\end{bmatrix} + \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\]
\[ = \begin{bmatrix}1 - x^2 & 0 \\ 0 & 1 - x^2\end{bmatrix} . . . (4)\]
Comparing (1) and (4), we get
(A + B)2 = A2 + B2
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