#### Please solve RD Sharma class 12 chapter Algebra of matrices exercise 4.3 question 51 maths textbook solution

$(A+B)^{2}=A^{2}+B^{2}$

Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.

Given:

$A=\left[\begin{array}{cc}0 & -x \\ x & 0\end{array}\right], B=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right], x^{2}=-1$

To prove:

$(A+B)^{2}=A^{2}+B^{2}$

Consider RHS

$\\\\ A^{2}=A A \\\\ B^{2}=B B$

Then,

$\\\\ \Rightarrow A^{2}+B^{2}=\left[\begin{array}{cc}0 & -x \\ x & 0\end{array}\right]\left[\begin{array}{cc}0 & -x \\ x & 0\end{array}\right]+\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] \\\\\\ A^{2}+B^{2}=\left[\begin{array}{cc}0(0)+(-x)(x) & 0(-x)+(-x) 0 \\ x(0)+0(x) & x(-x)+0(0)\end{array}\right]+\left[\begin{array}{ll}0(0)+1(1) & 0(1)+1(0) \\ 1(0)+0(1) & 1(1)+0(0)\end{array}\right]\\\\\\ A^{2}+B^{2}=\left[\begin{array}{cc}0-x^{2} & 0-0 \\ 0+0 & -x^{2}+0\end{array}\right]+\left[\begin{array}{ll}0+1 & 0+0 \\ 0+0 & 1+0\end{array}\right]\\\\\\ A^{2}+B^{2}=\left[\begin{array}{cc}-x^{2} & 0 \\ 0 & -x^{2}\end{array}\right]+\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \\\\\\ A^{2}+B^{2}=\left[\begin{array}{cc}-x^{2}+1 & 0+0 \\ 0+0 & -x^{2}+1\end{array}\right]\\\\\\ A^{2}+B^{2}=\left[\begin{array}{cc}-x^{2}+1 & 0 \\ 0 & -x^{2}+1\end{array}\right]$

Now put

$\\\\x^{2}=-1$

$\\\\ A^{2}+B^{2}=\left[\begin{array}{cc}-(-1)+1 & 0 \\ 0 & -(-1)+1\end{array}\right]=\left[\begin{array}{cc}1+1 & 0 \\ 0 & 1+1\end{array}\right] \\\\\\A^{2}+B^{2}=\left[\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right]$

Now taking LHS side

$\\\\(A+B)^{2}=(A+B)(A+B)\\\\ A+B=\left[\begin{array}{cc} 0 & -x \\ x & 0 \end{array}\right]+\left[\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right]=\left[\begin{array}{cc} 0+0 & -x+1 \\ x+1 & 0+0 \end{array}\right]=\left[\begin{array}{cc} 0 & -x+1 \\ x+1 & 0 \end{array}\right]\\\\\\ (A+B)^{2}=\left[\begin{array}{cc} 0 & -x+1 \\ x+1 & 0 \end{array}\right]\left[\begin{array}{cc} 0 & -x+1 \\ x+1 & 0 \end{array}\right]\\\\\\ (A+B)^{2}=\left[\begin{array}{cc} 0(0)+(-x+1)(x+1) & 0(-x+1)+(-x+1)(0) \\ (x+1)(0)+0(x+1) & (x+1)(-x+1)+0(0) \end{array}\right]\\\\\\ (A+B)^{2}=\left[\begin{array}{cc} 0+(-x+1)(x+1) & 0+0 \\ 0+0 & (x+1)(-x+1)+0 \end{array}\right]\\\\\\ (A+B)^{2}=\left[\begin{array}{cc} -x^{2}-x+x+1 & 0 \\ 0 & -x^{2}+x-x+1 \end{array}\right]$

$(A+B)^{2}=\left[\begin{array}{cc} -x^{2}+1 & 0 \\ 0 & -x^{2}+1 \end{array}\right]$

Now put

$x^2 = -1$

$\\\\ (A+B)^{2}=\left[\begin{array}{cc} -(-1)+1 & 0 \\ 0 & -(-1)+1 \end{array}\right]=\left[\begin{array}{cc} 1+1 & 0 \\ 0 & 1+1 \end{array}\right]\\\\\\ (A+B)^{2}=\left[\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right]\\\\\\$

Hence proved, LHS=RHS

$(A+B)^{2}=A^{2}+B^{2}$