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

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

Answers (1)

Answer: Hence proved,

(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}

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