#### Need solution for RD Sharma maths class 12 chapter Algebra of matrices exercise 4.3 question 10

Answer: Hence proved $A^{2}=0$

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: $A=\left[\begin{array}{cc}a b & b^{2} \\ -a^{2} & -a b\end{array}\right]$

Prove:$A^{2}=0$

Consider,

$A^{2}=A A\\\\ A^{2}=\left[\begin{array}{cc}a b & b^{2} \\ -a^{2} & -a b\end{array}\right]\left[\begin{array}{cc}a b & b^{2} \\ -a^{2} & -a b\end{array}\right]\\\\\\ =\left[\begin{array}{cc}a b \times a b+b^{2} \times\left(-a^{2}\right) & a b \times b^{2}+b^{2} \times(-a b) \\ -a^{2} \times a b+(-a b) \times(-a)^{2} & -a^{2} \times b^{2}+(-a b) \times(-a b)\end{array}\right]\\\\ \\=\left[\begin{array}{cc}a^{2} b^{2}-a^{2} b^{2} & a b^{3}-a b^{3} \\ -a^{3} b+a^{3} b & -a^{2} b^{2}+a^{2} b^{2}\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]\\\\ A^{2}=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]\\\\ A^{2}=0$

Hence, proved.

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