#### Please solve RD Sharma class 12 chapter Algebra of matrices exercise 4.3 question 31 maths textbook solution

Answer: Hence, proved $A^{3}-4 A^{2}+A=0$

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: $A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$

Prove:  $A^{3}-4 A^{2}+A=0$

Consider,$A^{2}=A A$

$A^{2}=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]=\left[\begin{array}{ll}4+3 & 6+6 \\ 2+2 & 3+4\end{array}\right]=\left[\begin{array}{cc}7 & 12 \\ 4 & 7\end{array}\right] \\\\ A^{3}=A^{2} A\\\\ A^{2} A=\left[\begin{array}{cc}7 & 12 \\ 4 & 7\end{array}\right]\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]\\\\ A^{3}=\left[\begin{array}{cc}14+12 & 21+24 \\ 8+7 & 12+14\end{array}\right]=\left[\begin{array}{ll}26 & 45 \\ 15 & 26\end{array}\right]$

Now putting value of $A^{3}, A^{2}$ and $A$ in the equation $A^{3}-4 A^{2}+A$we get,

$=\left[\begin{array}{cc}26 & 45 \\ 15 & 26\end{array}\right]-4\left[\begin{array}{cc}7 & 12 \\ 4 & 7\end{array}\right]+\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$

$=\left[\begin{array}{ll} 26 & 45 \\ 15 & 26 \end{array}\right]-\left[\begin{array}{ll} 28 & 48 \\ 16 & 28 \end{array}\right]+\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right] \\ =\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \\\\ =0 \\ \text { So, } A^{3}-4 A^{2}+A=0$

## Crack CUET with india's "Best Teachers"

• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support