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  • Need solution for RD Sharma maths class 12 chapter Algebra of matrices exercise 4.3 question 33

Need solution for RD Sharma maths class 12 chapter Algebra of matrices exercise 4.3 question 33

Answers (1)

Answer: A^{2}-5 A-14 I=0

Hint: I is identity matrix so, I=\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]

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}3 & -5 \\ -4 & 2\end{array}\right]

Now, we will find the matrix for A^2, we get

A^{2}=A A=\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]=\left[\begin{array}{cc}9+20 & -15-10 \\ -12-8 & 20+4\end{array}\right] \\\\ A^{2}=\left[\begin{array}{cc}29 & -25 \\ -20 & 24\end{array}\right] ... (i)

Now, we will find the matrix for 5A, we get

5 A=5\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right] \\\\ 5 A=\left[\begin{array}{cc}15 & -25 \\ -20 & 10\end{array}\right] ... ( ii)

So, substitute corresponding values from equation i & ii in eqn A^{2}-5 A-14 I

we get

=\left[\begin{array}{cc}29 & -25 \\ -20 & 24\end{array}\right]-\left[\begin{array}{cc}15 & -25 \\ -20 & 10\end{array}\right]-14\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] \\ \\\\ =\left[\begin{array}{cc}29 & -25 \\ -20 & 24\end{array}\right]-\left[\begin{array}{cc}15 & -25 \\ -20 & 10\end{array}\right]-\left[\begin{array}{cc}14 & 0 \\ 0 & 14\end{array}\right] \\\\\\ =\left[\begin{array}{cc}29-15-14 & -25+25-0 \\ -20+20-0 & 24-10-14\end{array}\right] \\\\\\ \left[\begin{array}{cc}0 & 0 \\ 0 & 0\end{array}\right] = 0

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