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Provide solution for rd sharma math class class 12 chapter Algebra of matrices exercise 4.3 question 32

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Answer: Hence, proved A^{2}-12 A-I=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}5 & 3 \\ 12 & 7\end{array}\right]

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

To show that A^{2}-12 A-I=0

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

A^{2}=A A=\left[\begin{array}{cc}5 & 3 \\ 12 & 7\end{array}\right]\left[\begin{array}{cc}5 & 3 \\ 12 & 7\end{array}\right]\\\\ A^{2}=\left[\begin{array}{ll}25+36 & 15+21 \\ 60+84 & 36+49\end{array}\right]\\\\A^{2}=\left[\begin{array}{cc}61 & 36 \\ 144 & 85\end{array}\right] .... (i)

Now, we will find the matrix for 12A, we get

12 A=12\left[\begin{array}{cc}5 & 3 \\ 12 & 7\end{array}\right] \\\\\ 12 A=\left[\begin{array}{cc}60 & 36 \\ 144 & 84\end{array}\right] \quad \ldots(i i)

So, substituting corresponding values from equation i & ii in A^{2}-12 A-I

we get

=\left[\begin{array}{cc} 61 & 36 \\ 144 & 85 \end{array}\right]-\left[\begin{array}{cc} 60 & 36 \\ 144 & 84 \end{array}\right]-\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\\\

\left[\begin{array}{cc} 61-60-1 & 36-36-0 \\ 144-144-0 & 85-84-1 \end{array}\right]

\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]=0 \\ A^{2}-12 A-I=0

Hence, matrix A is the root of the given equation.

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