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

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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