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Explain solution for rd sharma class 12 chapter Algebra of matrices exercise 4.3 question 50

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Answer: k=-4

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}-3 & 2 \\ 1 & -1\end{array}\right] and \ \ I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] and \ \ A^{2}+I=k A

Consider \\\\A^{2}=A A

\\\\ A^{2}=\left[\begin{array}{cc}-3 & 2 \\ 1 & -1\end{array}\right]\left[\begin{array}{cc}-3 & 2 \\ 1 & -1\end{array}\right]\\\\\\ =\left[\begin{array}{cc}(-3)(-3)+2(1) & (-3)(2)+2(-1) \\ 1(-3)+(-1)(1) & 1(2)+(-1)(-1)\end{array}\right]\\\\\\ =\left[\begin{array}{cc}9+2 & -6-2 \\ -3-1 & 2+1\end{array}\right] =\left[\begin{array}{cc}11 & -8 \\ -4 & 3\end{array}\right]

I is an identity matrix

Consider, 

\\\\ k A=k\left[\begin{array}{cc}-3 & 2 \\ 1 & -1\end{array}\right]

\\\\ k A=\left[\begin{array}{cc}-3 k & 2 k \\ k & -k\end{array}\right] ... (ii)

Substitute all values in equation A^{2}+I=k A, we get

\\\\ \Rightarrow\left[\begin{array}{cc}11 & -8 \\ -4 & 3\end{array}\right]+\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}-3 k & 2 k \\ k & -k\end{array}\right]\\\\\\ \Rightarrow\left[\begin{array}{cc}11+1 & -8+0 \\ -4+0 & 3+1\end{array}\right]=\left[\begin{array}{cc}-3 k & 2 k \\ k & -k\end{array}\right]\\\\\\ \Rightarrow\left[\begin{array}{cc}12 & -8 \\ -4 & 4\end{array}\right]=\left[\begin{array}{cc}-3 k & 2 k \\ k & -k\end{array}\right]

Since, corresponding entries of equal matrices are equal, So

\begin{array}{l} \Rightarrow-3 k=12 \\\\ \Rightarrow k=-\frac{12}{3} \\\\ \Rightarrow k=-4 \end{array}

Hence, value of k is -4

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