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

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

Answers (1)

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

Posted by

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