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Explain solution for rd sharma class 12 chapter Algebra of matrices exercise 4.3 question 43

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Answer: 

f(A)=\left[\begin{array}{ccc}6 & -2 & 6 \\ 0 & 4 & 4 \\ 1 & 1 & 4\end{array}\right]

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}{ccc}0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0\end{array}\right]\ \ \ and \ \ f(x)=x^{3}+4 x^{2}-x

Substitute x=A

Then, f(A)=A^{3}+4 A^{2}-A ... ( i)

Now we will find the matrix A^ 2

\begin {array}{ll}A^{2}=\left[\begin{array}{ccc}0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0\end{array}\right]\left[\begin{array}{ccc}0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0\end{array}\right] \\\\\\ A^{2}=\left[\begin{array}{lll}0+2+2 & 0-3-2 & 0+0+0 \\ 0-6+0 & 2+9+0 & 4+0+0 \\ 0-2+0 & 1+3+0 & 2+0+0\end{array}\right] \\\\ \\A ^{2}=\left[\begin{array}{ccc}4 & -5 & 0 \\ -6 & 11 & 4 \\ -2 & 4 & 2\end{array}\right] \end{}

 

Now we will find matrix A^ 3

 

\begin{array} {ll} A^{3}=A^{2} A=\left[\begin{array}{ccc}4 & -5 & 0 \\ -6 & 11 & 4 \\ -2 & 4 & 2\end{array}\right]\left[\begin{array}{ccc}0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0\end{array}\right] \\\\\\ A^{3}=\left[\begin{array}{ccc}0-10+0 & 4+15+0 & 8+0+0 \\ 0+22+4 & -6-33-4 & -12+0+0 \\ 0+8+2 & -2-12-2 & -4+0+0\end{array}\right] \\\\ \\A^{3}=\left[\begin{array}{ccc}-10 & 19 & 8 \\ 26 & -43 & -12 \\ 10 & -16 & -4\end{array}\right] \end{array}

Put the value of A, A^2, A^3 in equation i 

 

f(A)=A^{3}+4 A^{2}-A\\\\ =\left[\begin{array}{ccc}-10 & 19 & 8 \\ 26 & -43 & -12 \\ 10 & -16 & -4\end{array}\right]+4\left[\begin{array}{ccc}4 & -5 & 0 \\ -6 & 11 & 4 \\ -2 & 4 & 2\end{array}\right]-\left[\begin{array}{ccc}0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0\end{array}\right] \\\\\\ =\left[\begin{array}{ccc}-10 & 19 & 8 \\ 26 & -43 & -12 \\ 10 & -16 & -4\end{array}\right]+\left[\begin{array}{ccc}16 & -20 & 0 \\ -24 & 44 & 16 \\ -8 & 16 & 8\end{array}\right]-\left[\begin{array}{ccc}0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0\end{array}\right] \\\\\\ =\left[\begin{array}{ccc}-10+16-0 & 19-20-1 & 8+0-2 \\ 26-24-2 & -43+44+3 & -12+16-0 \\ 10-8-1 & -16+16+1 & -4+8-0\end{array}\right] \\\\\\ = \left[\begin{array}{ccc}6 & -2 & 6 \\ 0 & 4 & 4 \\ 1 & 1 & 4\end{array}\right]

 

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