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

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

Answers (1)

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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