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  • Provide solution for rd sharma math class 12 chapter Algebra of matrices exercise 4.3 question 41

Provide solution for rd sharma math class 12 chapter Algebra of matrices exercise 4.3 question 41 

Answers (1)

Answer: 

A^{2}-4 A+3 I_{3}=\left[\begin{array}{ccc}6 & -14 & 10 \\ -21 & 36 & -25 \\ -3 & 5 & -5\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}1 & 2 & 0 \\ 3 & -4 & 5 \\ 0 & -1 & 3\end{array}\right]

I_3 is identity matrix of size 3.

I_{3}=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] \ \ and \ \ \ 3 I_{3}=\left[\begin{array}{lll}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{array}\right]

Now we will find the matrix for A^2 we get

\begin {array}{ll}A^{2}=A A=\left[\begin{array}{ccc}1 & 2 & 0 \\ 3 & -4 & 5 \\ 0 & -1 & 3\end{array}\right]\left[\begin{array}{ccc}1 & 2 & 0 \\ 3 & -4 & 5 \\ 0 & -1 & 3\end{array}\right] \\\\\ \\A^{2}=\left[\begin{array}{ccc}1+6+0 & 2-8+0 & 0+10+0 \\ 3-12+0 & 6+16-5 & 0-20+15 \\ 0-3+0 & 0+4-3 & 0-5+9\end{array}\right] \\\\\\A^{2}=\left[\begin{array}{ccc}7 & -6 & 10 \\ -9 & 17 & -5 \\ -3 & 1 & 4\end{array}\right] \ \ \ \ ...(i) \end{}

Now, we will find the matrix for 4A, we get

\begin {array}{ll} 4 A=4\left[\begin{array}{ccc}1 & 2 & 0 \\ 3 & -4 & 5 \\ 0 & -1 & 3\end{array}\right] \\\\\ 4 A=\left[\begin{array}{ccc}4 & 8 & 0 \\ 12 & -16 & 20 \\ 0 & -4 & 12\end{array}\right] \ \ \ ...(ii) \end{}

Substituting corresponding values from equation i & ii in the given equation, we getA^{2}-4 A+3 I_{3} \\\\ =\left[\begin{array}{ccc}7 & -6 & 10 \\ -9 & 17 & -5 \\ -3 & 1 & 4\end{array}\right]-\left[\begin{array}{ccc}4 & 8 & 0 \\ 12 & -16 & 20 \\ 0 & -4 & 12\end{array}\right]+\left[\begin{array}{lll}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{array}\right] \\\\\\ =\left[\begin{array}{ccc}7-4+3 & -6-8+0 & 10-0+0 \\ -9-12+0 & 17+16+3 & -5-20+0 \\ -3-0+0 & 1+4+0 & 4-12+3\end{array}\right] \\\\\\ =\left[\begin{array}{ccc}6 & -14 & 10 \\ -21 & 36 & -25 \\ -3 & 5 & -5\end{array}\right]

Hence, 

A^{2}-4 A+3 I_{3}=\left[\begin{array}{ccc}6 & -14 & 10 \\ -21 & 36 & -25 \\ -3 & 5 & -5\end{array}\right]

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