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  • Explain solution for rd sharma class class 12 chapter Algebra of matrices exercise 4.3 question 34 math

Explain solution for rd sharma class class 12 chapter Algebra of matrices exercise 4.3 question 34 math

Answers (1)

Answer:

A^{2}-5 A-7 I=0, A^{4}=\left[\begin{array}{cc}39 & 55 \\ -55 & -16\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}{cc}3 & 1 \\ -1 & 2\end{array}\right]

Prove:A^{2}-5 A+7 I=0

Solution: I is identity matrix so

7 I=7\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}7 & 0 \\ 0 & 7\end{array}\right]

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

A^{2}=A A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\\\\ A^{2}=\left[\begin{array}{cc}9-1 & 3+2 \\ -3-2 & -1+4\end{array}\right]\\\\ A^{2}=\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right] ...(i)

Now, we will find the matrix for  5A, we get

5 A=5\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\\\\ 5 A=\left[\begin{array}{cc}15 & 5 \\ -5 & 10\end{array}\right] ... (ii)

So, substituting corresponding values from equation  i &  ii in
A^{2}-5 A+7 I

we get

\begin{array}{l} =\left[\begin{array}{cc} 8 & 5 \\ -5 & 3 \end{array}\right]-\left[\begin{array}{cc} 15 & 5 \\ -5 & 10 \end{array}\right]+\left[\begin{array}{ll} 7 & 0 \\ 0 & 7 \end{array}\right] \\\\ =\left[\begin{array}{cc} 8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7 \end{array}\right] \\\\ =\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]=0 \end{array}

\therefore A^{2}-5 A+7 I=0

Hence, proved.

We will find A^4 

A^{2}-5 A+7 I=0

Multiply both sides by A^2, we get

A^{2}\left(A^{2}-5 A+7 I\right)=A^{2}(0) \\\\ A^{4}-5 A^{2} A+7 I A^{2}=0 \\\\ A^{4}=5 A^{2} A-7 I A^{2}\\\\ A^{4}=5 A^{2} A-7 A^{2}

Now we will substitute the corresponding values we get

A^{4}=5\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]-7\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]\\\\ \\ A^{4}=5\left[\begin{array}{cc}24-5 & 8+10 \\ -15-3 & -5+6\end{array}\right]-7\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right] \\\\\\ A^{4}=5\left[\begin{array}{cc}19 & 18 \\ -18 & 1\end{array}\right]-7\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]\\\\\ A^{4}=\left[\begin{array}{cc}5 \times 19 & 5 \times 18 \\ -18 \times 5 & 1 \times 5\end{array}\right]-\left[\begin{array}{cc}7 \times 8 & 7 \times 5 \\ -5 \times 7 & 3 \times 7\end{array}\right] \\\\\\ A^{4}=\left[\begin{array}{cc}95 & 90 \\ -90 & 5\end{array}\right]-\left[\begin{array}{cc}56 & 35 \\ -35 & 21\end{array}\right]\\\\\\ A^{4}=\left[\begin{array}{cc}95-56 & 90-35 \\ -90+35 & 5-21\end{array}\right] \\\\\\A^{4}=\left[\begin{array}{cc}39 & 55 \\ -55 & -16\end{array}\right]

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