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### Answers (1)

Answer:

$A^{2}-5 A-14 I=\left[\begin{array}{cc}0 & 0 \\ 0 & 0\end{array}\right]=0\ \ and \ \ A^{3}=\left[\begin{array}{cc}187 & -195 \\ -156 & 148\end{array}\right]$

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:

$A=\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]$

First, we solve this   $A^{2}-5 A-14 I$

Where I is an identity matrix

Then,

$\\\\ \left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]-5\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]-14\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]\\\\ \\=\left[\begin{array}{cc}9+20 & -15-10 \\ -12-8 & 20+4\end{array}\right]-\left[\begin{array}{cc}15 & -25 \\ -20 & 10\end{array}\right]-\left[\begin{array}{cc}14 & 0 \\ 0 & 14\end{array}\right]$

$\\\\ =\left[\begin{array}{cc} 29 & -25 \\ -20 & 24 \end{array}\right]-\left[\begin{array}{cc} 15 & -25 \\ -20 & 10 \end{array}\right]-\left[\begin{array}{cc} 14 & 0 \\ 0 & 14 \end{array}\right]\\\\\\ =\left[\begin{array}{cc} 29-15-14 & -25+25-0 \\ -20+20-0 & 24-10-14 \end{array}\right]\\\\\\ =\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]=0\\\\\\ A^{2}-5 A-14 I=0\\\\$

Now,

$\\\\ A^{2}-5 A-14 I=0\\\\ A^{2}=5 A+14 I\\\\ A^{3}=A^{2} A=(5 A+14 I) A\\\\ A^{3}=5 A^{2}+14 A\\\\ A^{3}=A^{2} A=5 A^{2}+14 A$

As we know AI=A

$\\\\ A^{3}=5\left[\begin{array}{cc} 29 & -25 \\ -20 & 24 \end{array}\right]+14\left[\begin{array}{cc} 3 & -5 \\ -4 & 2 \end{array}\right]\\\\\\ A^{3}=\left[\begin{array}{cc} 145 & -125 \\ -100 & 120 \end{array}\right]+\left[\begin{array}{cc} 42 & -70 \\ -56 & 28 \end{array}\right]\\\\\\ A^{3}=\left[\begin{array}{cc} 145+42 & -125-70 \\ -100-56 & 120+28 \end{array}\right]\\\\\\ A^{3}=\left[\begin{array}{cc} 187 & -195 \\ -156 & 148 \end{array}\right]$

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