# Question #0ed08

May 5, 2016

$\left(x , y\right) = \left(- 4 , 1\right)$

#### Explanation:

Performing the multiplication, first, we have

${A}^{2} = {\left(\begin{matrix}3 & 2 \\ 1 & 1\end{matrix}\right)}^{2} = \left(\begin{matrix}3 & 2 \\ 1 & 1\end{matrix}\right) \left(\begin{matrix}3 & 2 \\ 1 & 1\end{matrix}\right) = \left(\begin{matrix}11 & 8 \\ 4 & 3\end{matrix}\right)$

$x A = x \left(\begin{matrix}3 & 2 \\ 1 & 1\end{matrix}\right) = \left(\begin{matrix}3 x & 2 x \\ x & x\end{matrix}\right)$

$y I = y \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right) = \left(\begin{matrix}y & 0 \\ 0 & y\end{matrix}\right)$

Then, summing, we have

${A}^{2} + x A + y I = \left(\begin{matrix}11 & 8 \\ 4 & 3\end{matrix}\right) + \left(\begin{matrix}3 x & 2 x \\ x & x\end{matrix}\right) + \left(\begin{matrix}y & 0 \\ 0 & y\end{matrix}\right)$

$= \left(\begin{matrix}3 x + y + 11 & 2 x + 8 \\ x + 4 & x + y + 3\end{matrix}\right) = \left(\begin{matrix}0 & 0 \\ 0 & 0\end{matrix}\right)$

Equating entries with corresponding indices, we get the following system of equations from the bottom row:

$\left\{\begin{matrix}x + 4 = 0 \\ x + y + 3 = 0\end{matrix}\right.$

From the first equation, we have $x = - 4$

Substituting this into the second equation, we get

$- 4 + y + 3 = 0$

$\implies y = 1$

If we substitute these values in for $x$ and $y$ we find that they fulfill the given equation, and thus we have $\left(x , y\right) = \left(- 4 , 1\right)$