# We have matrix A=[(2,x),(y,3)] and B=[(y,-3),(5,2x)], A,BinMM_(2,2) (CC). How you find x,y  such that (A+xI_2)(B+yI_2)=[(-25,59),(2x+y^2,20)] ?

Sep 24, 2017

Given: $A = \left[\begin{matrix}2 & x \\ y & 3\end{matrix}\right]$ and $B = \left[\begin{matrix}y & - 3 \\ 5 & 2 x\end{matrix}\right]$

Compute $A + x {I}_{2}$:

$A + x {I}_{2} = \left[\begin{matrix}2 & x \\ y & 3\end{matrix}\right] + x \left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]$

$A + x {I}_{2} = \left[\begin{matrix}2 & x \\ y & 3\end{matrix}\right] + \left[\begin{matrix}x & 0 \\ 0 & x\end{matrix}\right]$

$A + x {I}_{2} = \left[\begin{matrix}2 + x & x \\ y & 3 + x\end{matrix}\right] \text{ [1]}$

Compute $B + y {I}_{2}$

$B + y {I}_{2} = \left[\begin{matrix}y & - 3 \\ 5 & 2 x\end{matrix}\right] + y \left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]$

$B + y {I}_{2} = \left[\begin{matrix}y & - 3 \\ 5 & 2 x\end{matrix}\right] + \left[\begin{matrix}y & 0 \\ 0 & y\end{matrix}\right]$

$B + y {I}_{2} = \left[\begin{matrix}2 y & - 3 \\ 5 & 2 x + y\end{matrix}\right] \text{ [2]}$

Substitute equation [1] and equation [2] into the given equation $\left(A + x {I}_{2}\right) \left(B + y {I}_{2}\right) = \left[\begin{matrix}- 25 & 59 \\ 2 x + {y}^{2} & 20\end{matrix}\right]$:

$\left[\begin{matrix}2 + x & x \\ y & 3 + x\end{matrix}\right] \left[\begin{matrix}2 y & - 3 \\ 5 & 2 x + y\end{matrix}\right] = \left[\begin{matrix}- 25 & 59 \\ 2 x + {y}^{2} & 20\end{matrix}\right]$

Performing the multiplication:

$\left[\begin{matrix}5 x + 2 x y + 2 y & 2 {x}^{2} + x y - 3 x - 6 \\ 2 {y}^{2} + 5 x + 15 & 2 {x}^{2} + x y + 6 x\end{matrix}\right] = \left[\begin{matrix}- 25 & 59 \\ 2 x + {y}^{2} & 20\end{matrix}\right]$

We have two unknown values and we have 4 equation from which we may chose; let's use the two equations in column 2:

$2 {x}^{2} + x y - 3 x - 6 = 59 \text{ [3]}$
$2 {x}^{2} + x y + 6 x = 20 \text{ [4]}$

Subtract equation [4] from equation [3]:

$- 9 x - 6 = 39$

$- 9 x = 45$

$x = - 5$

Substitute -5 for x into equation [4]:

$2 {\left(- 5\right)}^{2} - 5 y + 6 \left(- 5\right) = 20$

$- 5 y = 0$

$y = 0$