Arithmetic Operation of Matrices

$\begin{aligned} \begin{pmatrix} x & y \\ 2 & 1 \end{pmatrix} \begin{pmatrix} x & 0 \\ y & x \end{pmatrix} & = 2 \begin{pmatrix} 10 & 6-x \\ 3 & 0 \end{pmatrix} + \begin{pmatrix} 5 & 2x \\ 4 & x \end{pmatrix} \\ \begin{pmatrix} \color{#3D99F6}{x^2 + y^2} & \color{#D61F06}{xy} \\ 2x + y & x \end{pmatrix} & = \begin{pmatrix} 2(10)+5 & 2(6-x)+2x \\ 2(3)+4 & 2(0)+x \end{pmatrix} \\ & = \begin{pmatrix} \color{#3D99F6}{25} & \color{#D61F06}{12} \\ 10 & x \end{pmatrix} \end{aligned}$

$\implies \begin{cases} \color{#3D99F6}{x^2 + y^2 = 25} & ...(1) \\ \color{#D61F06}{xy = 12} & ...(2) \end{cases}$

$\begin{aligned} (x+y)^2 & = \color{#3D99F6}{x^2} + 2\color{#D61F06}{xy} + \color{#3D99F6}{y^2} = \color{#3D99F6}{25} + 2(\color{#D61F06}{12}) = 49 \\ \implies x+y & = \boxed{\pm 7} \end{aligned}$

I'll just show you how my work goes

right side

$2\begin{pmatrix} 10 & 6-x \\ 3 & 0 \end{pmatrix}\quad +\quad \begin{pmatrix} 5 & 2x \\ 4 & x \end{pmatrix}\\ =\begin{pmatrix} 20 & 12-2x \\ 6 & 0 \end{pmatrix}\quad +\quad \begin{pmatrix} 5 & 2x \\ 4 & x \end{pmatrix}\\ =\begin{pmatrix} 25 & 12 \\ 10 & x \end{pmatrix}$

left side

$\begin{pmatrix} x & y \\ x & 0 \end{pmatrix}\begin{pmatrix} x & 0 \\ y & x \end{pmatrix}\\ =\begin{pmatrix} { x }^{ 2 }+{ y }^{ 2 } & xy \\ 2x+y & x \end{pmatrix}$

ending

$\begin{pmatrix} { x }^{ 2 }+{ y }^{ 2 } & xy \\ 2x+y & x \end{pmatrix}\quad =\quad \begin{pmatrix} 25 & 12 \\ 10 & x \end{pmatrix}\\ \\ { x }^{ 2 }+{ y }^{ 2 }\quad =\quad 25\\ 2x\quad +\quad y\quad =\quad 10\\ xy\quad =\quad 12\\ \\ { (x+y })^{ 2 }-2xy\quad =\quad { x }^{ 2 }\quad +\quad { y }^{ 2 }\\ { (x+y })^{ 2 }-2\times 12=25\\ { (x+y })^{ 2 }=49\\ x+y=7$