<s>[INST] Equations involving $2 \times 2$ Matrices There are two possible values of A in the solution of the matrix equation 
$\left[\begin{matrix}2A+1&-5\\-4&A\end{matrix}\right]^{-1}\cdot \left[\begin{matrix}A-5&B\\2A-2&C\end{matrix}\right]= \left[\begin{matrix}14&D\\E&F\end{matrix}\right]$ where $A,B,C,D,E,F \space all \space \in \mathbb{R}$. The absolute value of the difference of these two solutions is?
Options given :


*

*$\dfrac{8}{3}$

*$\dfrac{11}{3}$

*$\dfrac{1}{3}$

*$\dfrac{19}{3}$
I have first obtained the result of  $\left[\begin{matrix}2A+1&-5\\-4&A\end{matrix}\right]^{-1}$
$= \dfrac{1}{(A(2A+1)-20)} \cdot \left[\begin{matrix}A&5\\4&2A+1\end{matrix}\right]$
where $\left[\begin{matrix}A&5\\4&2A+1\end{matrix}\right] =$ adjoint($\left[\begin{matrix}2A+1&-5\\-4&A\end{matrix}\right]$)
Then the next step is to find 
$\dfrac{1}{(A(2A+1)-20)} \cdot \left[\begin{matrix}A&5\\4&2A+1\end{matrix}\right] \cdot \left[\begin{matrix}A-5&B\\2A-2&C\end{matrix}\right]= \left[\begin{matrix}14&D\\E&F\end{matrix}\right]$
$\implies \dfrac{1}{(A(2A+1)-20)} \cdot \left[\begin{matrix}A(A-5)+10(A-1)&AB+5C\\4(A-5)+2(A-1)(2A+1)&4B+C(2A+1)\end{matrix}\right]=\left[\begin{matrix}14&D\\E&F\end{matrix}\right]$
$\therefore \dfrac{A^2-5A+10A-10}{2A^2+A-20}=14$[From the above written equation]
$\implies 27A^2+23A-270=0$
whose curve is like this :

and the roots are :
$-10$, $-\dfrac{23}{27}$
The difference between the two is $-\dfrac{247}{27}$
Hence none of the options are matching. Where am I wrong and if Yes, please correct me and give the correct answer please. I request anyone. 
 [/INST] First keep in mind that if $x_{1,2}$ are the roots of a quadratic $ax^2+bx +c=0,\,a\neq 0$ then
$$|x_1-x_2|={\sqrt{\Delta}\over a}$$
Second you do not need to invert the first matrix just multiply both sides with it so the matrix equation is equivalent to
$$\begin{bmatrix} A-5 & B\\2A-2 & C\end{bmatrix}=\begin{bmatrix}2A+1 & -5\\-4 & A\end{bmatrix}\cdot\begin{bmatrix}14 & D\\E & F\end{bmatrix}$$
This is equivalent to
$$\begin{bmatrix} A-5 & B\\2A-2 & C\end{bmatrix}=\begin{bmatrix}28A+14-5E & (2A+2)D-5F\\-56+AE & -4D+AF\end{bmatrix}$$
We just identify the first column
$$\begin{cases} 5E=27A+19\\5EA=10A+270\end{cases}$$
Substituting $5E$ one gets
$$3A^2+A-30=0$$
This means $\Delta=361=19^2$ and 
$$|x_1-x_2|={19\over 3}$$
Your mistake is in the last transformation
$$A^2-5A+10A=28A^2+14A-280\iff 27A^2+9A-270\iff 3A^2+A-30=0$$
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