<s>[INST] Matrix Equation, Solving for Variables. I'm going through my exercises, and came across a problem that wasn't covered in our lectures. Here's the question:
$
\begin{align}
\begin{bmatrix}
a-b & b+c\\
3d+c & 2a-4d
\end{bmatrix}
\end{align}
=
$
$
\begin{align}
\begin{bmatrix}
8 & 1\\
7 & 6
\end{bmatrix}
\end{align}
$
What I have done so far is:
$
\begin{align}
\begin{bmatrix}
a-b-8 & b+c-1\\
3d+c-7 & 2a-4d-6
\end{bmatrix}
\end{align}
=
$
$
\begin{align}
\begin{bmatrix}
0 & 0\\
0 & 0
\end{bmatrix}
\end{align}
$

And the solving for variables,
$$
a-b-8 = 0
$$
$$
a-b = 8
$$
$$
a = \frac{8}{-b}
$$

$$
b+c-1=0
$$
$$
b+c=1
$$
$$
b=\frac{1}{c}
$$

$$
3d+c-7=0
$$
$$
3d+c=7
$$
$$
3d=\frac{7}{c}
$$
$$
d=\frac{7}{3c}
$$

$$
2a-4d-6=0
$$
$$
2a-4d=6
$$
$$
\frac{16}{-2b}-\frac{28}{12c}=6
$$

Am I going about this correctly? Or am I just doing this completely incorrect?
 [/INST] The best way is follow the solution given by @Alex Wertheim. Let me show you a tedious substitution method to solve this. From $a_{11}$ we have $$b=a-8$$ (dividing by $a$ or $b$ is invalid as we may have zeros). Now from $a_{12}$ entry we have $$c=-b+1=-a+9,$$ and using $a_{21}$ $$d=\dfrac{-c+7}{3}=\dfrac{a-2}{3}.$$ Finally from the last entry $a_{22}$ we have $$a-2d=a-\dfrac23(a-2)=3.$$ Find $a$ from this simple linear equation and substitute back to find $b, c$ and $d.$ 
</s>