How do I solve for matrix A in the equation y=mx+c?

[uperkurk]

A = \begin{bmatrix} 5 & 7 \\ 2 & 3 \end{bmatrix} \to A^{-1} =\frac{1}{ad - bc} \begin{bmatrix} 3 & -7 \\ -2 & 5 \end{bmatrix}

Which is what I wrote? To be honest whoever invented this crap could not have made it more complicated if they tried.

A^{-1} =\frac{1}{ad - bc} is where I go wrong because I get confused on my operators, yes I'm at uni, but I have dyscalculia and have never ever been good at remembering numbers or anything to do with maths. I failed GCSE math and the only reason I'm doing math now is because it's part of my computer science course so I have no option but to learn it.

 

[Muphrid]

Edit, if you want, though, here's another linear system:

\begin{align*} 5x+7y &= -4 \\ 2x - 3y &= 6\end{align*}

The matrix A should be

A = \begin{bmatrix} 5 & 7 \\ 2 & -3 \end{bmatrix}

with the negative sign on the 3.

 

[uperkurk]

The matrix A should be

A = \begin{bmatrix} 5 & 7 \\ 2 & -3 \end{bmatrix}

with the negative sign on the 3.

Ohhhh my bad I didn't realize that. anyway this thread has gone on way too long i'll let you go help some other people.

It's a shame it took so long to figure out what I was actually doing wrong I could have spend those 2 days on something else but not to worry still have 2 more days to revise.