Question

How do you solve `Ax=B` given `A^-1=((2, -1), (3, -2))` and `B=((2), (3))`?

Answer 1

I found: `x=((x_1),(x_2))=((1),(0))`

Explanation:

As in a normal equation we can multiply both sides for the same amount:
`color(red)(A^-1)Ax=color(red)(A^-1)B`
`Ix=A^-1B`
where `I` is the `4xx4` identity matrix giving;
`x=A^-1B`
multiplying the two matrices on the right we get:
`x=((4-3),(6-6))=((1),(0))`
where we get, for the incognitas' vector `x`, two solutions:
`x=((x_1),(x_2))=((1),(0))`