We have:

`A=((-1,5),(2,5)),` ` \ X=((x),(y))\ ` and `\ C=((4),(-2))`

To solve the system, we need the inverse of *A*, which we write as *A*^{-1}.

Swap leading diagonal:

`((5,5),(2,-1))`

Change signs of the other 2 elements:

`((5,-5),(-2,-1))`

Now we find the determinant of *A*:

`|A| = -5 - 10 = -15`

So

`A^-1` `=-1/15((5,-5),(-2,-1))` ` = ((-1/3,1/3),(2/15,1/15))` ` = ((-0.333,0.333),(0.133,0.067)) `

So the solution to the system is given by:

`X=A^-1C` `=((-0.333,0.333),(0.133,0.067))((4),(-2))` `=((-2),(0.4)) `

This answer means that we have found the solution `x = -2` and `y = 0.4`.

Is the solution correct?

We check it in the original set of equations:

`{:(-x+5y,=4),(2x+5y,=-2):}`

Substituting `x = -2` and `y = 0.4`, we get:

`−(−2) + 5×(0.4) = 2 + 2 = 4` [Checks OK]

`2×(−2) + 5×(0.4)` ` = −4 + 2` ` = −2` [Checks OK]

So the solution to the original system of equations is

`x = -2,\ \ y = 0.4`.