Question

How do I solve a system of equations using inverse matrices?

Answer 1

Consider a normal equation in `x` such as:
`3x=6`
To solve this equation you simply take the `3` in front of `x` and put it, dividing, below the `6` on the right side of the equal sign.
`x=6/3=3^-1*6=2`
at this point you can "read" the solution as: `x=2`.
With a system of `n` equations in `n` unknowns you do basically the same, the only difference is that you have more than 1 unknown (and equation) that can now be represented by matrices and by the inverse matrix in place of the coefficient to the -1 (in our example is `3^-1`).

You can use matrices and change your system in a matrix equation :
If you have the following system:

(where `a,b,c,d,e,g` are real numbers)
you can change it in a matrix equation:

Where `A` is the matrix of the coefficients of the unknowns, `U` is the column of the unknown and `B` is the column of the pure coefficients (without unknowns).

You can check that this representation with matrices represents the system by doing the multiplication `A*U` and setting it equal to `B` you'll get back your original system!!!

Now, to solve your matrix equation `A*U=B` you can multiply both sides by the inverse of `A`, i.e. `A^-1`

(Remembering that `I` is the identity matrix .

For example:

So:
`x=-3`
`y=5`