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

Dec 29, 2014

Consider a normal equation in $x$ such as:
$3 x = 6$
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.
$x = \frac{6}{3} = {3}^{-} 1 \cdot 6 = 2$
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 $A \cdot U$ and setting it equal to $B$ you'll get back your original system!!!

Now, to solve your matrix equation $A \cdot U = B$ you can multiply both sides by the inverse of $A$, i.e. ${A}^{-} 1$

(Remembering that $I$ is the identity matrix .

For example:

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