Solving System of Linear Equations: (lesson 5 of 5)

## Inverse Matrix Method

Suppose you are given an equation in one variable such as $4x = 10$. Then
you will find the value of $x$ that solves this equation by multiplying the
equation by the inverse of 4: $\color{blue}{\frac14} \cdot 4x = \color{blue}{\frac14} \cdot 10$,
so the solution will be $x = 2.5$.

Sometimes we can do something very similar to solve systems of linear
equations; in this case, we will use the inverse of the coefficient matrix. But
first we must check that this inverse exists! The conditions for the existence
of the inverse of the coefficient matrix are the same as those for using
Cramer's rule, that is

1. The system must have the same number of
equations as variables, that is, the coefficient matrix of the system must be
square.

2. The determinant of the
coefficient matrix must be non-zero. The reason, of course, is that the inverse
of a matrix exists precisely when its determinant is non-zero.

3. To use this method follow the
steps demonstrated on the following system:

$$
\begin{aligned}
-x + 3y + z &= 1 \\
2x + 5y &= 3 \\
3x + y - 2z &= -2
\end{aligned}
$$

Step 1: Rewrite the system using matrix multiplication:

$$
\left( {\begin{array}{*{20}{c}}
- 1\\
2\\
3
\end{array}\begin{array}{*{20}{c}}
3\\
5\\
1
\end{array}\begin{array}{*{20}{c}}
1\\
0\\
- 2
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
x\\
y\\
z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
1\\
3\\
- 2
\end{array}} \right)
$$

and writing the coefficient matrix as A, we have

$$
A\left( {\begin{array}{*{20}{c}}
x\\
y\\
z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
1\\
3\\
- 2
\end{array}} \right)
$$

Step 2: FInd the inverse of the coefficient matrix A. In this case the inverse is

$$
A^{ - 1} = \left( {\begin{array}{*{20}{c}}
- \frac{{10}}{9}\\
\frac{4}{9}\\
- \frac{{13}}{9}
\end{array}\begin{array}{*{20}{c}}
\frac{7}{9}\\
- \frac{1}{9}\\
\frac{10}{9}
\end{array}\begin{array}{*{20}{c}}
- \frac{5}{9}\\
\frac{2}{9}\\
- \frac{11}{9}
\end{array}} \right)
$$

Step 3: Multiply both sides of the equation (that you wrote in step #1) by the matrix $A^{-1}$.

**On the left you'll get**

$$
A^{ - 1}A \left( {\begin{array}{*{20}{c}}
x\\
y\\
z
\end{array}} \right),
\text{which is}
\left( {\begin{array}{*{20}{c}}
x\\
y\\
z
\end{array}} \right)
$$

**On the right, you get**

$$
{A^{ - 1}}\left( {\begin{array}{*{20}{c}}
1\\
3\\
{ - 2}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{ - \frac{{10}}{9}}\\
{\frac{4}{9}}\\
{ - \frac{{13}}{9}}
\end{array}\begin{array}{*{20}{c}}
{\frac{7}{9}}\\
{ - \frac{1}{9}}\\
{\frac{{10}}{9}}
\end{array}\begin{array}{*{20}{c}}
{ - \frac{5}{9}}\\
{\frac{2}{9}}\\
{ - \frac{{11}}{9}}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
1\\
3\\
{ - 2}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{21}\\
{ - 3}\\
{39}
\end{array}} \right)
$$

and so the solution is

$$
\left( {\begin{array}{*{20}{c}}
x\\
y\\
z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{21}\\
{ - 3}\\
{39}
\end{array}} \right)
$$