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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

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