# Solving a System of Linear Equations By Using an Inverse Matrix

## Problem 65

Consider the system of linear equations

\begin{align*}

x_1&= 2, \\

-2x_1 + x_2 &= 3, \\

5x_1-4x_2 +x_3 &= 2

\end{align*}

**(a)** Find the coefficient matrix and its inverse matrix.

**(b)** Using the inverse matrix, solve the system of linear equations.

(*The Ohio State University, Linear Algebra Exam*)

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## Solution.

### (a) Find the coefficient matrix and its inverse matrix.

The coefficient matrix is

\[ A:=\begin{bmatrix}

1 & 0 & 0 \\

-2 &1 &0 \\

5 & -4 & 1

\end{bmatrix}.\]
To find the inverse, we reduce the augmented matrix

\[ \left[\begin{array}{rrr|rrr}

1 & 0 & 0 & 1 &0 & 0 \\

-2 & 1 & 0 & 0 & 1 & 0 \\

5 & -4 & 1 & 0 & 0 & 1 \\

\end{array}\right]\]
Using the elementary row operation.

\begin{align*}

&\left[\begin{array}{rrr|rrr}

1 & 0 & 0 & 1 &0 & 0 \\

-2 & 1 & 0 & 0 & 1 & 0 \\

5 & -4 & 1 & 0 & 0 & 1 \\

\end{array}\right]
\xrightarrow[R_3-5R_1]{R_2+R_1}

\left[\begin{array}{rrr|rrr}

1 & 0 & 0 & 1 &0 & 0 \\

0 & 1 & 0 & 2 & 1 & 0 \\

0 & -4 & 1 & -5 & 0 & 1 \\

\end{array}\right]\\

& \xrightarrow{R_3+4R_2}

\left[\begin{array}{rrr|rrr}

1 & 0 & 0 & 1 &0 & 0 \\

0 & 1 & 0 & 2 & 1 & 0 \\

0 & 0 & 1 & 3 & 4 & 1 \\

\end{array}\right].

\end{align*}

Now that the left $3\times 3$ part became the identity matrix (this form is called the reduced row echelon form), the inverse matrix is

\[A^{-1}=\begin{bmatrix}

1 & 0 & 0 \\

2 &1 &0 \\

3 & 4 & 1

\end{bmatrix}.\]

### (b) Using the inverse matrix, solve the system of linear equations.

Using the coefficient matrix $A$ the given system can be written as the matrix equation

\[A \begin{bmatrix}

x_1 \\

x_2 \\

x_3

\end{bmatrix}

=

\begin{bmatrix}

2 \\

3 \\

2

\end{bmatrix}.\]
Multiplying it by the inverse matrix $A^{-1}$ on the left, we get

\begin{align*}

\begin{bmatrix}

x_1 \\

x_2 \\

x_3

\end{bmatrix}

&=A^{-1}

\begin{bmatrix}

2 \\

3 \\

2

\end{bmatrix}\\

&= \begin{bmatrix}

1 & 0 & 0 \\

2 &1 &0 \\

3 & 4 & 1

\end{bmatrix}

\begin{bmatrix}

2 \\

3 \\

2

\end{bmatrix}

=\begin{bmatrix}

2 \\

7 \\

20

\end{bmatrix}.

\end{align*}

Therefore the solution of the system is $x_1=2, x_2=7, x_3=20$.

## Comment.

Once you obtained the inverse matrix $A^{-1}$ in part (1), it’s better to calculate $AA^{-1}$ using your matrix.

If $AA^{-1}$ is not the identity matrix, you made a calculation mistake somewhere.

Also, in part (b), after finding the solution of the system, you may want to substitute your answer to the original system to see if they are actually the solution.

These extra (but not so time consuming) computations reduce careless mistakes in a exam.

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