# Solve the System of Linear Equations Using the Inverse Matrix of the Coefficient Matrix

## Problem 442

Consider the following system of linear equations

\begin{align*}

2x+3y+z&=-1\\

3x+3y+z&=1\\

2x+4y+z&=-2.

\end{align*}

**(a)** Find the coefficient matrix $A$ for this system.

**(b)** Find the inverse matrix of the coefficient matrix found in (a)

**(c)** Solve the system using the inverse matrix $A^{-1}$.

Contents

## Solution.

### (a) Find the coefficient matrix $A$ for this system.

The system can be written as

\[A\mathbf{x}=\mathbf{b},\]
where

\[A=\begin{bmatrix}

2 & 3 & 1 \\

3 &3 &1 \\

2 & 4 & 1

\end{bmatrix}\]
is the coefficient matrix of the system and

\[\mathbf{x}=\begin{bmatrix}

x \\

y \\

z

\end{bmatrix} \text{ and } \mathbf{b}=\begin{bmatrix}

-1 \\

1 \\

-2

\end{bmatrix}.\]

### (b) Find the inverse matrix of the coefficient matrix

We apply the elementary row operations to the augmented matrix $[A\mid I]$, where $I$ is the $3\times 3$ identity matrix.

\begin{align*}

&[A\mid I]= \left[\begin{array}{rrr|rrr}

2 & 3 & 1 & 1 &0 & 0 \\

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

2 & 4 & 1 & 0 & 0 & 1 \\

\end{array} \right]
\xrightarrow{\substack{R_2-R_1\\R_3-R_1}}

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

2 & 3 & 1 & 1 &0 & 0 \\

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

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

\end{array} \right]\\[6pt]
&\xrightarrow[\text{then } R_2 \leftrightarrow R_3]{R_1\leftrightarrow R_2}

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

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

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

2 & 3 & 1 & 1 &0 & 0

\end{array} \right]
\xrightarrow{R_3-2R_1}

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

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

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

0 & 3 & 1 & 3 & -2 & 0

\end{array} \right]\\[6pt]
&\xrightarrow{R_3-3R_2}

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

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

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

0 & 0 & 1 & 6 & -2 & -3

\end{array} \right].

\end{align*}

Now the left $3\times 3$ part has become the identity matrix.

So the matrix $A$ is invertible and the inverse is given by the right $3\times 3$ part.

Hence we obtain

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

-1 & 1 & 0 \\

-1 &0 &1 \\

6 & -2 & -3

\end{bmatrix}.\]

### (c) Solve the system using the inverse matrix $A^{-1}$.

As noted in (a), the system can be written using matrices as

\[A\mathbf{x}=\mathbf{b}.\]
Multiplying by the inverse $A^{-1}$ on the left, we have

\begin{align*}

A^{-1}A\mathbf{x}=A^{-1}\mathbf{b}\\

I\mathbf{x}=A^{-1}\mathbf{b}\\

\mathbf{x}=A^{-1}\mathbf{b}.

\end{align*}

Therefore the solution $\mathbf{x}$ of the system is given by

\begin{align*}

\mathbf{x}&=A^{-1}\mathbf{b}\\

&=\begin{bmatrix}

-1 & 1 & 0 \\

-1 &0 &1 \\

6 & -2 & -3

\end{bmatrix}\begin{bmatrix}

-1 \\

1 \\

-2

\end{bmatrix}\\[6pt]
&=\begin{bmatrix}

2 \\

-1 \\

-2

\end{bmatrix}.

\end{align*}

Thus, the solution of the system is

\[x=2, y=-1, z=-2.\]

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