We’re continuing to consider Gaussian elimination method. Earlier we’ve prepared several tutorials covering theoretical background and examples including application of matrix representation. Yet, all those examples suggested systems of linear equations which have a solution. That’s not always so, though. In your homework or linear algebra assignment you can face different situations. In fact, there are three possibilities: system of linear equations can have either unique solution, or infinitely many solutions, or no solution at all. We’ll discuss the last case in this section.

Let’s solve the following system of linear algebraic equations:

$$ \left\{ \begin{aligned}2x_1-x_2+3x_3=4\\-3x_1+2x_2+x_3=5\\-4x_1+2x_2-6x_3=1 \end{aligned}\right.$$

Here’s video version of this tutorial:

As you know, we can either work with system in equations form (as written above) or use matrix representation of the system. Let’s choose the second approach and consider matrix of coefficients for the given system:

$$A=\begin {pmatrix} 2&-1&3 \\-3& 2&1 \\-4& 2&-6 \end {pmatrix}, \vec{b}=\begin{pmatrix}4\\5\\1\end{pmatrix},\vec{x}=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}, \tilde{A}=\begin{pmatrix}2&-1&3&|&4 \\-3& 2&1&|&5 \\-4& 2&-6 &|&1 \end{pmatrix}$$

In these terms, the given system is represented in the following way:

$$A \vec{x}=\vec{b}$$

We intend to obtain our system in triangular (or echelon) form. Recall that we can swap rows of matrix, add or subtract them, multiply or divide by real non-zero number. Also note, that as we have non-zero right part of the system, i.e. our system is not homogeneous, we ought to do all the necessary transformations with the augmented matrix @$\tilde{A}@$ containing right parts of the equations and not @$A@$. It’s a common mistake to leave @$\vec{b}@$out of the consideration, so don’t forget about it when doing your algebra homework. Thus, we’re working with this matrix:

$$\tilde{A}=\begin{pmatrix}2&-1&3&|&4 \\-3& 2&1&|&5 \\-4& 2&-6 &|&1 \end{pmatrix}$$

First, let’s divide the first row by @$2@$. Other rows remain intact:

$$\begin{pmatrix}1&-\frac{1}{2}&\frac{3}{2}&|&2 \\-3& 2&1&|&5 \\-4& 2&-6 &|&1 \end{pmatrix}$$

Now we want to eliminate the first unknown @$x_1@$ from all the equations except the first. For that, we firstly subtract the first row multiplied by @$(-3)@$ from the second:

$$\begin{pmatrix}1&-\frac{1}{2}&\frac{3}{2}&|&2 \\0& \frac{1}{2}&\frac{11}{2}&|&11 \\-4& 2&-6 &|&1 \end{pmatrix}$$

Now, we multiply the second row by $2$ (in order to obtain coefficient @$1@$ in front of @$x_2@$):

$$\begin{pmatrix}1&-\frac{1}{2}&\frac{3}{2}&|&2 \\0& 1&\frac11&|&22 \\-4& 2&-6 &|&1 \end{pmatrix}$$

Also we subtract the first row times @$(-4)@$ from the third row:

$$\begin{pmatrix}1&-\frac{1}{2}&\frac{3}{2}&|&2 \\0& 1&\frac11&|&22 \\0&0&0 &|&9 \end{pmatrix}$$

Ok, we’ve successfully eliminated @$x_1@$ from the second and the third equation. Actually, there’s no need to proceed further. Let’s have a closer look at the third row of our matrix (which stands for the third equation). We obtained the equation @$0=9@$ which is obviously wrong. This means that given system has no solutions. Such systems are called inconsistent.

In the next section we’ll discuss the case when system of linear algebraic equations has infinitely many solutions. Generally,if you’re doing Gaussian elimination, you need to be attentive and check the system at each step. Such approach allows to avoid unnecessary calculations and saves your time, as in the example we’ve just considered.

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