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Rico Suave's question at Yahoo! Answers regarding using the Gauss-Jordan method

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MarkFL

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Feb 24, 2012
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Here is the question:

Solve the system of equation with Jordan Gauss method please?

2x+4y+z=1
x-2y-3z=2
x+y-z=-1

Find x= y= and z=

Also, find if this equation has 1)only one solution, 2)no solution or 3) an infinite number of solutions. please explain how you know the number of solutions for this problem.

Thank you for your help.
I have posted a link there to this thread so the OP can view my work.
 
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MarkFL

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Feb 24, 2012
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Hello Rico Suave,

We are given the system of equations:

\(\displaystyle 2x+4y+z=1 \)

\(\displaystyle x-2y-3z=2 \)

\(\displaystyle x+y-z=-1\)

The first thing we want to do is construct the augmented matrix using the coefficients:

\(\displaystyle \left[\begin{array}{ccc|c} 2&4&1&1\\ 1&-2&-3&2\\ 1&1&-1&-1 \end{array} \right]\)

Now, what we want to do is through a series of operations, obtain an augmented matrix of the form:

\(\displaystyle \left[\begin{array}{ccc|c} 1&0&0&x\\ 0&1&0&y\\ 0&0&1&z \end{array} \right]\)

As our first operation, we see that if we divide row 1 by 2, the operation denoted by \(\displaystyle \frac{R_1}{2}\), we will obtain the desired first element of 1 in this row:

\(\displaystyle \left[\begin{array}{ccc|c} 1&2&\frac{1}{2}&\frac{1}{2}\\ 1&-2&-3&2\\ 1&1&-1&-1 \end{array} \right]\)

Next, we observe that if we subtract the first row from the second and third rows, the operations \(\displaystyle R_2-R_1\) and \(\displaystyle R_3-R_1\), we obtain:

\(\displaystyle \left[\begin{array}{ccc|c} 1&2&\frac{1}{2}&\frac{1}{2}\\ 0&-4&-\frac{7}{2}&\frac{3}{2}\\ 0&-1&-\frac{3}{2}&-\frac{3}{2} \end{array} \right]\)

Next, we look at the second element in the second row, and observe that if we divide this row by $-4$, we will get a $1$ there as we desire, so our next operation is \(\displaystyle -\frac{R_2}{4}\):

\(\displaystyle \left[\begin{array}{ccc|c} 1&2&\frac{1}{2}&\frac{1}{2}\\ 0&1&\frac{7}{8}&-\frac{3}{8}\\ 0&-1&-\frac{3}{2}&-\frac{3}{2} \end{array} \right]\)

The next operations are then \(\displaystyle R_1-2R_2\) and \(\displaystyle R_3+R_2\) and we get:

\(\displaystyle \left[\begin{array}{ccc|c} 1&0&-\frac{5}{4}&\frac{5}{4}\\ 0&1&\frac{7}{8}&-\frac{3}{8}\\ 0&0&-\frac{5}{8}&-\frac{15}{8} \end{array} \right]\)

The next operation is \(\displaystyle -\frac{8R_3}{5}\)

\(\displaystyle \left[\begin{array}{ccc|c} 1&0&-\frac{5}{4}&\frac{5}{4}\\ 0&1&\frac{7}{8}&-\frac{3}{8}\\ 0&0&1&3 \end{array} \right]\)

And our final two operations are \(\displaystyle R_1+\frac{5R_3}{4}\) and \(\displaystyle R_2-\frac{7R_3}{8}\) so that we find:

\(\displaystyle \left[\begin{array}{ccc|c} 1&0&0&5\\ 0&1&0&-3\\ 0&0&1&3 \end{array} \right]\)

Thus, we may conclude that the given system of equations has the unique solution:

\(\displaystyle (x,y,z)=(5,-3,3)\)

If this system of equation had no solutions, referred to as an inconsistent system, then we would have, at some point during our series of operations, obtained a row of the form:

\(\displaystyle \left[\begin{array}{ccc|c} 0&0&0&a \end{array} \right]\)

where \(\displaystyle a\ne0\)

This row would be equivalent to the statement \(\displaystyle 0=a\), but this is never true, so it is impossible for there to be a solution.

If this system of equation had an infinite number of solutions, referred to as a dependent system, then we would have obtained a row of the form:

\(\displaystyle \left[\begin{array}{ccc|c} 0&0&0&0 \end{array} \right]\)

This would occur if one row is simply a multiple of another. This row is equivalent to the statement $0=0$ which is always true, and so an infinite number of solutions would be the result.

But, neither of these occurred, because the system we were given is independent, meaning it has one unique solution, which we found.