0
$\begingroup$

I'm trying to solve an equation system using Gauss-Jordan.

$$\begin {cases} 2x+4y = 6\\ 3x+6y = 5\end {cases}$$

So, first, the augmented matrix:

\begin{bmatrix} 2&4&5\\ 3&6&6\\ \end{bmatrix}

I want to reduce it to the "staggered reduced form" (how is that called?), so I start:

$$-f_1+f_2$$

\begin{bmatrix} 2&4&5\\ 1&2&1\\ \end{bmatrix}

$$-2f_2+f_1$$

\begin{bmatrix} 0&0&3\\ 1&2&1\\ \end{bmatrix}

$$f_1f_2$$

\begin{bmatrix} 1&2&1\\ 0&0&3\\ \end{bmatrix}

Then, I would transform that $3$ into a $1$ and somehow get rid of the $2$ and $1$ from the first row (not sure how) to complete the reduction, yes?

However, look at the last row: $0, 0, 3$. That's to say

$$0x+0y=3$$

Does that mean that the equation system has no solutions?

$\endgroup$
2
  • 5
    $\begingroup$ Your system is equivalent to $6x+12y=18$ and $6x+12y=10$. These cannot be both true at the same time. Your system has no solutions. $\endgroup$
    – JRN
    Apr 2, 2014 at 4:55
  • $\begingroup$ You have it almost. You just need to realize that there are no real numbers $x,y$ such that $0x+0y=3$. Therefore this no solution this system. $\endgroup$
    – user60887
    Apr 2, 2014 at 5:08

4 Answers 4

4
$\begingroup$

Divide the original equations by $2$ and $3$ respectively. Then they both specify the value of $x+2y$, but the values differ.

$\endgroup$
2
$\begingroup$

There are indeed no solutions. This makes sense: the two equations describe lines in $\mathbb{R^2}$. Note that they are parallel, so there can be no intersection points.

$\endgroup$
1
$\begingroup$

The lines are parallel, so there are no solutions.

The lines are: $y = -\frac{1}{2}x + \frac{3}{2}$, and $y = -\frac{1}{2}x + \frac{5}{6}$. Graph them and you will see they never intersect.

$\endgroup$
1
$\begingroup$

If you've had determinants, you find that the determinant of the coefficient matrix is $ \ \left| \begin{array}{cc} 2 & 4 \\ 3 & 6 \end{array} \right| \ = \ 0 \ , $ which is already a sign that you have either a dependent or inconsistent system. Row-reduction or any of the other techniques given by the other posters then establish that "Inconsistency, it is!"

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .