# Augmented matrix

Printable View

• July 13th 2010, 06:32 PM
arze
Augmented matrix
Determine whether there is a unique solution, no solution, or an infinite set of solutions with one or two parameters.
$x+y-2z=1$
$3x+y-4z=2$
$4x+2y-6z=5$

I'm supposed to use an augmented matrix to solve this.
$\left(\begin{array}{ccc}1 & 1 & -2\\3 & 1 & -4\\4 & 2 & -6\end{array}\right)$ (I don't know how to make the augmented part of the matrix.)
$r_1+r_2\rightarrow\left(\begin{array}{ccc}4&2&-6\\3&1&-4\\4&2&-6\end{array}\right)$
$r_1-r_3\rightarrow\left(\begin{array}{ccc}0&0&0\\3&1&-4\\4&2&-6\end{array}\right)$
When reducing the matrix, I get a row of zeros, so this means that there is a line of intersection for the planes, so an infinite set of solutions dependent on one parameter.
But the answer says that there is no solution.

I found the line to be: $x=\lambda,y=-\lambda,z=\lambda-\frac{1}{2}$

Thanks!

PS. I hope this is the right section. Sorry Mods if its not
• July 13th 2010, 06:41 PM
Ackbeet
The key is in the augmented matrix. When you do elementary row operations, you also have to do them on the solution vector. When you do that, notice what number corresponds to the row with all zeros in it.
• July 13th 2010, 06:43 PM
arze
It corresponds to 3.
• July 13th 2010, 06:44 PM
Ackbeet
That's what it corresponds to after the first elementary row operation. What about after the second?
• July 13th 2010, 06:50 PM
arze
-3
• July 13th 2010, 06:51 PM
Ackbeet
Try again. You have to subtract the third row element of the solution vector, not the third row element of the third column of the coefficient matrix.
• July 13th 2010, 07:00 PM
arze
$\left(\begin{array}{ccc}0&0&0\\3&1&-4\\4&2&-6\end{array}\left|\begin{array}{c}-2\\2\\5\end{array}\right)$
I attempted to get the augmented part in. I made a mistake in the previous post, is it correct now?
• July 13th 2010, 07:02 PM
Ackbeet
Excellent: that looks correct to me. Think about that top row. What does that top row correspond to in terms of an equation containing x, y, and z?
• July 13th 2010, 07:06 PM
arze
0x+0y+0z=-2? so it corresponds to the origin? Or does it mean that since 0 is not equal to -2 then there is no solution?
• July 13th 2010, 07:08 PM
Ackbeet
The latter is true. There is no way that 0x + 0y + 0z = -2. There are no values of x, y, and z that can make that true. Hence, there is no solution. This is typically what a linear system looks like that has no solution. Make sense?
• July 13th 2010, 07:13 PM
arze
Thank you! Now I realize where I was wrong, I thought that as long as the first three elements in the row were zero then the solution was a line.
• July 13th 2010, 07:14 PM
Ackbeet
You're very welcome. Have a good one!