1. ## 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

2. 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.

3. It corresponds to 3.

4. That's what it corresponds to after the first elementary row operation. What about after the second?

5. -3

6. 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.

7. $\left(\begin{array}{ccc}0&0&0\\3&1&-4\\4&2&-6\end{array}\left|\begin{array}{c}-2\\2\\5\end{array}\right)$
I attempted to get the augmented part in. I made a mistake in the previous post, is it correct now?

8. 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?

9. 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?

10. 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?

11. 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.

12. You're very welcome. Have a good one!