# Should be easy question on column space of a matrix

## Homework Statement

So the actual problem "Find the value of a for which the following system of linear equations has a solution"

2x + 4y + z = a
-4x -7y + 0 = 1
0 -1y -2z = 1

## The Attempt at a Solution

I thought one approach was to find a basis for the corresponding matrix and see what value of a would make that vector in the space formed by the basis. That is, see what value of a would make (a, 1, 1) in the range of the matrix.

But when I row reduce I get

2 4 1
0 1 2
0 0 0

So I want to say the 1st and 2cnd columns form a basis -> (2, -4, 0) and (4, -7, -1). But then some linear combination of these should equal (1, 0, -2). However, the first components of the basis appear to never be able to combine linearly to 1. That is there are no integers x,y such that 2x + 4y = 1 since -> x + 2y = 1/2.....? I feel like I must be making a really trivial mistake?!?!

## Answers and Replies

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gabbagabbahey
Homework Helper
Gold Member
Maybe you should show how you are row reducing. But first, ask yourself if there is any reason to think that there are any values of $a$ that would make it so there are no solutions.

HallsofIvy
Homework Helper

## Homework Statement

So the actual problem "Find the value of a for which the following system of linear equations has a solution"

2x + 4y + z = a
-4x -7y + 0 = 1
0 -1y -2z = 1

## The Attempt at a Solution

I thought one approach was to find a basis for the corresponding matrix and see what value of a would make that vector in the space formed by the basis. That is, see what value of a would make (a, 1, 1) in the range of the matrix.

But when I row reduce I get

2 4 1
0 1 2
0 0 0
Row reduce the augmented matrix
$$\begin{bmatrix}2 & 4 & 1 & a \\ -4 & -7 & 0 & 1 \\ 0 & -1 & -2 & 1\end{bmatrix}$$
instead. That will tell you immediately what a should be.

So I want to say the 1st and 2cnd columns form a basis -> (2, -4, 0) and (4, -7, -1). But then some linear combination of these should equal (1, 0, -2). However, the first components of the basis appear to never be able to combine linearly to 1. That is there are no integers x,y such that 2x + 4y = 1 since -> x + 2y = 1/2.....? I feel like I must be making a really trivial mistake?!?!
So you want x and y such that 2x+ 4y= 1, -4x- 7y= 0, and -y= -2? From the last equation, y= 2 so the first two equations becom 2x+ 8= 1 and -4x- 14= 0. Both of thos equations reduce to 2x= -7.

Hmm, I was being really dumb and thinking that a scalar multiple had to be an integer, since as I was saying in the OP there are no integer solutions to 2x = -7. Thanks!

uart
Hi Fractal. Did you yet solve this problem? x = -7/2 is not a solution, so it doesn't look like you have really understood this one yet.

Row reduce the augmented matrix that Halls' suggested.

- For all values of "a" except one, there are no solutions.

- For this one particular value of "a" there are infinite solutions, which you should express parametrically.

Thanks for the look over! So the x = -7/2 bit was just in reference to expressing the 3rd column as a linear combination of the other two (-7/2 times the first plus 2 times the 2cnd gives the 3rd). The motivation for my original post was motivated by getting weirded out by thinking that my row reduction wasn't giving a basis for the range. I think I was able to get it from there, but I don't have my work anymore. Thanks again!

uart
Thanks for the look over! So the x = -7/2 bit was just in reference to expressing the 3rd column as a linear combination of the other two (-7/2 times the first plus 2 times the 2cnd gives the 3rd). The motivation for my original post was motivated by getting weirded out by thinking that my row reduction wasn't giving a basis for the range. I think I was able to get it from there, but I don't have my work anymore. Thanks again!
Ok, I agree that "-7/2 times the first plus 2 times the 2nd gives the 3rd". But how exactly did that allow you to solve for "a"?

I think the easy way to solve it is to just row reduce the augmented matrix as suggested above. You should end up with something like,

\begin{bmatrix}1 & 0 & -7/2 & -7a/2 - 2 \\ 0 & 1 & 2 & 2a+1 \\ 0 & 0 & 0 & 2a+2\end{bmatrix}
from which you can conclude there are no solutions unless $2a + 2 = 0$.

Then you can subst in this value of "a" to express x,y and z parametrically. Like,

$x = 3/2 + 7t/2$
$y = -1 -2t$
$z = t$

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