3

I have the set of equations $$x+y-z=-5$$ $$x-y-z=-3$$ $$x+y-z=0$$ I'd just like to point out here that equation one and three are basically the same - just set equal to two different numbers. I don't know if that automatically means there are no solutions or not. Then, I put this into matrix form and got $$\begin{bmatrix} 1&1&-1\\1&-1&-1\\1&1&-1 \end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-5\\-3\\0\end{bmatrix}$$ (I'll refer to that first matrix as $A$.) I found the determinant of $A$ and it is equal to zero, meaning there is no inverse. I got really stuck here.

I know that to solve this has to do with row echelon form, null space, and column space. I put $A$ into row echelon form and got $$\begin{bmatrix}1&0&-1\\0&1&0\\0&0&0\end{bmatrix}$$ but translating this back out gave $$x-z=-5$$ $$y=-3$$ $$0=0$$ Trying to solve the first equation gave conflicting answers.

So, I guess I have three main questions here:

  1. How do you find the solution here (or figure out that there isn't a solution)?
  2. How do you calculate null space/column space? (Please note I've googled this and I just can't get it.)
  3. How does null space/column space relate to finding the solution here, if at all?

Thanks! Please keep in mind that I'm in 8th grade trying to figure this out, so the simpler the answer, the better. I am willing to try to figure out stuff, though. Any answers would be appreciated.

Edit: Yes, this particular system is unsolvable (thanks to Jack D'Aurizio and others), but I kind of wanted to know how to find a general way to calculate solutions or the absence of solutions when the determinant is zero and the matrix has no inverse. In other words, mainly questions two and three now.

auden
  • 1,842
  • 3
    Simply, the first and third equation are incompatible, hence the solutions set is empty. – Jack D'Aurizio Sep 20 '16 at 01:13
  • $1)$ In general you'll want to row reduce the augmented matrix. But in this specific case you could just make the substitution $t=x+y-z$ (unnecessary step, but it makes your conclusion a little more obvious). Then your first and third equations reduce to $t=-5$ and $t=0$ which are of course impossible to solve simultaneously. –  Sep 20 '16 at 01:14
  • You have an inconsistent set of equations. $x+y-z$ cannot equal two different things. – Alexis Olson Sep 20 '16 at 01:14
  • @JackD'Aurizio, okay, that makes sense. Thank you. However, I do want a bit more general of a solution (i.e., null space, or whatever). – auden Sep 20 '16 at 01:14
  • Look at the determinate of a square matrix to see if it is invertible (aka has a unique solution). If the determinate is null, then there is no solution. – mattapow Sep 20 '16 at 01:16
  • 1
    @user157116, indeed, but there can be solutions even if the matrix is not invertible. – auden Sep 20 '16 at 01:16
  • So you want an algorithmic way for stating the obvious? Ok, consider the row-echelon form of your matrix to get an equivalent system, in which one equation is $0=5$, clearly impossible. – Jack D'Aurizio Sep 20 '16 at 01:16
  • I deleted my comment. I missed the first and third row are the same. See Jack's response. – Phillip Hamilton Sep 20 '16 at 01:17
  • 2
    In the case of $det(A)=0$ it depends on the the vector term on the right hand as to whether there will be 0 or infinitely many solutions. See a similar question – mattapow Sep 20 '16 at 01:24

3 Answers3

4

In $Ax=b$ form, the will be at least one solution if and only if $b$ is in the column space of $A$.

If $A$ is a square matrix, there is a unique solution if and only if $\det(A) \ne 0$.

Putting these tests together we have for all square matrices $A$, $Ax=b$ has

  • no solution if $b$ is not in the column space of $A$.
  • a unique solution if $\det(A) \ne 0$.
  • infinitely many solutions if $b$ is in the column space of $A$ but $\det(A) = 0$.

A more general but also slightly more tedious (sometimes) method that will work for non-square matrices is to row reduce the augmented matrix $[A\mid b]$. If you get

  • a row with $[0\mid a]$ where the $0$ represents a row of zeros and $a\ne 0$, then there is no solution.
  • a pivot in each column and no rows with $[0\mid a]$ where $a\ne 0$, then there is a unique solution.
  • at least one column without a pivot and no rows with $[0\mid a]$ where $a\ne 0$, then there are infinitely many solutions.
  • Thank you so much! This is exactly what I needed! How then, would you calculate column space? – auden Sep 20 '16 at 01:30
  • The column space is the span of the columns, considered as vectors. I don't know what you mean by "calculate [the] column space." –  Sep 20 '16 at 01:34
  • Perhaps that was a poor choice of words. I mean how do you figure out if $b$ is in the column space of $A$? – auden Sep 20 '16 at 01:39
  • There are several methods. Some options include: do it the second way I list in my answer, take the exterior product of each column of $A$ and $b$ and check if its zero, or sometimes you can just look at it for a bit and see that $b$ is or couldn't possibly be a linear combination of the columns of $A$. –  Sep 20 '16 at 01:40
2

To turn Jack D'Aurizio's comment into an answer:

The first equation says that

$$x+y-z=-5$$

but the second equation says that

$$x+y-z=0$$

Since these cannot both be true (as it would imply $5=0$), there are no solutions. We may also note that if the determinant of the matrix $A$ of coefficients is $0$, that means that one expression can be expressed as a linear combination of the other two, and thus the system either has infinitely many solutions or no solutions.

Carl Schildkraut
  • 37,483
2

You did not put the linear system into row-echelon form correctly.

\begin{bmatrix}1&0&-1\\0&1&0\\0&0&0\end{bmatrix}

As you went about your work you dropped the solutions. Row-echelon form arises from the augmented matrix which includes \begin{bmatrix}-5\\-3\\0\end{bmatrix}

So the correct form begins from \begin{bmatrix} 1&1&-1&-5\\1&-1&-1&-3\\1&1&-1&0 \end{bmatrix}

You should end with a statement like $0=5$ (see Jack's comment)

Since you're also in 8th grade, here are some concepts:

1) A system of linear equations has either no solution, a unique solution, or infinitely many solutions.

2) For square matrix $A$, $det(A) \ne 0$ iff the system has a unique solution. Always.

3) If you have a statement like $x+y=2$ left after row-echelon form, this means that your system has infinitely many solutions. Naively, it's a line so how many points make up a line?

Phillip Hamilton
  • 1,191
  • Ah, that last bit explaining why a system would have infinitely many solutions makes a lot of sense. Good to know about the first part; thanks for pointing that out! – auden Sep 20 '16 at 01:38
  • No problem. If you're interested, buy a book specific to linear algebra. Something like Larson / Edwards / Falvo. The entire first couple chapters will just walk through the proofs for how elementary matrices, determinants, unique solutions, invertible matrices, and linear independence all play into one another. – Phillip Hamilton Sep 20 '16 at 01:44