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I am doing practice questions in my book and I came upon this True/False question:

If $\det(A) = 0$, then the linear system $Ax=b$, $b\neq 0$, has no solution.

The book is saying that the answer is false. But why is that? I thought the answer is true because of something like this

$ Ax = b$ $$\left(\begin{array}{ccc|c}1&0&0&1\\0&0&0&2\\0&0&1&3\end{array}\right)$$

When a matrix has its rref taken, the resulting matrix, when the determinant is zero, would always have a zero in its diagonal, right? This would result in a matrix with no solution because row 2 is impossible. Am I misunderstanding the question somehow? I am also confused by this question because I am not sure how augmented matrices work with square matrices because you can only find the determinant of a square matrix. Can someone please explain why the answer is false?

Bman72
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user2989964
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  • Note that the converse statement is true: if $Ax=b$ has no solution (and $A$ is square) then certainly $\det(A)=0$. But the book is right: the statement in the question is not true. – Marc van Leeuwen Oct 15 '14 at 12:24

4 Answers4

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The statement is not true. For it to be true, it would have to be true for any combination of $A$ and $b$ where $\det(A)=0$ and $b\neq 0$, but that is not the case, since the equation

$$\begin{bmatrix}1 & 0\\0&0\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}1\\0\end{bmatrix}$$ clearly has a solution

5xum
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  • Hello, I don't understand what you are saying here. b cannot equal zero, but you have a b = 0 in the second row. – user2989964 Oct 15 '14 at 06:46
  • @user2989964 Yes, but that does not mean $b$ is zero. The statement $b\neq 0$ means that $b$ is not equal to the all zero vector, not that all of the components of $b$ are zero. – 5xum Oct 15 '14 at 06:47
  • Oh, that's what always confused me. So is this similar to the 0 vector? Where it can also have different values other than zero? – user2989964 Oct 15 '14 at 06:50
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    When they state $b$ equals $0$, they mean that every number in $b$ is $0$. $(1\ \ 0)$ is not $0$. Generally speaking, when we speak about something being equal to $0$, we talk about the neutral element, which is the only element $e$ such that for all $a$ (here, for all vectors), $a+e=a$. – Traklon Oct 15 '14 at 06:50
  • @user2989964 $0$ is used to mark the zero vector in vector spaces, meaning that in $\mathbb R^n$, a vector is not equal to $0$ if at least one of its components is not zero. – 5xum Oct 15 '14 at 06:51
  • @Traklon But wouldn't a + e = a mean that all values of the elements in e would have to equal to zero? Doesn't this violate the b cannot equal zero clause? – user2989964 Oct 15 '14 at 06:59
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    @user2989964 That is exactly the case. That is why only the vector with all values set to $0$ is equal to the $0$ vector. The vector $(1,0)$ I used is not the neutral element for addition, so it satisfies the demand "$b\neq 0$." – 5xum Oct 15 '14 at 07:02
  • @5xum Thank you for being so patient with me, but I'm still a bit confused by this. When is b = 0? b in the question is a vector, right? How does the vector(1,0) exactly satisfy the demand? – user2989964 Oct 15 '14 at 07:10
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    @user2989964 $b=(1,0)$ satisfies the demand $b\neq 0$, not $b=0$. I have given you an example of a matrix $A$ and vector $b$ such that $\det A=0$ is true, $b\neq 0$ is also true, and $Ax=b$ has a solution. – 5xum Oct 15 '14 at 07:12
  • @5xum I was confusing myself the whole time. Thank you for your help! – user2989964 Oct 15 '14 at 07:18
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In your example, if you choose $b$ equal to one of the colums of $A$, then the system $Ax=b$ is solvable.

If $\det(A)=0$ then the system $Ax=b$ is never uniquely solvable, though. So you have the alternatives: no solution or many solutions (depends on the underlying field).

daw
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Excluding the already mentioned cases where we had one line filled with zeros in the complete matrix, we don't need to have a zero line in order to have $\textrm{det}(A)=0$. Just take the cases where one line is a multiple from the other, for example: $$ \begin{bmatrix}a & b\\ka&kb\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}c\\kc\end{bmatrix} $$ This is a system which has $\textrm{det}(A)=0$ and also has infinite solutions as long $a$ and $b$ are not both zeros (except in the case $c$ is also zero).

Décio Soares
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  • Correct me if I'm wrong but wouldn't taking the rref (in the process of looking for the solution) result in having a zero in the diagonal? And in the case of an rref, that would mean a line of zeroes. – user2989964 Oct 15 '14 at 07:03
  • @user2989964 So what? The system $Ax=b$ given in this answer has a solution, thus proving the statement wrong. – 5xum Oct 15 '14 at 07:14
  • I need to change the way I think about these. Thanks! – user2989964 Oct 15 '14 at 07:28
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If $\det A=0$, then there are indeed vectors $b$ for which $Ax=b$ does not have a solution.

But, there are also values of $b$ for which it DOES have a solution. Say $a$ is an arbitrary vector and $b=Aa$. Then clearly the system $Aa=b$ possesses at least one solution, namely $x=a$.

Yiorgos S. Smyrlis
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