Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have performed Gaussian elimination on this matrix to reduce it to

$$ \left[ \begin{array}{@{}ccc|c@{}} -3&-1&2 & 1 \\ 0& \frac{-5}{3}& \frac{10}{3} & \frac{8}{3} \\ 0&0& a+2 & b + \frac{6}{5} \\ \end{array} \right] $$

I thought that setting $a$ equal to $-2$ and having $b$ not equal to $-\frac{6}{5}$ would be the answer to this problem, but it apparently isn't. Could someone please explain why?

share|cite|improve this question
Yes, you did, thank you. – user1709173 Feb 8 '13 at 21:14
How about using Cramer's rule? – Lazar Ljubenović Feb 8 '13 at 21:17
Tags go to the tags, not the title. – Asaf Karagila Feb 8 '13 at 21:19
up vote 2 down vote accepted

Given a system of linear equations represented by the matrix equation: $\mathbf{A}\vec{x}=\vec{b}$, there is no unique set of solutions for $\det{\mathbf{A}}=0$.

Therefore, in your case:

$$\begin{bmatrix}-3 & -1 & 2 \\ 0 & -\frac{5}{3} & \frac{10}{3} \\ 0 & 0 & a+2\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}=\begin{bmatrix}1 \\ \frac{8}{3} \\ b+\frac{6}{5}\end{bmatrix}$$

So we are interested in the case when:

$$\begin{vmatrix}-3 & -1 & 2 \\ 0 & -\frac{5}{3} & \frac{10}{3} \\ 0 & 0 & a+2\end{vmatrix}=0 \implies (5a+10)=0\implies a =-\frac{10}{5}=-2$$

share|cite|improve this answer
Edit: As a follow-up, could I ask when the matrix will have no solution? Will it be when the bottom row is entirely 0s? – user1709173 Feb 8 '13 at 21:14
So the asnwer to the question is? – 1015 Feb 8 '13 at 21:28

there is no solution when the matrix is $\textbf{inconsistent}$. This means you will have a zero row in your reduced matrix corresponding to a non-zero entry of the desired solution eg.

$$ \left[ \begin{array}{@{}ccc|c@{}} -3&-1&2 & 1 \\ 0& \frac{-5}{3}& \frac{10}{3} & \frac{8}{3} \\ 0&0& 0 & \text{any non-zero} \\ \end{array} \right] $$

this is because the third row would imply $0*x+0*y+0*z = 0 = c \ne 0$ which is obviously false

share|cite|improve this answer

Answering the title of the post: when the rank of the augmented matrix is different (greather than) from the rank of the matrix of coefficients (Rouchè-Capelli Theorem).

The augmented matrix is the matrix that you obtain when you append the column $\vec b$ to $A$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.