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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?

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Can you please check that I transcribed your problem correctly? Regards –  Amzoti Feb 8 '13 at 21:12
    
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

3 Answers 3

up vote 1 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:

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

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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
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@Shaktal: Where did the $5$ come from? Regards –  Amzoti Feb 8 '13 at 21:19
    
So the asnwer to the question is? –  1015 Feb 8 '13 at 21:28
    
@Shaktal: Of course you could have also used the last row with $a+2$. Regards –  Amzoti Feb 8 '13 at 21:29

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

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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 $A$.

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