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I am resolving the following matrix:

 1  2 1|1
-1  4 3|2
 2 -2 a|3

Where I have to find all the values of $a$ so that the system can have a unique solution. Doing elementary row operations I ended up with:

1 2  1 |1
0 1 2/3|1/3
0 0 2+a|4

Where I see that $a$=2. Using this for back substitution: $z$=1, $y$=-1/3 and $x$=-2/3 The problem is that this values does not satisfy the system (last equation of the original matrix)

Can someone explain the right procedure to find all the values of $a$? What I am doing wrong?

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You need to not have a zero appear in the bottom right cell, so $a \neq -2$. For any other value of $a$ you have a unique solution because you can divide that line by $2+a$ –  Ross Millikan Sep 2 '12 at 21:52
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2 Answers 2

up vote 2 down vote accepted

There is a mistake in your Gaussian Elimination. The $1/3$ should be $1/2$.

Start with the original system. Add $r_1 + r_2$ to get $(0, 6, 4 \mid 3)$ in the second row. Divide by $6$ to get $(0, 1, 2/3 \mid 1/2)$.

And to double check: the solution is: $ z = 1, y = -1/6, x = 1/3$.


Wordy update: The system is consistent if there is no row in the reduced echelon form of the augmented system of the form $(0\ 0 \dots 0 \mid b)$. In this case, if $a \neq -2$ the system is consistent. A linear system has free variables, if there is a row in the reduced echelon form of the augmented system that looks like $(0\ 0\dots 0 \mid 0)$. So the system in hand has no free variables. A consistent system with no free variables has a unique solution. (and for the records, a consistent system with $1$ or more free variables has infinitely many solutions.)

Alternatively, and easily: a linear system $Ax = b \neq 0$ has a unique solution if the coefficient matrix $A$ is square and its determinant $\neq 0$. The determinant (up to multiplying by a factor $=6$ from the second row) can be easily read off from the reduced matrix, as the product of the pivots: $1\times 1\times (2+a) = 2+a$. Hence the system has a unique solution if $\det \neq 0 \iff a \neq -2$.

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many many thanks. Did you know a website where there are problems like this? @Jennifer –  Edwardo Sep 2 '12 at 18:07
    
I'm not sure. There are lot of linear algebra websites and books. Try this one: numbertheory.org/book –  user2468 Sep 2 '12 at 18:09
    
ok, thanks. @Jennifer –  Edwardo Sep 2 '12 at 18:12
    
@Edwardo: this doesn't answer the original question, namely what values of $a$ give a unique solution. The solution will depend upon what $a$ is. –  Ross Millikan Sep 2 '12 at 21:54
    
@Edwardo answer updates. There are alternative and equivalent characterizations of the invertibility of a matrix and uniqueness of solution. For example, a square matrix with non-zero pivots is invertible (which is Ross's first comment below your question). Please read e.g. here on Wikipedia. –  user2468 Sep 4 '12 at 14:09
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Simply compute the determinant of $\begin{pmatrix}1 & 2 &1 \\-1 & 4 &3\\2 & -2 & a\end{pmatrix}$, which is $6a+12$, so $a\in(-\infty,-2)\cup(-2,\infty)$

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@Ross Millikan Sorry, fixed it. –  chaohuang Sep 2 '12 at 22:41
    
(this comment for @Edwardo's benefit) To put it differently, the system has a unique solution whenever $\det \neq 0 \iff a \neq -2$. –  user2468 Sep 4 '12 at 13:54
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