solutions of a linear equation system

The following matrix is given over $\mathbb R$

A=$\begin{pmatrix} 1 & -1 & -1 \\ 1 & 0 & a \\ 1 & a & 0 \end{pmatrix}$

The linear equation System $Ax=\begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}$ has exactly one solution for all $a$ except $a=-2$ which has no solution. Is it right?

-

Almost correct.

• At $a = -2$, we have no solution, so you are correct on that.
• At $a = 0$, our augmented RREF matrix gives:

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

This means we have infinite solutions as:

$x = 1, y = -1 + z$, where $z$ is a free variable.

• At all other $a$, we have a single unique solution.
-
RREF, solutions or lack thereof for what values of $a$...? (Popular assigned problem-type presently!) +1 for the assistance! – amWhy Oct 26 '13 at 0:55
@amWhy: There does seem to be an awful rash of those lately! Maybe instructors are running out of other problems and they can add variables all over the matrix! :-) – Amzoti Oct 26 '13 at 0:57
this is the reduced form: $$\begin{bmatrix} 1 & 0 & 0 & 2(a+1)/(a+2) \\ 0 & 1 & 0 & 1/(-a-2) \\ 0 & 0 & 1 & 1/(-a-2) \end{bmatrix}$$ a=-2 will be not defined (division by 0) and therefore i've to handle this case separately. But it seems for all other cases $Rank(A)=Rank(A´)=3$ (exactly one solution). But for a=0 there are of course infinite solutions. I am unable to see this regarding the reduced form !? – Evgeniya Oct 26 '13 at 13:50
Sometimes, you need to look at several things and you always start with the determinant which is $-a(a+2)$. That shows the two points of interest $0$ and $-2$. Then, I do the general RREF and then handle the special cases as sometimes those are not always evident from the reduced form. Regards – Amzoti Oct 26 '13 at 13:58
my current basic math course will not give me the term "determinant" :/ – Evgeniya Oct 26 '13 at 14:58

We see that $det(A)=-a(a+2)$. A matrix $A$ is invertible if and only if $det(A)\neq 0$. When $a=0$ there exist an infinite number of solutions. When $a=-2$ there exists no solution.

-
if a=0 then $$A=\begin{pmatrix} 1 & -1 & -1\\1 & 0 & 0\\1 & 0 & 0 \end{pmatrix}$$ the augmented matrix will be $$B=\begin{pmatrix} 1 & -1 & -1 & | & 2\\1 & 0 & 0 & | & 1\\1 & 0 & 0 & | & 1\end{pmatrix}$$ the row reduced form: $$\begin{pmatrix} 1 & 0 & 0 & | & 1\\0 & 1 & 1 & | & -1\\0 & 0 & 0 & | & 0\end{pmatrix}$$ $Rank(A)=Rank(B)=2 \ne n$ => exactly one solution ??? – Evgeniya Oct 25 '13 at 23:52
Amzoti cleared the issue. – 1233dfv Oct 26 '13 at 0:04
We would have $x=1$ and $y+z=-1$ which implies an infinite number of solutions. – 1233dfv Oct 26 '13 at 0:07
i am wrong: Rank(A)=Rank(B)=2≠n => infinite solutions – Evgeniya Oct 26 '13 at 0:11
No worries. You had me looking through my Linear Algebra - Friedberg textbook to make sure that $a=0$ would imply an infinite number of solutions. – 1233dfv Oct 26 '13 at 0:21