for which values of "a" does the system have one, zero, or infinite solutions? [SOLVED]

Question

Question:

I'm having trouble solving this, I've looked everywhere for a similar question but this one doesn't seem to follow the rules of other questions. after Gauss-Jordan elimination I end up with.

$$ \left[ \begin{array}{ccc|c} 1&2&1&2\\ 0&-6&1&-3\\0&0&2-a^2&a-2 \end{array} \right] $$

the issue I'm really having is finding when the system has infinite solutions. Please help

[EDIT] This has been solved, there are NO values of "a" for which the system has infinite solutions.

Answer 1

The deteminant which is equal to$$-6 (2-a^2) $$ must be zero. this gives $a=\pm \sqrt{2} $.

but in this case $a-2\neq 0$.

we conclude that

if $a\neq \pm \sqrt{2} $, there is one solution

if $a=\pm \sqrt {2} $, there is no solution.

Answer 2

Let $A=\begin{pmatrix} 1 &2 &1 \\ 2&-2 &3 \\ 1 &2 & 3-a^{2} \end{pmatrix}$ and $B=\begin{pmatrix} 2\\ 1\\ a \end{pmatrix}$.

$detA=6(a^2-2)$.

Case 1: $det(A) \neq 0$
Then the system has unique solution $\Leftrightarrow a\in \mathbb{R}-\left \{ \pm \sqrt{2} \right \}$

Case 2: $det(A)=0 \Leftrightarrow a=\left \{\pm \sqrt{2} \right \}$
We have $d=\begin{vmatrix} 1 &2 \\ 2&-2 \end{vmatrix}=-6 \neq 0$, so $rank(A)=2$
But $d_{char}=\begin{vmatrix} 1 &2 &2 \\ 2&-2 & 1\\ 1&2 &\pm \sqrt{2} \end{vmatrix} \neq 0$, hence the system has no solution

Answer 3

Note: JMoravitz's comment on the row echelon form gave all necessary clues for 1. and 3.

1. There is no solution if there is an inconsistent equation, i.e. all zero coefficents for the unknowns left vs. non-zero inhomogenity right. Only possible in the last row, iff $$ 2-a^2 = 0 \wedge a-2 \ne 0 \iff \\ a = \pm\sqrt{2} \wedge a \ne 2 $$ So this is the case for $a =\pm\sqrt{2}$.

2. There is one solution if the matrix has full rank, which is the case for $$ 2-a^2\ne 0 \iff \\ a \ne \pm\sqrt{2} $$

3. There are infinite many solutions if there is at least one empty row with zero inhomogenity to be consistent. This is only possible for the last row, iff $$ 2-a^2 = 0 \wedge a-2=0 \iff \\ a = \pm\sqrt{2} \wedge a=2 $$ which is not possible.