How to solve for unknown variables in a matrix? 
How do I solve this problem? Ideally, I think $$a^2-4=1$$ is the best option to solve for a I set the entire expression equal to 1. Then to solve for b I just put $b=1$. Is this right? I feel like that's really wrong.
So update I got this

I suppose my answer is just b does not equal 0 and a does not equal +- sqrt 4
 A: There's a fundamental result for this kind of problems:

A non-homogenous linear system $AX=B$ has solutions if & only if the matrix $A$ and the augmented matrix $A|B$ have the same rank. When this condition is satisfied, the common rank is the codimension of the affine subspace of solutions.

Therefore, if $(a^2-4)b\ne 0$, the matrix $A$ has rank $3$, which is also the maximum rank of  the augmented matrix, and there is a unique solution.
If $a=\pm 2$, $A$ has rank $2$, but the augmented matrix may have rank $3$. If it has rank $2$, the set of solutions has dimension $1$.
To determine the solutions, write the augmented matrix in reduced row echelon form. The set of solutions is the last column.
A: This question is asking for ALL pairs of a and b for which the system is consistent, not just one.
You are likely better off trying to consider what makes the system inconsistent. Try doing Gaussian elimination on the matrix, and take good notes along the way. Any time you are dividing by a term containing a or b, determine what value of a or b would make that term zero (thus dividing by zero). These are values where the system is inconsistent.
Hope that helps.
A: Sorry for the late reply.
The system can be written as $$x_{1}+6x_{2}-9x_{3}=9$$
$$(a^2-4)x_{2}+x_{3}=-7$$
$$x_{3}b=7.$$
Clearly the last equation is only consistent if $b\neq0.$ So this will give $x_{3}=\frac{7}{b}.$ Now we substitute this in the second equation. This gives $$(a^2-4)x_{2}=-7-x_{3}=-7-\frac{7}{b}=-\frac{7b+7}{b} (\star)$$
Two cases to consider:

*

*If $a\neq2,-2$ then we have $x_{2}=-\frac{7b+7}{b(a^2-4)}$ and then we can use the first equation to obtain $x_{1}$ and this will give a unique solution.

*What happens if $a=2,-2$? Well in this case we have from $(\star)$ that $$0=-\frac{7b+7}{b}$$
and this forces $b=-1$ so the numerator vanishes. So in this case both equations (2) and (3) are satisfied and substituting into equation (1) we get that $x_{1}=-6x_{2}+9x_{3}+9=-6x_{2}+9(x_{3}+1)=-6x_{2}+9(-7+1)=-6x_{2}-54$ for $x_{3}=\frac{7}{b}=\frac{7}{-1}=-7$ and there are infinitely many solutions (since any $x_{2}$ satisfies the second equation because it vanishes for $a=\pm 2$).
