For what values of "a" would the system have an infinite solutions?

Question

If consider $$\left[ \begin{array}{ccc|c} 2&-1&3&b_1\\ a&1&0&b_2\\2&1&1&b_3 \end{array} \right]$$

a) If $$b_1,b_2,b_3$$ are not three zero. For what values of "a" would the system have an infinite solutions and which would have to satisfy b1,b2 and b3?

b)Suppose that $$b_1=b_2=b_3=0$$ and determine for which values of a the system has non-trivial solutions

I saw this problem on the page but i have a question, could this be resolved without using determinants? And what would be the complete process? I never taken the Linear Algebra class,but i'm interested in the subject; I only know some things about it.I think you could use RREF but i don't know to apply it.I could be like this or it's incorrect? $$\left[ \begin{array}{ccc|c} a&1&0&b_2\\ 0&\frac{a-2}{a}&1&\frac{ab_3-2b_2}{a}\\0&0&\frac{3a-2}{a-2}&\frac{b_1a-2b_1+ab_3+2b_3-4b_2}{a-2} \end{array} \right]$$

Answer 1

The $b$'s are irrelevant to the question of infinite number of solutions. The matrix reduces to $$\begin{pmatrix} 1&0&1\\ 0&1&-1\\ 0&0&1-a\end{pmatrix}$$

So only $a=1$ gives an infinity of solutions.

Answer 2

If we just reduce this matrix a bit to something easier to work with we have $$\left[ \begin{array}{ccc|c} 2&-1&3&b_1\\ 0&-4(a-1)&0&ab_1+4b_2-3ab_3\\0&2&-2&b_3-b_1 \end{array} \right]$$

It's clear to see that $-4(a-1)=0 \implies a=1$. The rest is irrelevant since you only asked for what values of $a$. But if you wanted to see for what values of $b$ it has infinite solution we have $b_1 = 3b_3 -4b_2$