# Proving a system of equations has solutions based on knowing the determinant

Once again I'm facing a problem I'm not sure I can solve. I did some messing around with the system of equations and got the formula that I wanted, but I'm unsure about the assumptions I made.

If you could care to explain me if I did something wrong and if there's a better simpler or more clearer answer I would deeply appreciate it for I really don't get what they intended to say in the solution.

The question is the following: Given the system of equations: \left\{ \begin{align} a_1 x + b_1y &= c_1\\ a_2 x + b_2y &= c_2 \end{align} \right. , knowing that the following determinant \begin{vmatrix} a_1 & b_1\\ a_2 & b_2 \end{vmatrix} equals zero. And that $a_1 \ne 0$ . Our system has solutions if and only if $c_2 = \frac{a_2}{a_1}c_1$

I put the system in the following form \begin{pmatrix} a_1 & b_1 & | & c_1 \\ a_2 & b_2 & | & c_2 \\ \end{pmatrix}

Solving it towards this: \begin{pmatrix} a_1 & b_1 & | & c_1 \\ 0 & b_2 - \frac{a_2}{a_1}b_1 & | & c_2 - c_1\frac{a_2}{a_1} \\ \end{pmatrix}

I know that the system isn't homogeneous (trivial solution if I remember correctly) because of the determinant, and therefore I assumed that $b_2 - \frac{a_2}{a_1}b_1 = 0$ (I'm unsure about this) and by that I got that $c_2 - c_1\frac{a_2}{a_1} = 0 \iff c_2 = c_1\frac{a_2}{a_1}$.

Their solution is at 1H.5) https://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/1.-vectors-and-matrices/part-a-vectors-determinants-and-planes/problem-set-1/MIT18_02SC_SupProbSol1.pdf but I find it rather confusing.

• It's a little short, but correct: the condition $b_2-\frac{a_2}{a_1}b_1=0$ is satisfied, since it is equivalent to $\frac{b_2a_1-a_2b_1}{a_1}b_1=\frac{\det A}{b_1}=0$ . – Bernard Aug 19 '18 at 13:23

I used Cramer's rule. if $D=0$ then $D_1=0$$,\hspace{5pt}$$D_2 =0$
$$\begin{vmatrix} c_1 & b_1\\ c_2 & b_2 \end{vmatrix}=0$$ $$\begin{vmatrix} a_1 & c_1\\ a_2 & c_2 \end{vmatrix}=0$$
• This isn't complete, is it? It shows that if $D=0$ and the system has a solution, then $D_2=0,$ but it doesn't show that if $D=D_2=0$ then the system has a solution. You need the hypothesis $a_1\neq0$ for that. Otherwise, you could have $a_1=a_2=0$ and $D_1\neq0$ – saulspatz Aug 19 '18 at 13:43