Can the system of equations be extracted from its solution?

Question

While I was solving the secondary school exam of 2014 I came across a question that states:

After solving those equations: $a_{1}x + b_{1}y = c_{1}$ and $a_{2}x + b_{2}y = c_{2}$, we found that x = $\frac{-7}{\begin{vmatrix} 3 & 1 \\ 1 & -2 \end{vmatrix}}$ and y = $\frac{-21}{\begin{vmatrix} 3 & 1 \\ 1 & -2 \end{vmatrix}}$, then $c_{1} = .....$ and $c_{2} = .....$

The answer is $c_{1} = 6$ and $c_{2} = -5$ in the ministry's model answers on its website with no steps available ,but it was probably solved by Cramer's rule.

Is this answer right and does $c_{1}$ and $c_{2}$ have only one value?

Answer 1

Presumably the denominator $\left|\matrix{3 & 1\cr 1 & -2}\right|$ indicates that the coefficient matrix $\pmatrix{a_1 & b_1\cr a_2 & b_2} = \pmatrix{3 & 1\cr 1 & -2}$. Then we have $$ \pmatrix{c_1 \cr c_2} = \pmatrix{3 & 1\cr 1 & -2} \pmatrix{x\cr y\cr} = \pmatrix{3 & 1\cr 1 & -2} \pmatrix{1\cr 3\cr} = \pmatrix{6\cr -5\cr}$$ agreeing with the ministry's answer.

Answer 2

In view of the main determinant:

$$\begin{vmatrix}3 & 1\\1 & -2\end{vmatrix}$$

one may infer that the system is

$$\begin{cases}3x_1+x_2&=&c_1\\1x_1-2x_2&=&c_2\end{cases}$$

Using Cramer's rule (https://en.wikipedia.org/wiki/Cramer's_rule), the two numerators can be written under the form:

$$\begin{vmatrix}c_1 & 1\\c_2 & -2\end{vmatrix}=-7 \ \ \text{and} \ \ \begin{vmatrix}3 & c_1\\1 & c_2\end{vmatrix}=-21$$

This can be transformed into a linear system:

$$\begin{cases}-2c_1-c_2&=&-7\\-c_1+3c_2&=&-21\end{cases}$$

Solving this system gives $c_1=6$ and $c_2=-5$, as indicated.