# Justifying a solution for a system of equations in linear algebra

I'm trying to understand how to solve systems of linear equations using matrices, however I'm faced with the following problem. Given the following system:

$$\begin{cases} x+2y+3z-3w=a \\ 2x-5y-3x+12w=b \\ 7x+y+8x+5w=c \end{cases}$$

What steps do I need to do to confirm and justify that it only has a single solution, which is:

$$37a +13b = 9c$$

Appreciate all hints! Thanks.

• Your question is confusing since the equation $37a + 13b = 9c$ is not a "solution" to the system of equations that you presented Dec 5, 2020 at 16:57
• thanks for the quick reply @BenGrossmann. maybe I'm misinterpreting the problem badly. The problem I'm facing asks to "Justifiably show that the system admits solution if and only if 37a + 13b = 9c". Dec 5, 2020 at 17:08
• Saying that "the system has a unique solution if and only if X is true" is different from saying "X is the unique solution to the system". Dec 5, 2020 at 22:20
• Correct @BenGrossmann. Was indeed my bad. sorry for the misunderstanding. Dec 6, 2020 at 10:33

$$\left( \begin{matrix} 1 & 2 & 3 & -3 & | & a\\ 2 & -5 & -3 & 12 & | & b \\ 7 & 1 & 8 & 5 & | & c \\ \end{matrix} \right) >>{\text{R2=R2-2R1, R3=R3-7R1}}>>$$ $$\left( \begin{matrix} 1 & 2 & 3 & -3 & | & a\\ 0 & -9 & -9 & 18 & | & b-2a \\ 0 & -13 & -13 & 26 & | & c-7a \\ \end{matrix} \right)>>{\text{R3=R3+(- \frac{13} {9} R2)}}>>$$
$$\left( \begin{matrix} 1 & 2 & 3 & -3 & | & a\\ 0 & -9 & -9 & 18 & | & b-2a \\ 0 & 0 & 0 & 0 & | & c-7a + [- \frac{13} {9}(b-2a)]\\ \end{matrix} \right)$$
$$c-7a + [- \frac{13} {9}(b-2a)] =0$$ $$-\frac{37}{9}a-\frac{13}{9}b+c=0$$ $$-37a -13b +9c=0$$ $$37a+13b=9c$$