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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.

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  • $\begingroup$ Your question is confusing since the equation $37a + 13b = 9c$ is not a "solution" to the system of equations that you presented $\endgroup$ Dec 5, 2020 at 16:57
  • $\begingroup$ 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". $\endgroup$
    – nunob
    Dec 5, 2020 at 17:08
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    $\begingroup$ 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". $\endgroup$ Dec 5, 2020 at 22:20
  • $\begingroup$ Correct @BenGrossmann. Was indeed my bad. sorry for the misunderstanding. $\endgroup$
    – nunob
    Dec 6, 2020 at 10:33

1 Answer 1

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You can get your answer using simple elementary row operations:
$$\left( \begin{matrix} 1 & 2 & 3 & -3 & | & a\\ 2 & -5 & -3 & 12 & | & b \\ 7 & 1 & 8 & 5 & | & c \\ \end{matrix} \right) >>{\text{$R2=R2-2R1, R3=R3-7R1$}}>>$$ $$\left( \begin{matrix} 1 & 2 & 3 & -3 & | & a\\ 0 & -9 & -9 & 18 & | & b-2a \\ 0 & -13 & -13 & 26 & | & c-7a \\ \end{matrix} \right)>>{\text{$R3=R3+(- \frac{13} {9} R2)$}}>>$$

$$\left( \begin{matrix} 1 & 2 & 3 & -3 & | & a\\ 0 & -9 & -9 & 18 & | & b-2a \\ 0 & 0 & 0 & 0 & | & c-7a + [- \frac{13} {9}(b-2a)]\\ \end{matrix} \right)$$
Now, notice that the last row of the matrix is a zero row, thus the system has a solution only if the right side of the equation is equal to zero. thus-
$$c-7a + [- \frac{13} {9}(b-2a)] =0$$ $$-\frac{37}{9}a-\frac{13}{9}b+c=0$$ $$-37a -13b +9c=0$$ $$37a+13b=9c$$

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