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I have this system of linear equations with parameter:

$ ax + 4y + z =0 $

$2y + 3z = 1$

$3x -cz=-2$

What I did was to put those equations into a matrix and transform that matrix it into a triangular matrix. Then I got these results on the right side: $ \frac{-c+10}{ac-15}$, $\frac{3+(c-6)a}{2ac-30}$, $\frac{2a-6}{ac-15}$. However, I don't know what to do now.

Thanks for any help!

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From the first equation we get $$z=-ax-4y$$ so we get with the third equation $$-3ax-10y=1$$ and with the last one: $$3x+acx+4yc=-2$$

From the second equation above $$y=-\frac{3}{10}ax-\frac{1}{10}$$ so we obtain finally

$$(15-ac)x=2c-1$$ Can you finish?The solution is given by $$\left\{ x=-2\,{\frac {c-5}{ac-15}},y=1/2\,{\frac {ac-6\,a+3}{ac-15}}, z=2\,{\frac {a-3}{ac-15}} \right\} $$ It remains to consider the case $$ac-15=0$$

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  • $\begingroup$ Thanks, but I'm trying to solve it using Maple, so I'm trying to use matrices. Is there any ways using matrices, for an example via rank of the matrix? $\endgroup$ – J. Doe May 12 at 19:33
  • $\begingroup$ It is better to use the elimination method.(so do I) $\endgroup$ – Dr. Sonnhard Graubner May 12 at 19:35
  • $\begingroup$ So if ac-15=0, then it has no solution? $\endgroup$ – J. Doe May 12 at 19:42
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This is a situation where Cramer's rule works quite well.

Our system is of the form $A\vec{x}=\vec{b}$ where \begin{align*} A &= \left[\begin{array}{rrr} a & 4 & 1 \\ 0 & 2 & 3 \\ 3 & 0 & -c \end{array}\right] & \vec{x} &= \left[\begin{array}{r} x \\ y \\ z \end{array}\right] & \vec{b} &= \left[\begin{array}{r} 0 \\ 1 \\ -2 \end{array}\right] \end{align*} Note that $$ \det(A)=2\cdot(15-ac) $$ This means that the system has a unique solution if and only if $ac\neq 15$.

Now, assuming $ac\neq 15$, we define \begin{align*} A_1 &= \left[\begin{array}{rrr} 0 & 4 & 1 \\ 1 & 2 & 3 \\ -2 & 0 & -c \end{array}\right] & A_2 &= \left[\begin{array}{rrr} a & 0 & 1 \\ 0 & 1 & 3 \\ 3 & -2 & -c \end{array}\right] & A_3 &= \left[\begin{array}{rrr} a & 4 & 0 \\ 0 & 2 & 1 \\ 3 & 0 & -2 \end{array}\right] \end{align*} Note that $A_i$ is $A$ with the $i$th column replaced by $\vec{b}$. According to Cramer's rule, the solution to the system is given by \begin{align*} x &= \frac{\det(A_1)}{\det(A)} = \frac{2 \, {\left(c - 5\right)}}{15-ac} & y &= \frac{\det(A_2)}{\det(A)} = \frac{a {\left(c - 6\right)} + 3}{2 \, {\left(a c - 15\right)}} & z &= \frac{\det(A_3)}{\det(A)} = \frac{2 \, {\left(a - 3\right)}}{a c - 15} \end{align*}

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