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I am having problems solving the system of linear equations as a function of $\lambda$ and $\mu$ for $\lambda , \mu \in \mathbb R$.

$$\left[\matrix{9&2&4\cr 3 &0&-3 \cr \lambda &2 &9}\right]x=\left[\matrix{\mu \cr 8 \cr 6}\right]$$

I never get an equation where I get an explicit expression of $x_1, x_2$ or $x_3$. Probably just basics, but can somebody give me the right approach?

3 Answers 3

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rewriting your system in the form $$x_1-x_3=\frac{8}{3}$$ $$9x_1+2x_2+4x_3=\mu$$ $$\lambda x_1+2x_2+9x_3=6$$ multiplying the first by $4$ and adding to the second we get $$13x_1+2x_2=\frac{32}{3}+\mu$$ $I$ multiplying the first equation by $9$ and adding to the third $$9x_1+\lambda x_1+2x_2=\frac{72}{3}+6$$ $II$ multiplying $I$ by $$(-1)$$ and adding to $II$ we get $$-4x_1+\lambda x_1=-\frac{32}{3}-\mu+\frac{72}{3}+6$$ simplifying this equation you will get $$x_1$$ and so on.

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Hint: Let $x = [x_1,x_2,x_3]^T$, then the problem is equivalent to:

$$9x_1 +2x_2+4x_3 = \mu$$ $$3x_1 -3x_3 = 8$$ $$\lambda x_1 +2x_2+9x_3 = 6.$$

Now, apply Gaussian-Elimination and be careful when dividing by expressions containing $\lambda$.

Can you complete it from here?

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    I do know the process of Gaussian elimination when there aren't two arbitrary variables $\mu$ and $\lambda$. My problem is that I keep ending up with a equation consisting of $x_1$ and $\lambda$ for example where I can't solve explicitly for $x_1$.. – 2017-01-30
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    The solutions will always depend on $\lambda$ and $\mu$. – 2017-01-30
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We have $$9x_1 +2 x_2 +4x_3 = \mu \tag {1} $$ $$3x_1-3x_3 =8 \tag {2} $$ $$\lambda x_1 + 2x_2 +9x_3 =6 \tag {3} $$

$$(1)-(3) \equiv (9-\lambda)x_1 -5x_3 =\mu-6 \tag {4} $$ Solving $(2) $ and $(4) $, we get, $$x_1 = \frac {3\mu-58}{12-3\lambda} \text { and } x_3 = \frac {3\mu + 8\lambda-90}{12-3\lambda} $$

Hope you can take it from here.