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The following linear system is as follows:

$$ax+y+0=a$$ $$x+y+z=1$$ $$0+y+az=b$$

Find $a$ and $b$ such that the linear system has: 1. No solution 2. Exactly one solution 3. Infinitely many solutions

The row-echelon form of the augmented matrix of the linear system is this after working it out using Gauss Elimination:

$$ \left[ \begin{array}{ccc|c} 1&1&1&1\\ 0&1&a&b\\0&0&a^2-2a&ab-b \end{array} \right] $$

For this to have no solution, $$a^2-2a=0$$ Therefore: $$a=2, a=0$$

However, I will need to substitute it into $ab-b$ and ensure that $ab-b$ is not 0. $$a=0, ab-b=-b$$ $$a=2, ab-b=b$$

Therefore: $$a=0, b\in\mathbb R-(0)$$ $$a=2, b\in\mathbb R-(0)$$

However, I do need some help with regards to having exactly one solution and infinitely many solutions, thanks!

2 Answers 2

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from the first equation we get $$y=a-ax$$ and from the second we get $$z=1-x-a+ax$$ thus we get $$a-a x+(1-x-a+ax)=b$$ simplifying we have $$x(-a-a +a^2)=b+a^2-2a$$ can you finish?

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    I used your equations to obtain $x$, $y$ and $z$ in terms of $a$ and $b$. $$x=1-b$$ $$y=ab$$ $$z=b(1-a)$$ Any hints on how do I continue? – 2017-02-15
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    and where is $\alpha$? – 2017-02-15
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    Oh that $a$ is the same $a$, for the first equation – 2017-02-15
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    now i got $$x(-2a+a^2)=b+a^2-2a$$ – 2017-02-15
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    I simplified the 3rd equation to obtain $x=1-b$ – 2017-02-15
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    this is wrong you can not do this – 2017-02-15
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    note that $$x=\frac{a^2-2a+b}{a^2-2a}$$ if $$a^2-2a\ne 0$$ – 2017-02-15
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    As in your third equation, if I simplify that I will obtain $x=1-b$. Sorry I am not sure about this but can I ask why is this wrong? – 2017-02-15
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Well, we have that:

$$ \begin{cases} \text{a}\cdot\text{x}+\text{y}+0=\text{a}\\ \\ \text{x}+\text{y}+\text{z}=1\\ \\ 0+\text{y}+\text{a}\cdot\text{z}=\text{b} \end{cases}\tag1 $$

Now, simplifying and solving $\text{a}$ out of the first equation:

$$ \begin{cases} \text{a}=\frac{\text{y}}{1-\text{x}}\\ \\ \text{y}=1-\text{x}-\text{z}\\ \\ \text{y}+\text{a}\cdot\text{z}=\text{b} \end{cases}\tag2 $$

So, for the third equation we get:

$$1-\text{x}-\text{z}+\frac{1-\text{x}-\text{z}}{1-\text{x}}\cdot\text{z}=1-x+\frac{\text{z}^2}{x-1}=\text{b}\space\Longleftrightarrow\space\text{z}=\pm\sqrt{\text{x}-1}\cdot\sqrt{\text{b}+\text{x}-1}\tag3$$

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    Do I need to solve for $x$ in terms of $a$ and $b$ in order to continue? – 2017-02-15
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    @Derp Maybe that is a good start – 2017-02-15