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$$2x-y+z=0, 3x+2y-z=0,x+4y+3z=0$$ $$ \begin{vmatrix} 2 & -1 & 1 \\ 3 & 2 & -1 \\ 1 & 4 & 3 \\ \end{vmatrix} $$

By reducing row $$R_1=R_1-2R_3\\R_2=R_3-3R_3$$

we get

\begin{vmatrix} 0 & -9 & -5 \\ 0 & -10 & -10 \\ 1 & 4 & 3 \\ \end{vmatrix} Expanding along C3 and solving the determinant we get 40

But how to solve further ?

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    The discriminant in non-zero, hence there's only the trivial solution. – 2017-02-15
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    @Bernard yes the solution is trivial but can you explain how to find the values of x, y and z. – 2017-02-15
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    ‘Trivial’ in this context means $(0,0,0)$. – 2017-02-15

3 Answers 3

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Once all equations are equal to zero then:

$$\begin{vmatrix} 0 & -9 & -5 \\ 0 & -10 & -10 \\ 1 & 4 & 3 \\ \end{vmatrix}=\begin{vmatrix} 0 & -9 & -5 \\ 0 & 1 & 1 \\ 1 & 4 & 3 \\ \end{vmatrix}$$

And now $R_1=R_1+9R_2$ and get

$$\begin{vmatrix} 0 & 0 & 4 \\ 0 & 1 & 1 \\ 1 & 4 & 3 \\ \end{vmatrix}$$

The above system is equivalent to

$$4z=0\to z=0\\ y+z=0\to y=0\\ x+4y+3z=0\to x=0$$

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    the answer in my book is given x=1, y= -1 and z = 2. But i dont know how to get the values of x, y and z – 2017-02-15
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    I'm afraid that solution is wrong. Plug it into the system and see that it doesn't work. Did you write everything right? – 2017-02-15
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    @ Arnaldo yes question is correct, I think the answer might be printed wrong. but can you please explain how to find the values of x, y and z. – 2017-02-15
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    @bappy: please, take another look in the answer – 2017-02-15
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    @bappy: Is it clear? – 2017-02-16
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    @ Arnaldo yes very clear – 2017-02-21
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adding the first and the second equation we get $$5x+y=0$$ multiplying the second by $3$ and adding to the first we have $$10x+10y=0$$ or $$x+y=0$$ subtracting both equations we obtain $$4x=0$$ thus $$x=0$$ and $$y=0$$ and $$z=0$$

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    @ Dr. Sonnhard Graubner your answer is wrong. The answers are given in my book x=1, y= -1 and z= 2. – 2017-02-15
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    if $$2+1+2=0$$ then is my solution is wrong – 2017-02-15
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Your linear system $Ax=0$ only have the trivial solution $x=0$ since $null(A)$ is an empty matrix, i.e., the null space is empty, so you can't find a linear combination which generate $x$ from a non-existant base. If the solution given from your book is right, may you didn't write up the problem ok.