Solve the following system of homogeneous linear equation. $$2x-y+z=0, 3x+2y-z=0,x+4y+3z=0$$
$$
        \begin{vmatrix}
        2 & -1 & 1 \\
        3 & 2 & -1 \\
        1 & 4 &  3 \\
        \end{vmatrix}
$$
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 ?
 A: 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
$$4z=0\to z=0\\
y+z=0\to y=0\\
x+4y+3z=0\to x=0$$
A: 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 $$z=0$$
A: 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.
