Solve the following system of homogeneous linear equation.

Question

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

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 ?

Answer 1

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

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

Answer 2

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 $$z=0$$