Solve this

Solution:

Here,

$x+y+z=0$             ....(1)

$x-y-5 z=0$          ....(2)

$x+2 y+4 z=0$      ....(3)

The given system of homogeneous equations can be written in matrix form as follows:

$\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & -1 & -5 \\ 1 & 2 & 4\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$

$A X=O$

Here,

$A=\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & -1 & -5 \\ 1 & 2 & 4\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $O=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$

Now,

$|A|=\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & -1 & -5 \\ 1 & 2 & 4\end{array}\right|$

$=1(-4+10)-1(4+5)+1(2+1)$

$=6-9+3$

$=0$

$\therefore|A| \neq 0$

So, the given system of homogeneous equations has non-trivial solution.

Substituting $z=k$ in eq. (1) and eq, (2), we get

$x+y=-k$ and $x-y=5 k$

$A X=B$

Here,

$\mathrm{A}=\left[\begin{array}{cc}1 & 1 \\ 1 & -1\end{array}\right], \mathrm{X}=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{c}-k \\ 5 k\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}1 & 1 \\ 1 & -1\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}-k \\ 5 k\end{array}\right]$

$\begin{aligned}|A|=&\left|\begin{array}{cc}1 & 1 \\ 1 & -1\end{array}\right| \\ &=(1 \times-1-1 \times 1) \\ &=-2 \end{aligned}$

So, $A^{-1}$ exists.

We have

$\operatorname{adj} A=\left[\begin{array}{cc}-1 & -1 \\ -1 & 1\end{array}\right]$

$A^{-1}=\frac{1}{|A|}$ adj $A$

$\Rightarrow A^{-1}=\frac{1}{-2}\left[\begin{array}{cc}-1 & -1 \\ -1 & 1\end{array}\right]$

$X=A^{-1} B$

$\Rightarrow\left[\begin{array}{l}x \\ y\end{array}\right]=\frac{1}{-2}\left[\begin{array}{cc}-1 & -1 \\ -1 & 1\end{array}\right]\left[\begin{array}{c}-k \\ 5 k\end{array}\right]$

$=\frac{1}{-2}\left[\begin{array}{l}k-5 k \\ k+5 k\end{array}\right]$

Thus, $x=2 k, y=-3 k$ and $z=k$ (where $k$ is any real number) satisfy the given system of equations.