Question

x + y + z = 0
x − y − 5z = 0
x + 2y + 4z = 0

Answer 1

Here,
x + y + z = 0             ...(1)
x − y − 5z = 0           ...(2)
x + 2y + 4z = 0         ...(3)

The given system of homogeneous equations can be written in matrix form as follows:
\[\begin{bmatrix}1 & 1 & 1 \\ 1 & - 1 & - 5 \\ 1 & 2 & 4\end{bmatrix} \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}\]
\[AX = O\]
Here, 
\[A = \begin{bmatrix}1 & 1 & 1 \\ 1 & - 1 & - 5 \\ 1 & 2 & 4\end{bmatrix}, X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}\text{ and }O = \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}\]
Now,
\[\left| A \right| = \begin{vmatrix}1 & 1 & 1 \\ 1 & - 1 & - 5 \\ 1 & 2 & 4\end{vmatrix}\]
\[ = 1\left( - 4 + 10 \right) - 1\left( 4 + 5 \right) + 1(2 + 1)\]
\[ = 6 - 9 + 3\]
\[ = 0\]
\[\therefore\left| A \right|\neq0\]
So, the given systemof homogeneous equations has non-trivial solution.
Substituting z=k in eq. (1) and eq, (2), we get
\[x + y = - k\text{ and }x - y = 5k\]
\[AX = B\]
Here,
\[A=\begin{bmatrix}1 & 1 \\ 1 & - 1\end{bmatrix}, X=\binom{x}{y}\text{ and }B = \binom{ - k}{5k}\]
\[ \Rightarrow \begin{bmatrix}1 & 1 \\ 1 & - 1\end{bmatrix}\binom{x}{y} = \binom{ - k}{5k}\]
\[\left| A \right|=\begin{vmatrix}1 & 1 \\ 1 & - 1\end{vmatrix}\]
\[ =\left( 1 \times - 1 - 1 \times 1 \right)\]
\[ =-2\]
\[So, A^{- 1}\text{ exists }. \]
We have
\[adjA=\begin{bmatrix}- 1 & - 1 \\ - 1 & 1\end{bmatrix}\]
\[ A^{- 1} =\frac{1}{\left| A \right|}adjA\]
\[ \Rightarrow A^{- 1} = \frac{1}{- 2}\begin{bmatrix}- 1 & - 1 \\ - 1 & 1\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow \binom{x}{y} = \frac{1}{- 2}\begin{bmatrix}- 1 & - 1 \\ - 1 & 1\end{bmatrix}\binom{ - k}{5k}\]
\[ = \frac{1}{- 2}\binom{k - 5k}{k + 5k}\]
\[\text{ Thus,} x=2k,y=-3k\text{ and }z=k\left( \text{ wherekis any real number } \right)\text{ satisfy the given system of equations. }\]