Question

x + y − z = 0
x − 2y + z = 0
3x + 6y − 5z = 0

Answer 1

x + y − z = 0                ...(1)
x − 2y + z = 0              ...(2)
3x + 6y − 5z = 0          ...(3)

The given system of homogeneous equations can be written in matrix form as follows:
\[\begin{bmatrix}1 & 1 & - 1 \\ 1 & - 2 & 1 \\ 3 & 6 & - 5\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}\]
\[AX = O\]
Here,
\[A = \begin{bmatrix}1 & 1 & - 1 \\ 1 & - 2 & 1 \\ 3 & 6 & - 5\end{bmatrix}, X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}\text{ and }O = \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}\] 
Now, 
\[\left| A \right| = \begin{vmatrix}1 & 1 & - 1 \\ 1 & - 2 & 1 \\ 3 & 6 & - 5\end{vmatrix}\]
\[ = 1\left( 10 - 6 \right) - 1\left( - 5 - 3 \right) - 1\left( 6 + 6 \right)\]
\[ = 4 + 8 - 12\]
\[ = 0\]
So, the given systemof homogeneous equations has non-trivial solution.
Substituting z=k in eq. (1) & eq. (2), we get
\[x+y=k \text{ and }x-2y=-k\]
\[ \Rightarrow \begin{bmatrix}1 & 1 \\ 1 & - 2\end{bmatrix}\binom{x}{y} = \binom{k}{ - k}\]
\[AX=B\]
Here,
\[A=\begin{bmatrix}1 & 1 \\ 1 & - 2\end{bmatrix},X=\binom{x}{y}\text{ and }B=\binom{k}{ - k}\]
\[\left| A \right| = \begin{vmatrix}1 & 1 \\ 1 & - 2\end{vmatrix} = - 3\]
\[\text{ So,} A^{- 1}\text{ exists .} \]
\[adj A = \begin{bmatrix}- 2 & - 1 \\ - 1 & 1\end{bmatrix}\]
\[ A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ \Rightarrow A^{- 1} = \frac{1}{- 3}\begin{bmatrix}- 2 & - 1 \\ - 1 & 1\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow \binom{x}{y} = \frac{1}{- 3}\begin{bmatrix}- 2 & - 1 \\ - 1 & 1\end{bmatrix}\binom{k}{ - k}\]
\[ \Rightarrow \binom{x}{y} = \frac{1}{- 3}\binom{ - 2k + k}{ - k - k}\]
\[\text{ Thus,}x=\frac{k}{3}, y=\frac{2k}{3}\text{ and }z=k\left( \text{ wherekis any real number }\right)\text{ satisfy the given system of equations.}\]