Question

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

Answer 1

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

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