Question

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

Answer 1

Here,
2x − y + 2z = 0                 ...(1)
5x + 3y − z = 0                 ...(2)
x + 5y − 5z = 0                 ...(3)
The given system of homogeneous equations can be written in matrix form as follows:
\[\begin{bmatrix}2 & - 1 & 2 \\ 5 & 3 & - 1 \\ 1 & 5 & - 5\end{bmatrix} \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}\]
\[AX = O\]
Here, 
\[A = \begin{bmatrix}2 & - 1 & 2 \\ 5 & 3 & - 1 \\ 1 & 5 & - 5\end{bmatrix}, X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}\text{ and }O = \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}\]
Now,
\[ \left| A \right| = \begin{vmatrix}2 & - 1 & 2 \\ 5 & 3 & - 1 \\ 1 & 5 & - 5\end{vmatrix}\]
\[ = 2\left( - 15 + 5 \right) + 1\left( - 25 + 1 \right) + 2(25 - 3)\]
\[ = - 20 - 24 + 44\]
\[ = 0\]
\[\therefore\left| A \right|\neq 0\]
So, the given systemof homogeneous equations has non-trivial solution.
\[\text{ Substituting z=k in eq. }(1)\hspace{0.167em} \text{ and eq. }(2),\text{ we get }\]
\[2x - y = - 2k \text{ and }5x + 3y = k\]
\[AX = B\]
Here,
\[A=\begin{bmatrix}2 & - 1 \\ 5 & 3\end{bmatrix}, X=\binom{x}{y}\text{ and }B = \binom{ - 2k}{k}\]
\[ \Rightarrow \begin{bmatrix}2 & - 1 \\ 5 & 3\end{bmatrix}\binom{x}{y} = \binom{ - 2k}{k}\]
\[\left| A \right|=\begin{vmatrix}2 & - 1 \\ 5 & 3\end{vmatrix}\]
\[ =\left( 3 \times 2 + 1 \times 5 \right)\]
\[ =11\]
\[So, A^{- 1}\text{ exists . }\]
We have
\[adjA=\begin{bmatrix}3 & 1 \\ - 5 & 2\end{bmatrix}\]
\[ A^{- 1} =\frac{1}{\left| A \right|}adjA\]
\[ \Rightarrow A^{- 1} = \frac{1}{11}\begin{bmatrix}3 & 1 \\ - 5 & 2\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow \binom{x}{y} = \frac{1}{11}\begin{bmatrix}3 & 1 \\ - 5 & 2\end{bmatrix}\binom{ - 2k}{k}\]
\[ = \frac{1}{11}\binom{ - 6k + k}{10k + 2k}\]
\[\text{ Thus },x=\frac{- 5k}{11},y=\frac{12k}{11}\text{ and }z=k\left( \text{ where k is any real number }\right)\text{ satisfy the given system of equations.}\]