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Question
Solve each of the following system of homogeneous linear equations.
x + y − 2z = 0
2x + y − 3z = 0
5x + 4y − 9z = 0
Solution
Given: x + y − 2z = 0
2x + y − 3z = 0
5x + 4y − 9z = 0
\[D = \begin{vmatrix}1 & 1 & - 2 \\ 2 & 1 & - 3 \\ 5 & 4 & - 9\end{vmatrix}\]
\[ = 1( - 9 + 12) - 1( - 18 + 15) - 2(8 - 5)\]
\[ = 0\]
So, the system has infinitely many solutions . Putting z = k in the first two equations, we get
\[x + y = 2k\]
\[2x + y = 3k\]
Using Cramer's rule, we get
\[x = \frac{D_1}{D} = \frac{\begin{vmatrix}2k & 1 \\ 3k & 1\end{vmatrix}}{\begin{vmatrix}1 & 1 \\ 2 & 1\end{vmatrix}} = \frac{- k}{- 1} = k\]
\[y = \frac{D_2}{D} = \frac{\begin{vmatrix}1 & 2k \\ 2 & 3k\end{vmatrix}}{\begin{vmatrix}1 & 1 \\ 2 & 1\end{vmatrix}} = \frac{- k}{- 1} = k \]
\[z = k\]
Clearly, these values satisfy the third equation .
Thus,
\[x = y = z = k \left[ k \in R \right]\]
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