Question

Solve each of the following system of homogeneous linear equations.
x + y − 2z = 0
2x + y − 3z = 0
5x + 4y − 9z = 0

Answer 1

Given: x + y − 2z = 0
2x + y − 3z = 0
5x + 4y − 9z = 0

$$\begin{gathered} D=\left|\begin{array}{ccc}1& 1& -2\\ 2& 1& -3\\ 5& 4& -9\end{array}\right| \\ =1(-9+12)-1(-18+15)-2(8-5) \\ =0 \\ \mathrm{So},\mathrm{the}\mathrm{system}\mathrm{has}\mathrm{infinitely}\mathrm{many}\mathrm{solutions}.\mathrm{Putting}z=k\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{two}\mathrm{equations},\mathrm{we}\mathrm{get} \\ x+y=2k \\ 2x+y=3k \\ \mathrm{Using}\mathrm{Cramer}\text{'}\mathrm{s}\mathrm{rule},\mathrm{we}\mathrm{get} \\ x=\frac{D_1}{D}=\frac{\left|\begin{array}{cc}2k& 1\\ 3k& 1\end{array}\right|}{\left|\begin{array}{cc}1& 1\\ 2& 1\end{array}\right|}=\frac{-k}{-1}=k \\ y=\frac{D_2}{D}=\frac{\left|\begin{array}{cc}1& 2k\\ 2& 3k\end{array}\right|}{\left|\begin{array}{cc}1& 1\\ 2& 1\end{array}\right|}=\frac{-k}{-1}=k \\ z=k \\ \mathrm{Clearly},\mathrm{these}\mathrm{values}\mathrm{satisfy}\mathrm{the}\mathrm{third}\mathrm{equation}. \\ \mathrm{Thus}, \\ x=y=z=k\left[k\in R\right] \end{gathered}$$