Question

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

Answer 1

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

$$\begin{gathered} D=\left|\begin{array}{ccc}3& 1& 1\\ 1& -4& 3\\ 2& 5& -2\end{array}\right|=0 \\ \mathrm{The}\mathrm{system}\mathrm{has}\mathrm{infinitely}\mathrm{many}\mathrm{solutions}.\mathrm{Putting}z=k\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{two}\mathrm{equations},\mathrm{we}\mathrm{get} \\ 3x+y=-k \\ x-4y=-3k \\ \mathrm{Solving}\mathrm{these}\mathrm{equations}\mathrm{by}\mathrm{Cramer}\text{'}\mathrm{s}\mathrm{rule},\mathrm{we}\mathrm{get} \\ x=\frac{D_1}{D}=\frac{\left|\begin{array}{cc}-k& 1\\ -3k& -4\end{array}\right|}{\left|\begin{array}{cc}3& 1\\ 1& -4\end{array}\right|}=-\frac{7k}{13} \\ y=\frac{D_2}{D}=\frac{\left|\begin{array}{cc}3& -k\\ 1& -3k\end{array}\right|}{\left|\begin{array}{cc}3& 1\\ 1& -4\end{array}\right|}=\frac{8k}{13} \\ z=k \\ \Rightarrow x=-\frac{7k}{13},y=\frac{8k}{13}\mathrm{and}z=k \\ \mathrm{or}x=-7k,y=8k\mathrm{and}z=13k \\ \mathrm{Clearly},\mathrm{these}\mathrm{values}\mathrm{satisfy}\mathrm{the}\mathrm{third}\mathrm{equation}. \\ \mathrm{Thus}, \\ x=-7k \\ y=8k\left[k\in R\right] \\ z=13k \end{gathered}$$