Question

x + y + z = 0
x − y − 5z = 0
x + 2y + 4z = 0

Answer 1

Here,
x + y + z = 0 ...(1)
x − y − 5z = 0 ...(2)
x + 2y + 4z = 0 ...(3)

$$\begin{gathered} \text{The given system of homogeneous equations can be written in matrix form as follows:} \\ \left[\begin{array}{ccc}1& 1& 1\\ 1& -1& -5\\ 1& 2& 4\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right] \\ AX=O \\ \mathrm{Here}, \\ A=\left[\begin{array}{ccc}1& 1& 1\\ 1& -1& -5\\ 1& 2& 4\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\mathrm{and}O=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right] \\ \mathrm{Now}, \\ \left|A\right|=\left|\begin{array}{ccc}1& 1& 1\\ 1& -1& -5\\ 1& 2& 4\end{array}\right| \\ =1\left(-4+10\right)-1\left(4+5\right)+1(2+1) \\ =6-9+3 \\ =0 \\ \text{∴}\left|A\right|\text{≠0} \\ \text{So, the given system}\mathrm{of\; homogeneous\; equations}\text{has non-trivial solution.} \\ \text{Substituting}\mathit{\text{z}}\text{=}\mathit{\text{k}}\text{in eq. (1) and eq, (2), we get} \\ x+y=-k\mathrm{and}x-y=5k \\ AX=B \\ \mathrm{Here}, \\ \text{A=}\left[\begin{array}{cc}1& 1\\ 1& -1\end{array}\right]\text{, X=}\left[\begin{array}{c}x\\ y\end{array}\right]\text{and B}=\left[\begin{array}{c}-k\\ 5k\end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}1& 1\\ 1& -1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}-k\\ 5k\end{array}\right] \\ \left|A\right|\text{=}\left|\begin{array}{cc}1& 1\\ 1& -1\end{array}\right| \\ \text{=}\left(1\times -1-1\times 1\right) \\ \text{=-2} \\ \mathrm{So},A^{-1}\mathrm{exists}. \\ \text{We have} \\ \text{adj}\mathit{\text{A}}\text{=}\left[\begin{array}{cc}-1& -1\\ -1& 1\end{array}\right] \\ {\mathit{\text{A}}}^{-1}\text{=}\frac{1}{\left|A\right|}\text{adj}\mathit{\text{A}} \\ \Rightarrow A^{-1}=\frac{1}{-2}\left[\begin{array}{cc}-1& -1\\ -1& 1\end{array}\right] \\ X=A^{-1}B \\ \Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{-2}\left[\begin{array}{cc}-1& -1\\ -1& 1\end{array}\right]\left[\begin{array}{c}-k\\ 5k\end{array}\right] \\ =\frac{1}{-2}\left[\begin{array}{c}k-5k\\ k+5k\end{array}\right] \\ \text{Thus,}\mathit{\text{x}}\text{=2}\mathit{\text{k}}\text{,}\mathit{\text{y}}\text{=-3}\mathit{\text{k}}\text{and}\mathit{\text{z}}\text{=}\mathit{\text{k}}\left(\text{where}\mathit{\text{k}}\text{is any real number}\right)\text{satisfy the given system of equations.} \end{gathered}$$