Question

x + y − z = 0
x − 2y + z = 0
3x + 6y − 5z = 0

Answer 1

x + y − z = 0 ...(1)
x − 2y + z = 0 ...(2)
3x + 6y − 5z = 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& -2& 1\\ 3& 6& -5\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& -2& 1\\ 3& 6& -5\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& -2& 1\\ 3& 6& -5\end{array}\right| \\ =1\left(10-6\right)-1\left(-5-3\right)-1\left(6+6\right) \\ =4+8-12 \\ =0 \\ \text{So, the given system}\mathrm{of}\mathrm{homogeneous}equations\mathrm{h}\text{as non-trivial solution.} \\ \text{Substituting}\mathit{\text{z=k}}\text{in eq. (1) \& eq. (2), we get} \\ \mathit{\text{x}}\text{+}\mathit{\text{y}}\text{=}\mathit{\text{k}}\text{and}\mathit{\text{x}}\text{−2}\mathit{\text{y}}\text{=-}\mathit{\text{k}} \\ \Rightarrow \left[\begin{array}{cc}1& 1\\ 1& -2\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}k\\ -k\end{array}\right] \\ \mathit{\text{AX}}\text{=}\mathit{\text{B}} \\ \text{Here,} \\ \mathit{\text{A}}\text{=}\left[\begin{array}{cc}1& 1\\ 1& -2\end{array}\right]\text{,}\mathit{\text{X}}\text{=}\left[\begin{array}{c}x\\ y\end{array}\right]\text{and}\mathit{\text{B}}\text{=}\left[\begin{array}{c}k\\ -k\end{array}\right] \\ \left|A\right|=\left|\begin{array}{cc}1& 1\\ 1& -2\end{array}\right|=-3 \\ \mathrm{So},A^{-1}\mathrm{exists}. \\ \mathrm{adj}A=\left[\begin{array}{cc}-2& -1\\ -1& 1\end{array}\right] \\ {A}^{-1}=\frac{1}{\left|A\right|}\mathrm{adj}A \\ \Rightarrow A^{-1}=\frac{1}{-3}\left[\begin{array}{cc}-2& -1\\ -1& 1\end{array}\right] \\ X=A^{-1}B \\ \Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{-3}\left[\begin{array}{cc}-2& -1\\ -1& 1\end{array}\right]\left[\begin{array}{c}k\\ -k\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{-3}\left[\begin{array}{c}-2k+k\\ -k-k\end{array}\right] \\ \text{Thus,}\mathit{\text{x}}\text{=}\frac{k}{3}\mathit{\text{, y}}\text{=}\frac{2k}{3}\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}$$