Question

3x − y + 2z = 0
4x + 3y + 3z = 0
5x + 7y + 4z = 0

Answer 1

Here,
3x − y + 2z = 0 ...(1)
4x + 3y + 3z = 0 ...(2)
5x + 7y + 4z = 0 ...(3)


$$\begin{gathered} \text{The given system of homogeneous equations can be written in matrix form as follows:} \\ \left[\begin{array}{ccc}3& -1& 2\\ 4& 3& 3\\ 5& 7& 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}3& -1& 2\\ 4& 3& 3\\ 5& 7& 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}3& -1& 2\\ 4& 3& 3\\ 5& 7& 4\end{array}\right| \\ =3\left(12-21\right)+1\left(16-15\right)+2(28-15) \\ =-27+1+26 \\ =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) \& eq. (2), we get} \\ 3x-y=-2k\mathrm{and}4x+3y=-3k \\ AX=B \\ \mathrm{Here}, \\ \text{A=}\left[\begin{array}{cc}3& -1\\ 4& 3\end{array}\right]\text{, X=}\left[\begin{array}{c}x\\ y\end{array}\right]\text{and B}=\left[\begin{array}{c}-2k\\ -3k\end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}3& -1\\ 4& 3\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}-2k\\ -3k\end{array}\right] \\ \left|A\right|\text{=}\left|\begin{array}{cc}3& -1\\ 4& 3\end{array}\right| \\ \text{=}\left(3\times 3+4\times 1\right) \\ \text{=13} \\ \mathrm{So},A^{-1}\mathrm{exists}. \\ \text{We have} \\ \text{adj}\mathit{\text{A}}\text{=}\left[\begin{array}{cc}3& 1\\ -4& 3\end{array}\right] \\ {\mathit{\text{A}}}^{-1}\text{=}\frac{1}{\left|A\right|}\text{adj}\mathit{\text{A}} \\ \Rightarrow A^{-1}=\frac{1}{13}\left[\begin{array}{cc}3& 1\\ -4& 3\end{array}\right] \\ X=A^{-1}B \\ \Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{13}\left[\begin{array}{cc}3& 1\\ -4& 3\end{array}\right]\left[\begin{array}{c}-2k\\ -3k\end{array}\right] \\ =\frac{1}{13}\left[\begin{array}{c}-6k-3k\\ 8k-9k\end{array}\right] \\ \text{Thus, x=}\frac{-9k}{13}\text{,y=}\frac{-k}{13}\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}$$