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)

The given system of homogeneous equations can be written in matrix form as follows:
\[\begin{bmatrix}3 & - 1 & 2 \\ 4 & 3 & 3 \\ 5 & 7 & 4\end{bmatrix} \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}\]
\[AX = O\]
Here, 
\[A = \begin{bmatrix}3 & - 1 & 2 \\ 4 & 3 & 3 \\ 5 & 7 & 4\end{bmatrix}, X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}\text{ and }O = \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}\]
Now, 
\[\left| A \right| = \begin{vmatrix}3 & - 1 & 2 \\ 4 & 3 & 3 \\ 5 & 7 & 4\end{vmatrix}\]
\[ = 3\left( 12 - 21 \right) + 1\left( 16 - 15 \right) + 2(28 - 15)\]
\[ = - 27 + 1 + 26\]
\[ = 0\]
\[\therefore\left| A \right|\neq0\]
So, the given systemof homogeneous equations has non-trivial solution.
Substituting z=k in eq. (1) & eq. (2), we get
\[3x - y = - 2k\text{ and }4x + 3y = - 3k\]
\[AX = B\]
Here,
\[A=\begin{bmatrix}3 & - 1 \\ 4 & 3\end{bmatrix}, X=\binom{x}{y}\text{ and }B = \binom{ - 2k}{ - 3k}\]
\[ \Rightarrow \begin{bmatrix}3 & - 1 \\ 4 & 3\end{bmatrix}\binom{x}{y} = \binom{ - 2k}{ - 3k}\]
\[\left| A \right|=\begin{vmatrix}3 & - 1 \\ 4 & 3\end{vmatrix}\]
\[=\left( 3 \times 3 + 4 \times 1 \right)\]
\[=13\]
\[\text{ So, }A^{- 1}\text{ exists .} \]
We have
\[adjA=\begin{bmatrix}3 & 1 \\ - 4 & 3\end{bmatrix}\]
\[ A^{- 1} =\frac{1}{\left| A \right|}adjA\]
\[ \Rightarrow A^{- 1} = \frac{1}{13}\begin{bmatrix}3 & 1 \\ - 4 & 3\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow \binom{x}{y} = \frac{1}{13}\begin{bmatrix}3 & 1 \\ - 4 & 3\end{bmatrix}\binom{ - 2k}{ - 3k}\]
\[ = \frac{1}{13}\binom{ - 6k - 3k}{8k - 9k}\]
\[\text{ Thus, }x=\frac{- 9k}{13},y=\frac{- k}{13}\text{ and }z=k\left( \text{ wherekis any real number }\right)\text{ satisfy the given system of equations. }\]