Solve this

Question:

$3 x-y+2 z=0$

$4 x+3 y+3 z=0$

$5 x+7 y+4 z=0$

Solution:

Here,

$3 x-y+2 z=0$               ....(1)

$4 x+3 y+3 z=0$         .....(2)

$5 x+7 y+4 z=0$         ....(3)

The given system of homogeneous equations can be written in matrix form as follows:

$\left[\begin{array}{ccc}3 & -1 & 2 \\ 4 & 3 & 3 \\ 5 & 7 & 4\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$

$A X=O$

Here,

$A=\left[\begin{array}{ccc}3 & -1 & 2 \\ 4 & 3 & 3 \\ 5 & 7 & 4\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $O=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$

Now,

$|A|=\left|\begin{array}{ccc}3 & -1 & 2 \\ 4 & 3 & 3 \\ 5 & 7 & 4\end{array}\right|$

$=3(12-21)+1(16-15)+2(28-15)$

$=-27+1+26$

$=0$

$\therefore|A| \neq 0$

So, the given system of homogeneous equations has non-trivial solution.

Substituting $z=k$ in eq. (1) \& eq. (2), we get

$3 x-y=-2 k$ and $4 x+3 y=-3 k$

$A X=B$

Here,

$\mathrm{A}=\left[\begin{array}{cc}3 & -1 \\ 4 & 3\end{array}\right], \mathrm{X}=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{l}-2 k \\ -3 k\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}3 & -1 \\ 4 & 3\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}-2 k \\ -3 k\end{array}\right]$

$|A|=\left|\begin{array}{cc}3 & -1 \\ 4 & 3\end{array}\right|$

$=(3 \times 3+4 \times 1)$

$=13$

So, $A^{-1}$ exists.

We have

$\operatorname{adj} A=\left[\begin{array}{cc}3 & 1 \\ -4 & 3\end{array}\right]$

$A^{-1}=\frac{1}{|A|}$ adj $A$

$\Rightarrow A^{-1}=\frac{1}{13}\left[\begin{array}{cc}3 & 1 \\ -4 & 3\end{array}\right]$

$X=A^{-1} B$

$\Rightarrow\left[\begin{array}{l}x \\ y\end{array}\right]=\frac{1}{13}\left[\begin{array}{cc}3 & 1 \\ -4 & 3\end{array}\right]\left[\begin{array}{l}-2 k \\ -3 k\end{array}\right]$

$=\frac{1}{13}\left[\begin{array}{c}-6 k-3 k \\ 8 k-9 k\end{array}\right]$

Thus, $\mathrm{x}=\frac{-9 k}{13}, \mathrm{y}=\frac{-k}{13}$ and $z=k$ (where $k$ is any real number ) satisfy the given system of equations

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