Question

The solution of the system of equations $6x-9y-20z=-4;4x-15y+10z=-1;2x-3y-5z=-1$ is such that :

• A. $2x=5y=3z$
• B. $2x=3y=5z$
• C. $5x=2y=3z$
• D. $x=y\ne z$

Answer 1

The correct option is B $2x=3y=5z$

Given
$6x-9y-20z=-44x-15y+10z=-12x-3y-5z=-1$
Changing in to matrix form,
$AX=B\Rightarrow X=A^{-1}B\cdots \left(i\right)$
Here,
$A=\left[\begin{array}{ccc}6& -9& -20\\ 4& -15& 10\\ 2& -3& -5\end{array}\right],B=\left[\begin{array}{c}-4\\ -1\\ -1\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$

Now,
$\left|A\right|=-90$
Cofactor matrix of $A=\left[\begin{array}{ccc}105& 40& 18\\ 15& 10& 0\\ -390& -140& -54\end{array}\right]$

Now $A^{-1}=\frac{1}{\left|A\right|}adjA=\frac{1}{\left|A\right|}[\text{cofactor matrix of A}{]}^T$
$=\frac{1}{90}\left[\begin{array}{ccc}105& 15& -390\\ 40& 10& -140\\ 18& 0& -54\end{array}\right]$

From $\left(i\right)$ we get,
$X=A^{-1}B=\frac{1}{-90}\left[\begin{array}{ccc}105& 15& -390\\ 40& 10& -140\\ 18& 0& -54\end{array}\right]\left[\begin{array}{c}-4\\ -1\\ -1\end{array}\right]$

$=-\frac{1}{90}\left[\begin{array}{c}-45\\ -30\\ -18\end{array}\right]$

$=\left[\begin{array}{c}\frac{1}{2}\\ \frac{1}{3}\\ \frac{1}{5}\end{array}\right]$

$\therefore \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}\frac{1}{2}\\ \frac{1}{3}\\ \frac{1}{5}\end{array}\right]$

So $x=\frac{1}{2},y=\frac{1}{3},z=\frac{1}{5}\Rightarrow 2x=3y=5z=1$