Question

If $\left|\begin{array}{ccc}x& x+y& x+y+z\\ 2x& 3x+2y& 4x+3y+2z\\ 3x& 6x+3y& 10x+6y+3z\end{array}\right|$ = $64$ then find x.

Answer 1

Applying $R_3-2R_2$ and $R_2-2R_1$, we get

$\left|\begin{array}{ccccc}x& & x+y& & x+y+z\\ 0& & x& & 2x+y\\ -x& & -y& & 2x-z\end{array}\right|=64$

Now apply $R_1+R_3$

$\left|\begin{array}{ccccc}0& & x& & 3x+y\\ 0& & x& & 2x+y\\ -x& & -y& & 2x-z\end{array}\right|=64$

Applying $R_1-R_2$, we get

$\left|\begin{array}{ccccc}0& & 0& & x\\ 0& & x& & 2x+y\\ -x& & -y& & 2x-z\end{array}\right|=64$

$⟹x\left(x^2\right)=64⟹x^3=64⟹x=4$