Question

Evaluate. $\left|\begin{array}{ccc}\frac{1}{z}& \frac{1}{z}& -\frac{(x+y)}{z^2}\\ -\frac{(y+z)}{x^2}& \frac{1}{x}& \frac{1}{x}\\ -\frac{y(y+z)}{x^{2}z}& \frac{x+2y+z}{xz}& -\frac{y(x+y)}{x{z}^2}\end{array}\right|$

Answer 1

Multiply $C_1,C_2,C_3$ by x,y and z respectively and hence divide by xyz.

$\therefore △=\frac{1}{xyz}$ $\left|\begin{array}{ccc}\frac{x}{z}& \frac{y}{z}& -\frac{(x+y)}{z}\\ -\frac{(y+z)}{x}& \frac{y}{x}\frac{z}{x}& \\ -\frac{y(y+z)}{xz}& \frac{x+2y+z}{xz}& -\frac{y(x+y)}{xz}\end{array}\right|$

Apply $C_1+C_3+C_3$ then $C_1$ becomes a column of zeros.

$\therefore △=0$