Question

Find x,y,z using cramers rule, if x-y+z=4, 2x+y-3z=0 and x+y+z=2

Answer 1

Answer:

The solution is

x=2, y=-1, z=1

Step-by-step explanation:

Given system of equations is

x-y+z=4

2x+y-3z=0

x+y+z=2

$\Delta=\left|\begin{array}{ccc}1&-1&1\\2&1&-3\\1&1&1\end{array}\right|$

$\Delta=1(1+3)+1(2+3)+1(2-1)$

$\Delta=4+5+1$

$\Delta=10$

$\Delta_x=\left|\begin{array}{ccc}4&-1&1\\0&1&-3\\2&1&1\end{array}\right|$

$\Delta_x=4(1+3)+1(0+6)+1(0-2)$

$\Delta_x=16+6-2$

$\Delta_x=20$

$\Delta_y=\left|\begin{array}{ccc}1&4&1\\2&0&-3\\1&2&1\end{array}\right|$

$\Delta_y=1(0+6)-4(2+3)+1(4-0)$

$\Delta_y=6-20+4$

$\Delta_y=-10$

$\Delta_z=\left|\begin{array}{ccc}1&-1&4\\2&1&0\\1&1&2\end{array}\right|$

$\Delta_z=1(2-0)+1(4-0)+4(2-1)$

$\Delta_z=2+4+4$

$\Delta_z=10$

By cramer's rule,

$x=\frac{\Delta_x}{\Delta}$

$x=\frac{20}{10}$

$x=2$

$y=\frac{\Delta_y}{\Delta}$

$y=\frac{-10}{10}$

$y=-1$

$z=\frac{\Delta_z}{\Delta}$

$z=\frac{10}{10}$

$z=1$

The solution is

x=2, y=-1, z=1