Answer to Question #323617 in Linear Algebra for tee

Question

1.use Cramer's rule to solve the equation below (10 marks)

y-z =2

3x+2y+z=4

5x+4y=1

Answer 1

$\Delta=\begin{vmatrix} 0& 1&-1 \\ 3& 2&1\\ 5&4&0 \end{vmatrix}=0+(-1)^3(-5)-(-1)^4(12-10)=3$

$\Delta_{x}=\begin{vmatrix} 2& 1&-1 \\ 4& 2&1\\ 1&4&0 \end{vmatrix}=(-1)^22(-4)+(-1)^3(-1)+(-1)^4(-1)(16-2)=-8+1-14=-21$

$\Delta_{y}=\begin{vmatrix} 0& 2&-1 \\ 3& 4&1\\ 5&1&0 \end{vmatrix}=(-1)^32(-5)+(-1)^4(-1)(3-20)=27$

$\Delta_{z}=\begin{vmatrix} 0& 1&2 \\ 3& 2&4\\ 5&4&1 \end{vmatrix}=(-1)^3(3-20)+(-1)^42(12-10)=21$

x=-21/3=-7

y=27/3=9

z=21/3=7