Discussion Forum

If x=a  ,y=b,z=c is the solutions of the system of simultaneous linear equations x+y+z=4 , x-y+z=2 , x+2y+2z=1 , then ab+bc+ca=

Answer:

Explanation:

Given that,

x+y+z=4, x-y+z=2 and x+2y+2z=1

$\triangle$= $\begin{bmatrix}1 & 1&1 \\1 & -1&1\\1&2&2 \end{bmatrix}=1(-2-2)-1(2-1)+(2+1)]$

  =  -4-1+3=2

$\triangle_{1}=\begin{bmatrix}4 & 1&1 \\1 & -1&1\\1&2&2 \end{bmatrix}=4(-2-2)-1(4-1)+1(4+1)$

  =$4(-4)-3+5=-16-3+5=-14$

$\triangle_{2}=\begin{bmatrix}1 &4&1 \\1 & 2&1\\1&1&2 \end{bmatrix}$

$\triangle_{2}=1(4-1)-4(2-1)+1(1-2)=3-4-1=2$

$\triangle_{3}=\begin{bmatrix}1 &1&4 \\1 & -1&2\\1&2&1 \end{bmatrix}$

$\triangle_{3}=$  1(-1-4)-1(1-2)+4(2+1)=(-5)+1+12=8$

 $x=\frac{\triangle_{1}}{\triangle}=\frac{-14}{2}=7$, $y=\frac{\triangle_{2}}{\triangle}=\frac{-2}{-2}=1$

and $z=\frac{\triangle_{3}}{\triangle}=\frac{-8}{2}=-4$

 So, a=7, b=1, c=-4

Now, ab+bc+ca

 = $7(1)+1(-4)+(-4)(7)=7-4-28=-25$