Question

If $x=a, y=b, z=c$ is the solution of the system of simultaneous linear equations $x+y+z=4$, $x-y+z=2, x+2 y+2 z=1$, then $a b+b c+c a=$

• A. 0
• B. -25
• C. 1
• D. -4

Answer 1

Answer: (B)

Given that,
$x+y+z=4$, $ x-y+z=2$ and $x+2 y+2 z=1$
$\left.\Delta=\begin{vmatrix}1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 2 & 2\end{vmatrix}=l(-2-2)-1(2-1)+(2+1)\right]$
$=-4-1+3=-2$
$\Delta_{1}=\begin{vmatrix}4 & 1 & 1 \\ 2 & -1 & 1 \\ 1 & 2 & 2\end{vmatrix}4(-2-2-1(4-1)+1(4+1)$
$=4(-4)-3+5=-16-3+5=-14$
$\Delta_{2}=\begin{vmatrix}1 & 4 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2\end{vmatrix}$
$\Delta_{2}=1(4-1)-4(2-1)+1(1-2)=3-4-1=-2$
$\Delta_{3}=\begin{vmatrix}1 & 1 & 4 \\ 1 & -1 & 2 \\ 1 & 2 & 1\end{vmatrix}$
$\Delta_{3}=1(-1-4)-1(1-2)+4(2+1)=(-5)+1+12=8$
$x=\frac{\Delta_{1}}{\Delta}=\frac{-14}{-2}=7$,
$y=\frac{\Delta_{2}}{\Delta}=\frac{-2}{-2}=1$
and $z=\frac{\Delta_{3}}{\Delta}=\frac{8}{-2}=-4$
So, $a=7, b=1, c=-4$
Now, $a b+b c+c a$
$=7(1)+1(-4)+(-4)(7)=7-4-28=-25$