Question

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

• A. $0$
• B. $-25$
• C. $1$
• D. $-4$

Answer 1

Answer: (B)

$-25$

$x+y+z=4$

$x-y+z=2$

$x+2y+2z=1$

$(x=a,y=b,z=c$ is a solution)

$D=\left|\begin{array}{ccc}1& 1& 1\\ 1& -1& 1\\ 1& 2& 2\end{array}\right|=1(-4)-1\left(1\right)+1\left(3\right)$

$=2$

$D_1=\left|\begin{array}{ccc}4& 1& 1\\ 2& -1& 1\\ 1& 2& 2\end{array}\right|=4(-4)-1\left(3\right)+1\left(5\right)$

$=-14$

$D_2=\left|\begin{array}{ccc}1& 4& 1\\ 1& 2& 1\\ 1& 1& 2\end{array}\right|=1\left(3\right)-4\left(1\right)+1(-1)$

$=-2$

$D_3=\left|\begin{array}{ccc}1& 1& 4\\ 1& -1& 2\\ 1& 2& 1\end{array}\right|=1(-5)-1(-1)+4\left(3\right)$

$=8$

$a=\frac{D_1}{D}=\frac{-14}{2}=-7$, $b=\frac{D_2}{D}=\frac{-2}{2}=-1$

$c=\frac{D_3}{D}=\frac{8}{2}=4$

$ab+bc+ca=7-4-28=-25$

B is correct.