TS EAMCET 5 May 2018 Shift 1 Solved Paper

The system of equation is given as, $x+y+z=4$ $x-y+z=2$ $x+2y+2z=1$ From above, the determinant ∆ is calculated as, $\Delta=\begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 2 & 2 \end{vmatrix}$ $=1(-2-2)-1(2-1)+(2+1)$ $=-2$ The determinant $\Delta_{1}$ is calculated as, $\Delta_{1}=\begin{vmatrix} 4 & 1 & 1 \\ 2 & -1 & 1 \\ 1 & 2 & 2 \end{vmatrix}$ $=4(-2-2)-1(4-1)+1(4+1)$ $=-14$ The determinant $\Delta_{2}$ is calculated as, $\Delta_{2}=\begin{vmatrix} 1 & 4 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{vmatrix}$ $=1(4-1)-4(2-1)+1(1-2)$ $=-2$ The determinant $\Delta_{3}$ is calculated as, $\Delta_{3}=\begin{vmatrix} 1 & 1 & 4 \\ 1 & -1 & 2 \\ 1 & 2 & 1 \end{vmatrix}$ $=1(-1-4)-1(1-2)+5(2+1)$ $=8$ The value of x is calculated as, $x=\frac{\Delta_{1}}{\Delta}$ $=\frac{-14}{-2}$ $=7$ The value of y is calculated as, $y=\frac{\Delta_{2}}{\Delta}$ $=\frac{-2}{-2}$ $=1$ The value of z is calculated as, $z=\frac{\Delta_{3}}{\Delta}$ $=\frac{8}{-2}$ $=-4$ So, the values of, $a=7, b=1$ and $c=-4$ The value of the given expression is calculated as, $ab+bc+ca$ $=7(1)+1(-4)+(-4)7$ $=-25$