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,
Δ=|

1	1	1
1	−1	1
1	2	2

|
=1(−2−2)−1(2−1)+(2+1)
=−2
The determinant Δ1 is calculated as,
Δ1=|

4	1	1
2	−1	1
1	2	2

|
=4(−2−2)−1(4−1)+1(4+1)
=−14
The determinant Δ2 is calculated as,
Δ2=|

1	4	1
1	2	1
1	1	2

|
=1(4−1)−4(2−1)+1(1−2)
=−2
The determinant Δ3 is calculated as,
Δ3=|

1	1	4
1	−1	2
1	2	1

|
=1(−1−4)−1(1−2)+5(2+1)
=8
The value of x is calculated as,
x=

Δ1

Δ

=

−14

−2

=7
The value of y is calculated as,
y=

Δ2

Δ

=

−2

−2

=1
The value of z is calculated as,
z=

Δ3

Δ

=

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