TS EAMCET 6-Aug-2021 Shift 2 Question Paper

Given, [

1

−1

2

][

1−1

2

][

x

y

z

]=[

5

−5

10

] ⇒‌‌[

1	−1	2
−1	1	−2
2	−2	4

][

x

y

z

]=[

5

−5

10

]
⇒‌‌[

x−y+2z

−x+y−2z

2x−2y+4z

]=[

5

−5

10

] i.e
x−y+2z‌‌=5. . . (i)
−x+y−2z‌‌=−5. . . (ii)
2x−2y+4z‌‌=10. . . (iii)
From Eqs. (i), (ii) and (iii), we observe that the solution lies on a plane not parallel to any of the coordinate plane.