AP EAMCET Engineering 21 Apr 2019 Shift 1 Paper

Consider the equation. x2−25x+24=0
x2−x−24x+24=0
x(x−1)−24(x−1)=0
x=1,24
Consider the matrix and substitute 1 for k
A=[

1	2	1
3	2	3
1	1	1

]

Calculate the determinant.

|A|‌‌=1(2−3)−2(3−3)+1(3−2)
‌‌=−1+0+1
‌‌=0

k=1 is not possible because the given matrix A is singular
Substitute 24 for k
A=[

1	2	1
3	2	3
1	1	24

]

Calculate the determinant
[A]‌‌=1(48−3)−2(27−3)+1(3−2)
  ‌‌=45−138+1
  ‌‌=−92
The adjoint of the matrix is calculated as
adjA‌‌=[

45	−69	1
−47	23	1
4	0	−4

]
  ‌‌=[

45	−47	4
−69	23	0
1	1	−4

]
The inverse of the matrix is calculated as
A−1=‌

1

|A|

adjA
=−‌

1

92

[

45	−47	4
−69	23	0
1	1	−4

]