AP EAMCET Engineering 20 Apr 2019 Shift 2 Paper

Consider the matrix,
(

2	2	1
1	3	1
1	2	2

)−x(

1	0	0
0	1	0
0	0	1

)
It is solved as,
(

2	2	1
1	3	1
1	2	2

)−x(

1	0	0
0	1	0
0	0	1

)=[

2−x	2	1
1	3−x	1
1	2	2−x

]
The matrix is singular. Therefore,
|

2−x	2	1
1	3−x	1
1	2	2−x

|=0
Apply operation C1→C1+C2+C3
|

5−x	2	1
5−x	3−x	1
5−x	2	2−x

|=0
(5−x)|

1	2	1
1	3−x	1
1	2	2−x

|=0
Apply operation R2→R2−R1 and R3→R3−R1
(5−x)|

1	2	1
0	1−x	0
0	0	1−x

|=0
(5−x)(1−x)2=0
x=1,5,5
Thus, the sum of the required value of x is,
Sum= 1 + 1 + 5 = 7