TS EAMCET 2017 Code A Solved Paper

Consider the given matrix A.
A=[

1	0	1
0	2	0
1	−1	4

]
The transpose of matrix A is,
AT=[

1	0	1
0	2	−1
1	0	4

]
The value of (A+AT) is calculated as
A+AT=[

2	0	2
0	4	−1
2	−1	8

]
The value of (A−AT) is calculated as
A−AT=[

0	0	0
0	0	1
0	−1	0

]
The matrix A can be calculated as,
A=‌‌

1

2

(A+AT)+‌‌

1

2

(A−AT)
A=[

1	0	1
0	2	−0.5
1	−0.5	4

]+[

0	0	0
0	0	0.5
0	−0.5	0

]
A=B+C
The matrix C is,
C=[

0	0	0
0	0	0.5
0	−0.5	0

]=−CT