To find M^1 we use (1). As already noted,
2. Also adj M is given to us. Hence
1
1 1 1
1 1
8 6 2
2 2
5 3 1
M adjM
1 / 2 1 / 2 1 / 2
4 3 1
5 / 2 3 / 2 1 / 2
(5)
Putting this into (4), we get
1 / 2 1 / 2 1 / 2 1
4 3 1 1
5 / 2 3 / 2 1 / 2 1
(6)
which gives 1, 1 and 1 .Hence
3. Therefore (D) is correct.
For (B), we already know M adjM I 23.
So,adjM M 2 ^1.
Taking inverses of both the sides, we get
1 1
2
adjM M
(7)
As for the second term, we apply (1) with M
replaced by M^1. Since detM 2, det(M^1 )
1
2
.
Therefore
1 1 1 1 1
2 2
adjM M M
(8)
Adding (7) and (8), we see that (B) is correct
too.
Finally, for (C), we apply (1) again with M
replaced by M^2. Since
2 2
detM 2, det(M ) ( 2) 4. Therfore
adj M M I^2 ^2 4 3 (9)
We take determinants of both the sides. We
already know detM^2 4. The determinant
of 4 I 3 is 43 64 .Therefore
detadjM^2 equals 16 and not 81. So, (C) is
false.
SECTION-
13.Sol: Given
1 1 1 1 1
(1, 0, 0), , , 0 and , ,
2 2 3 3 3
A B C
Hence,
(^1) ˆ (^1) ˆ (^2) ˆ (^1) ˆ (^1) ˆ
and
2 2 3 3 3
AB i j AC i j k
So, area
1 1 1 2 1
2 2 2 3 4
AB AC
1
2 2 3
now
SECTION-
13.Sol:
14.Sol:
(6 )^23 0.
4
14.Sol: By Bayes theorem
1 2
1 2
2
( )
( / )
( )
P E E
P E E
P E
(1)
The sample space S has 29 elements in all
and they are equally likely. Therefore, the
probabilities of events are proportional to the
numbers of elements in the respective subsets.
Hence
1 2
1 2
2
( / )
E E
P E E
E
(2)
It is easier to find E 2. As each of the 9 entries
of a matrix A S has only 0 and 1 as possible
values, their sum will equal 7 if and only if
exactly 7 of the entries are 1 and the remaining
2 are 0. This gives
2
(^9) 9.
36
2 2
E
(3)
For a matrix A E 2 , the determinant will