matrices and determinants
DESCRIPTION
maths for classes 12 and higher very exhaustiveTRANSCRIPT
-
LEVEL - IModel Question :Problem based on operation on two matricesOne correct answer MCQ
1. If
0214222322360
2330164
7243
bcxy
czbbax
yzx
Then value of
a+b+c+x+y+ z is(A) -16 (B) - 11(C) 20 (D) - 52
2. If Y=
4123
and
2301
2 YX , then the sum of element of X is equal to
(A) - 5 (B) -6(B) - 10 (D) -12
3. Write the following as a single matrix : 752057420512
321
(A) 151221 (B) 101221 (C) 101521 (D) 151241
4. If
5207
YX and
3003
YX then the sum of the elements of the matrix 3X -4Y
is equal to (A) 14 (B) 16 (C) 12 (D) 25.5 If A and B are square matrices of the same order, then which of the following are not true ? (A) ''' BAAB (B) ''' ABAB (C) 0AB 0AB If 0A or 0B (D) AB=0 If A = I or B =I
6. If A =
0110
and B =
0
0i
i, then
(A) IA 2 (B) IA 2
(C) IB 2 (D) IB 2
-
7. If abccabbca
111
1 and 2
2
2
111
2
ccbbaa
, then which of the following is not true ?
(A) 21 (B) 21
(C) 21 2 (D) 12 28. For the equations 4955,232,132 zyxzyxzyx
(A) there is only one solution (B) there exist infinitely many solutions (C) there is no solution (D) the equations are consistent.
9. Statement 1 : The determinants
2
2
2
111
1
1
ccbbaa
andabccababca
are not identical.
because Statement 2 : The first two columns in both the determinants are identical and third columns are different. (A) Statement -1 is true, statement=2 is true ; statement -2 is a correct explanation for statement -1 (B) Statement -1 is true, statement-2 is true; statement-2 is not a correct explanation for statement-1 (C) Statement -1 is true, statement- is false. (D) Statement-1 is false, statement -2 is true.
10. Statement :1 If thenii
A
721212
det (A) is real .
Statement : 2 If A = ,2221
1211
aaaa
A ija being complex numbers then det (A) is always real.
(A) Statement -1 is true, statement=2 is true ; statement -2 is a correct explanation for statement -1 (B) Statement -1 is true, statement-2 is true; statement-2 is not a correct explanation for statement-1 (C) Statement -1 is true, statement- is false. (D) Statement-1 is false, statement -2 is true.
Comprehension-1 321,.123012001
andUUifUA
are columns matrices satisfying
andAUAU
032
,001
21 UAU ,132
3
is 33 matrix whose columns are
321 ,, UUU then answer the following questions
-
11. The value of [U] is(A) 3 (B) -3(C) 3/2 (D) 2
12. The sum of the elements of 1U is(A) -1 (B) 0(C) 1 (D) 3 Comprehension
(A) 1 (B) 0(C) 2 (D)
13. The value of [3 2 0]
023
is
(A) [5] (B)
25
(C) [4) (D)
23
MatrixMatch TypeThis section contains 2questions. Each question contains statementsgiven in two column which have to be matched. Statements (A, B, C, D)in Column I have to be matched with statements (p, q, r, s) in Column II.The answers to these questions have to be appropriately bubblesas illustrated in the following example.If the correct matches are Ap, As, Bq, Br, Cp, Cq and Ds,then the correctly bubbled 4 4 matrix should be as follows :
p q r s
p q r s
p
p
p
q
q
q
r
r
r
s
s
s
A
B
C
D
14 Column-I Column-II
(A) Idempotent (p)
2 2 41 3 4
1 2 3
(B) Orthogonal (q)
4 1 43 0 43 1 3
(C) Unitary (r)
1 i i 12 2
1 i 1 i2 2
(D) Involutory (s)cos sinsin cos
-
PQ15 The matrix X is equal to such that 3A - 2B + X = 0 , where
2312
;3124
BA
(A)
31416
(B)
53416
(C)
5344
(D)
8344
16 Let and
02
tan2
tan0
I is the identity matrix of order 2. Then ( I -A).
cossinsincos
is equal to (A) 2I+2A (B) 2I -2B
(C) I - A (D) I +A
17. If
7101
A and
1001
I , then find k so that kIAA 82 .
(A) 7 (B) 8(C) 10 (D) none of these.
18 If A= '
2'
1 21&
21,
00
0AAQAAifQ
cbcaba
Then 21.QQ is equal to
(A) 3I (B) 3O(C) A (D) None of these.
19 If
221321121
,212101654
,131201321
CBA then A - 2B + 3C is equal to
-
(A)
311696261410
(B)
315696261410
(C)
315696261810
(D)
3156156261810
20. . The system of linear equations 023,02,0 yxzyxzyx (A) has no solution (B) has a unique solution (C) is consistent (D) an infinitely many solutions.
21 . The value of satisfying 0
11142782cos
323sin
is
(A) n (B) 6
n
(C) 62 n (D) 6
1 nn
22 Statement 1 : If A is a matrix of order ,22 then ladjA = A . because
Statement 2 : A = TA . A) Statement -1 is true, statement=2 is true ; statement -2 is a correct explanation for statement -1 (B) Statement -1 is true, statement-2 is true; statement-2 is not a correct explanation for statement-1 (C) Statement -1 is true, statement- is false. (D) Statement-1 is false, statement -2 is true.
23 Statement -1 : If areaaa n..............., 21 in G.P.. ( 0ia for all i ) 0
logloglogloglogloglogloglog
876
543
21
nnn
nnn
nnn
aaaaaaaaa
because Statement : The three elements in any row of the determinant are in A.P. (A) Statement -1 is true, statement=2 is true ; statement -2 is a correct explanation for statement -1 (B) Statement -1 is true, statement-2 is true; statement-2 is not a correct explanation for statement-1 (C) Statement -1 is true, statement- is false. (D) Statement-1 is false, statement -2 is true.
-
Comprehension - 2 is an imaginary cube root of unity..
24 is an imaginary cube root of unity . 1
11
45
43
53
(A) 3 (B) - 3 (C) 21 (D) 21
25 A root of the polynomial
xx
x
11
1
2
2
2
is
(A) 1 (B) 0 (C) 2 (D)
26 The value of Nnnn
nn
nn
?1
11
2
2
2
=
(A) (B) 2 (C) 1 (D) 0
MatrixMatch TypeThis section contains 2questions. Each question contains statementsgiven in two column which have to be matched. Statements (A, B, C, D)in Column I have to be matched with statements (p, q, r, s) in Column II.The answers to these questions have to be appropriately bubblesas illustrated in the following example.If the correct matches are Ap, As, Bq, Br, Cp, Cq and Ds,then the correctly bubbled 4 4 matrix should be as follows :
p q r s
p q r s
p
p
p
q
q
q
r
r
r
s
s
s
A
B
C
D
27 If C & D are symmetric matrices of order 3 3Column-I Column-II(A) CD + DC (p) symmetric(B) C + D (q) Antisymmetric(C) CD DC(D) C D
-
Answers1.(A) 2.(A) 3.(C). 4(A). 5.(B)(C)(D) 6.(B) (C) 7.(B)(C)(D) 8(C) (D) 9.(D) 10.(D)11.(A) 12(B) 13(A) 14() 15(B) 16(D) 17(A) 18(B) 19.(A) 20(A) (D) 21.(A) (D) 22.(B)(A) (D)23. (A) 24.(AQ) 25.( B) 26(D) 27.(C)
-
Level - IM.Q Solution1. As the given matrices are equal, therefore, their corresponding elements must be equal. Comparing
the corresponding elements we get)....(03 ix )....(64 iiz )....(2372 iiiyy
),.....(264 ivxx )....(31 va )....(220 vic )...(423 viibb ).....(213 viiib )...(02 ixcz
From (i) andyiiifromx 5)(,3 from (ii) z =2From (V) a = -2, from (viii) b = - 7 and from (vi) c- 1Thus a = - 2, b = -7, c = - 1, x = -3, y = - 5, z = 2
2.
4123
2301
22 YYXX
2422
42132031
1211
2422
21X
3. 33
057420512
321
5.32)2()1(1)7(30.22.1[ 31]0.34)2(5.1
75231019752057420512
321
4.
3003
5207
2
2
YXYXX
XYXYX
82010
35020037
-
4105
82010
21X
Again
1102
2204
21
2204
35020037
3003
5207
)()(2
Y
YXYXY
5. )(),(),( dca
6
1001
0110
01102A
1001
00
00
00
2
22
ii
ii
ii
B )(),( cb
7. 111
1
111
2
2
2
2
2
2
ccbbaa
abcccabcbbabcaa
abcabccabbca
21
)(),(),( dcb
8. 0023112
111, Here
)(),( dc9. ANSWER = D
10. clearly 9721212
ii
A Which is real SI is not current
-
11.
zyx
U Given
001
123012001
001
1
zyx
AU
001
2302
1
zyxyx
x
02302
1
zyxyx
x
Solving we get 12
1
zyx
12
1
1U
41
2
032
22 UAU
and
31
2
132
33 UAU
Hence U=
341112
221U
12. Adj 3369357
021
UandU
369357
021
311
UadjUU
-
Sum of elements of 01 U
13.
023
341112
221023
023
023 U
= 583023
441
14. P,Q,R,S be the matrices corresponding to the options (P) (Q) , (R), (S) respectivelyP,Q,S, have real elements
TT
TT
TT
SS
QQ
PP
22
2
2
).........(..........PP
APPPP
PP
TT
).........(........... 3
2
2
DIQQQQ
QIQ
TT
)...(..........
.
2
22
22
2
CIRR
IRRIRRR
T
))((........... 22
22
2
CBISSSS
SSIS
T
-
15
2312
;3124
BA
Now 3A- 2B + X =0X = - 3A + 2B
2312
23124
3
= 53416
4624
93612
16.
222
2
2
2
12
2/tan12/tan2sin,
11
2/12/tan1cos
tt
tt
t
where t2tan
Also
11
00
1001
tt
tt
AI
And.
11
00
1001
tt
tt
AI
Now
cossinsincos
AI
2
2
2
22
2
11
12
12
11
11
tt
tt
tt
tt
tt
2
2
2
2
22
2
2
2
22
2
2
2
11
12
12
11
11
12
12
11
tt
tt
tt
ttt
ttt
tt
tt
tt
AItt
1
1
Hence,
cossinsincos
AIAI
P.Q solutions
-
17 Given , A =
71
01
49801
7101
71012 AAA
Again, 8A+ kI = 8
k
kk
kk
56808
00
56808
1001
7101
k
kkIAA
56808
49801
82
k 81 and 56 + k = 49 7 k
18
1121
21
00
0
AAAAA
cbcaba
A
000000000
000000000
21
00
0
00
0
21
1
cbcaba
cbcaba
Q
Similarly
00
0
21 1
1
cbcaba
AAQ 321 QQQ
19 A - 2B + 3C = 0
663963363
424202
12108
131201321
311696261410
-
20 Here ,
106
,21
321111
BA
If equations have no solution, then |A| = 0 and (adj A). 0BCalculating |A| = 303
Also, adj 01022010
..110220
110
BAadjA
da ,10
21 0162cos143142cos1123sin10 0202cos203sin10 022cos23sin 02sin42sin43sin 23
0sin0sin3sin4sin4 23 or
da
21sin
22 In fact adj A is got by keeping the diagonal elements as they are and changing the sign of theother two elements.
b a d
23.
rmarmarma
rmarmarmrmrmarma
log)7(loglog6loglog5loglog4loglog3loglog2log
log1loglog)(loglog1log
rrrrrr
rnarnarna
log3log3log3log3log3log3
log1loglogloglog1log Using 122( RRR and )233 RRR
= 0Hence = (A)
-
24. 3211454
443
5353
45
43
53
1111
1
11
1CCCC
.,,110
1212
43
22
etcas
= 12222
22
10012
CCR
= 32322 12 3122212 24222
25. 32112
22
22
2
2
2
1111
1
11
1CCCC
xxxx
x
x
x
0
011
1 222
xxxx
x
When x =0 0 is a root
26. nnnnnnn
nn
nn
nn
42223
2
2
2
11
11
110.111 3 nn = 0 answer D27.
-
28.
)(
05
05
CAnswer
yxx
yy
x
-
Level - II28 If A=
05y
xA and A = thenAT ,
(a) x=o. y = 5 (b) x + y = 5(c) x = y (d) none of these
29 If
126523
,652132
BA and A +B - C =0, then C =
(a)
778655
(b)
5344
11
(c)
776855
(d)
577855
30 If A and B are square matrices of order 3, then(A) adj (AB) = adjA + adj B (B) (A + B)1 = A1 + B1(C) AB = 0 |A| = 0 or |B| = 0 (d) AB = 0 |A| = 0 and |B| = 0
31. The inverse of the matrix iscb
aB
101001
(a) =
101001
cbaca (b) =
100001
cba
(c) =
101001
baca (d)
10010
1c
baca
32. The values of and for which the system of equations. zyxxyxzyx 2,1032,6 have no solution are
(A) 3 (B) 10 (C) 3 (D) 3
MQ solution
-
33. If rqp TTT ,, are the thth qp , and thr terms of an A.P ., then 11141 rqp
TTT rqp
cannot be equal to
(A) 1 (B) - 1 (C) 0 (D) p + q + r
34 Statement 1 : If A = 22
22
22
xcaxbcbxacaxbcxbxcabbxaccxabxa
and xabaxcbcx
B
, then .2BA
(A) Statement -1 is true, statement=2 is true ; statement -2 is a correct explanation for statement -1 (B) Statement -1 is true, statement-2 is true; statement-2 is not a correct explanation for statement-1 (C) Statement -1 is true, statement- is false. (D) Statement-1 is false, statement -2 is true.
35 Let A be a 2 2 matrix with real entries. let I be the 2 2 identity matrix. Denote by tr(A), the sumof diagonal entries of A. Assume that A2 = I.Statement1: : If A I and A I, then det A = 1Statement2: : If A I and A I, then tr(A) 0.
(A) Statement -1 is true, statement=2 is true ; statement -2 is a correct explanation for statement -1 (B) Statement -1 is true, statement-2 is true; statement-2 is not a correct explanation for statement-1 (C) Statement -1 is true, statement- is false. (D) Statement-1 is false, statement -2 is true.
III. For a given square matrix A, if there exists a matrix B such that AB = BA = I, then B is calledinverse of A. Every nonsingular square matrix passes inverse and it exists if |A| 0.
A1 = adj(A)det (A) adj A = |A| (A
1).
36 Let, a matrix A = 2 31 2
, then it will satisfy the equation
(A) A2 4A + I = 0 (B) A2 + 4A + I = 0(C) A2 4A 5I = 0 (D) A2 4A + 5I = 0
37 Let, a matrix A = 2 31 2
, then AA1 will be
(A) 2 31 2
(B) 3 2
2 1
(C) 1 2
2 3
(D) 2 3
1 2
-
38 Let matrix A = 3 21 1
satisfies the equation AA2 + aA + bI = 0, then the value of 4b
3
a
x . cosxdxequals
(A) a ba b
(B) a 2ba b
(C) a 4b4a b
(D) a 4b4a b
MatrixMatch TypeThis section contains 2questions. Each question contains statementsgiven in two column which have to be matched. Statements (A, B, C, D)in Column I have to be matched with statements (p, q, r, s) in Column II.The answers to these questions have to be appropriately bubblesas illustrated in the following example.If the correct matches are Ap, As, Bq, Br, Cp, Cq and Ds,then the correctly bubbled 4 4 matrix should be as follows :
p q r s
p q r s
p
p
p
q
q
q
r
r
r
s
s
s
A
B
C
D
39. Let
1tan
tan1x
xA
COLUM I COLUM -II
(A) 1A (P)
1tan
tan1x
x
(B) 1AdjA (Q)
1tantan1
xx
(C) AdjAAdj (R)
xxxx2cos12sin
sin2cos121
(D) AAdj 2 (S)
xxxx2cos12sin
2sin2cos1
-
PQ solution
40 If A is
42476
268
is a singular matrix , then =
(A) 3 (B) 4(C) 2 (D) 5
41 If A = ,12100
,0201
B then
(A) AB = O, BA =O (B) OBAOAB ,(C) OBAOAB , (D) OBAOAB ,
42. If
5005
A and
dcba
B , then AB =
(A) B (B) 5B(C) 5B (D) none of these.
43 If A =
00
0
abacbc
A and
2
2
2
cbcacbcbabacaba
B , then AB =
(A) A (B) B(C) I (D) O
44. If F (X) =
yy
yyyGxx
xx
cos0sin010
sin0cos)(,
1000cossin0sincos
then [F(x). 1]yG is equal to
(A) F (X) G(-Y) (B) 11 yGxF(C) 11 xfyG (D) G (-y) F(-x)
45 IF
then,1sin1
sin1sin1sin1
lies in the interval
(A) [2,3] (B) [3,4](C) [2,4] (D) (2,4)
-
46 The determinant
cossincossincossin
2cossincos
is equal to
(A) (B) (C) and (D) Neither nor
47. If , CBA then the value of
0tan)cos(tan0sincossinsin
ABAAB
BCBA
is
(A) 0 (B) 1(C) 2 sinB tan A cosC (D) None of these.
48. Suppose x, y, z are positive and none of x,y, z is 1. If
thenzyxyxzxzy
yy
xx
,sincossinlog1logloglog1
2 is independent of
(A) x (B) y(C) y and z only (D) z
49. If
124
321612
33
22
nnznrrnnyr
nnxrSr , then the value
n
rrS
1 is independent of
(A) x (B) y(C) n (D) z
50 Statement 1 : If
012101210
, |A| = 0
BecauseStatement 2 : The value of the determinant of a skew symmetric matrix is zero
(A) Statement -1 is true, statement=2 is true ; statement -2 is a correct explanation for statement -1 (B) Statement -1 is true, statement-2 is true; statement-2 is not a correct explanation for statement-1 (C) Statement -1 is true, statement- is false. (D) Statement-1 is false, statement -2 is true.
-
51. Statement 1 : The inverse of the matrix A=
113111
111
Does not exist
becauseStatement 2 : |A| =0
(A) Statement -1 is true, statement=2 is true ; statement -2 is a correct explanation for statement -1 (B) Statement -1 is true, statement-2 is true; statement-2 is not a correct explanation for statement-1 (C) Statement -1 is true, statement- is false. (D) Statement-1 is false, statement -2 is true.
Paragraph Let
;..............;..............
22222121
11212111
bxaxaxabxaxaxa
nn
nn
nnnmnn bxaxaxa ....2211 be a system of n linear equations in n unknowns. Then this can bewritten in the matrix form as
AX = B Where A =
nn
ij
b
bbb
B
x
xxx
Xnma
.
.
.
.
.
.;][3
2
1
3
2
1
Then (I) If |A| 0, the system is consistent, and has a unique solution given by BAX 1 (II) If |A| =0 and (adj A) B =0, then the system is consistent and has infinitely many solutions. (III) If (A) = 0 and (adj A ) B 0, then the system is inconsistent.
52. The system of equations 1,52,132 zyxzyxzyx has(A) a unique solution (B) infinitely many solutions(C) no solutions (D) finite number of solutions.
53 Let 1023,1223,42 kzyxzyxzyx . The value of k in the above system ofequations so that system does not have a unique solution is(A) 2 (B) 3(C) -1 (D) -2
-
54. If yxzyxzyx 2,1032,6 , the values of and , for which the systemhas infinitely many solutions is(A) 9,3 (B) 10,3 (C) 10,2 (D) 3,10
ANSWERS
28(C) 29(B) 30 (C) 31(A) 32 (A)(B) 33 (A)(B) (D) 34 (A) 35 (C) 36 (A) 37 (A) 38(C)39 (A)-(R) ;(B)-(S); (C)-(P);(D)-(Q) 40 (A) 41(B)42 (B) 43 (D) 44(D) 45(C) 46 (C) 47 (A)48 (A)(B)(D) 49 (A) (B) (C) (D) 50 (C) 51 (A) 52 (A) 53 (B) 54(B) 55 (B) 56 (A) 57 (D)58 (C) 59(B) 60(A) (C) (D) 61 (C) 62(D) 63 (C) 64 (C) 65(A) 66 (A) (B)(C) (D)67(A) (B)(C) (D) 68 (B)69 (D) 70 (A)71(D)
-
ANSWERS & SOLUTIONS
LEVEL -II M.Q29. A+B - C = 0
)(0
50
5
BAnswer
yxx
yy
x
30. 00)( AABC or |B| =0
31. adj A =
101001
cbaca
|A| = 1 {Upper triangular matrix }
)(101001
1 AAnswercbac
aA
32. The required conditions are |A| = 0 and (adj A ) B=0
021
321111
and
000
106
110213
1262
i.e., 00362 and 0.6 - 10 + 0
10,3
-
33. ,1 dpaTp
Similarly, rqandTT
111
111rqp
dradqadpa
0111111111
rqpddd
rqprdqdpd
rqpaaa
)(),(),( DBAAnswer
34.
35 Answer (C)
36
2132
21322A
=
74127
43226634
20000
1001
1001
O
37.
AIAAIA
IAAAAAIAA
IAA
444
404
11
1121
2
2
2132
4004
2132
1001
41A
=
2132
-
38. Clearly a = -4, b = 1
4
4
3 0cos xdxx
Also 014444
4
ba
aba Answer (C)
39
1tantan1
)(1tan
tan1)()();();()();()(
xx
Aadjx
xA
QDCSBRA
xxxxxx
xx
xAAadjA 2
2
21
coscossincossincos
1tantan1
tan11
xxxx
2cos12sin2sin2cos1
21
RA )(
Ax
xAdjAAdj
1tantan1
xxxxA
AdjAAdjAAdjAdjA
2cos12sin2sincos1
21
sec21
PCandSB ;
QDxx
1tantan1
2
40 0A ( A is singular)
042
476268
A
)(3
6020020483612856
AAnswer
-
41.
002500
00000
BA
AB
Answer (B)
42.
dcba
AB5050
0505 Answer (B)
43. 0000000000
33
AB
Answer - (D)
44. ifxFyGyGxF 111 .)().({ 0.0 xGxf
Here , 1.1 yGxf
Also , xFxxxx
xF
1000cossin0sincos
1
and
yGyy
yxyG
cos0sin010
sin0cos1
xFyGygxf 111. xFyG 1
45.
)(:
4,2sin22
1sin1sinsinsinsin112
22
CAnswer
-
46
cossincossincossinsinsincossinsin
cossincossincossin
coscossincoscos 22
21
Apply 321 cos RRR 22111 sin inRRRand
)(000CAnswer
47.
0)tan(sincos)tan(cossin0tancos
tan0sincossin0
ABCACBAC
ABCB
Answer = (A)
48.
0
sin)cos(sinloglogloglogloglog
loglog1
2
zyxyxzyxzyx
yx Using
abba log
loglog
Answer (A) ,(B) ,(d)
49.
0113232
11
124
3216
12
33
22
3
1
3
2
1
2
1
1
nnznnnnynn
nnxnn
nnznrr
nnyr
nnxr
Sn
r
n
r
n
rn
rr
( as 1C and 3C are same)Answer (A) ,(B) (C) (D)
50. Because the value of the determinants of skew - symmetric matrix of odd orderis zeroAnswe (C)
51 Answer (A)
-
52
0113211212111211
312
AA The solution is unique
Answer (A)53. If the system does not have a unique solution the value of the determinant of coefficients =0
023
231112
k
54. The required conditions are 0A and (Adj A) B = 0
,021
321111
and
000
106
110213
1262
i.e., 00362 and 0.6 - 10 + 010,3 Answer - (B)
-
Level -IIIMQ55 If A is the diagonal matrix diag ( ndddd ..............., 321 ) then ,, NnAn is
(A) diag nndndndnd ........,, 321 (B) diag nnnnn dddd ....,.........,, 321(C) diag 1131211 ....,.........,, nnnnn dddd (D) none of these.
56 If
abba
1tantan1
1tantan1
, then
(A) a = 2sin,2cos ba (B) a =1, b= 1(C) 2cos,2sin ba (D) none of these.
57 If A and B are any aa matrices , then det (A+B) = 0 implies(A) det A + det B = 0 (B) det A = 0 or det B = 0(C) det A =0 and B = 0 (D) None of these.
58 If
23213
xxx
A is a symmetric matrix, then x =
(A) 4 (B) 3(C) -4 (D) -3
59 If
13321210
aA and 1A
2/12/32/534
2/12/12/1c then
(A) 2/1,2 ca (B) 1,1 ca(C) 1,1 ca (D) 2/1,2/1 ca
60. If f(x) is a polynomial satisfying
xf
xffxfxf
11
21
21
and f (2) = 17 , then f (5)
(A) 126 (B) 626(C) 124 (B) 624
-
61. The number of distinct real roots of 0sincoscoscossincoscoscossin
xxxxxxxxx
in the interval isx 440
(A) 0 (B) 2(C) 1 (D) 3
LEVE L IIIP.Q62 For what value of x, the matrix,
isx
xx
142142223
singular
(A) x =1,2 (B) x =0,2(C) x= 0,1 (D) x =0,3
63 If the trace of the matrix : A =
6402132114235201
2
2
xx
xx
is o then x is equal to
(A) (-2, 3) (B) (2,-3)(C) (-3, 2) (D) (3,-2)
64 If A and B are square matrices of order 3, then
(A) adj (AB) = adjA+adjB (B) 111 BABA(C) 0|| AOAB or |B| (D) OAOAB or B =O
65 If andA
496121132
thenB ,103122131
-
(A) AB = BA (B) BAAB
(C) BAAB 21
(D) none of these.
66 If A=
1000sincos0cossin
then which of the following are true ?
(A) 1TA (B) 11 A
(C) adjAA 1 (D) 1TAA
67 If matrix ,
bacacbcba
A where a.b. c are real positive number , abc = 1 and ,1AAT
then which of the following is/are true.(A) a + b+ c =1 (B) 1222 cba(C) ab + bc+ ca =0 (D) 4333 cba
68 Statement - 1 : If a, b, c are real positive numbers with abc = 1 and ,1TAA where A =
bacacbcba
, then 4333 cba because
Statement - 2 : 0111
cba(A) Statement -1 is true, statement=2 is true ; statement -2 is a correct explanation for statement -1 (B) Statement -1 is true, statement-2 is true; statement-2 is not a correct explanation for statement-1 (C) Statement -1 is true, statement- is false. (D) Statement-1 is false, statement -2 is true.
r us called the rank of the matrix. A if there exists at least one nonzero minor or order r and everyminor of order r +1 of the matrix equals zero. Again if A is a square matrix and I is the unit matrixof the same order then the equation |a-xI| = 0 is called the characteristic equation and the rootsof the characteristic equation are called eigenvalues of the matrix.A.
Let
442331
311A
69 The characteristic equation of A is(A) 045183 xxx (B) 0496 23 xxx(C) 04518 23 xxx (D) 08203 xx
-
70 The inverse of matrix A is
(A) 281
25 AI (B) AI 8
125
(C) 261
25 AI (D) None of these.
71 The sum of elements of the matrix A is(A) 5/3 (B) 5/4(C) 5/2 (D) None of these.
ANSWERS
55 (B)56 (A) 57 (D)58 (C)59 (B) 60(A) (C) (D) 61 (C) 62(D) 63 (C) 64 (C) 65(A) 66 (A) (B)(C) (D)67(A) (B)(C) (D) 68 (B)69 (D) 70 (A)71(D)
Level - IIIM.Q Solutions55. Given A = diag ndddd .......,, 321 Now AAA 2
nn d
dd
d
d
dd
d
.......000...............................................................0...........000.........000........00
.......000...............................................................0...........000.........000........00
3
2
1
3
2
1
-
56.
adjAA
A 11tan
tan1tan11
1tantan1 1
2
=
1tantan1
tan11
2
abba
1tantan1
1tantan1
abba
1tantan1
tan11
1tantan1
2
abba
2
2
2 tan1tan2tan2tan1
tan11
cos
tan1tan1
2
2
a and sin
tan1tan2
2 b
57. A is symmetric AA 1
23213
421323
xxx
xxxx
4321 xxxHence (c) is the correct answer
58. Det (A+B) cannot be expressed in terms of det A and det B . Hence the given equation givesno inference.Hence (d) is the correct answer
59. If we must have IAA 1
(3.1)th entry of 01 AA = (3,1)th entry of 1AA
02514
213 a
2121
210 c
044 a and c+1 =0 a =1 and c = - 1.Hence (b) is the correct answer
-
60.
x
fxfx
fxfxfx
fx
fxfxf 11112
nxxf 1Now, as f(2) = 17
62651 4 fxxfAnswer (A),(C) (D)
61
3211sinsincoscoscossincos
cos2sincos2sincos2sinRRRRgU
xxxxxx
xxxxxx
0cos2sincossin0cos
0cossincos001
xxxxx
xxx
322sin CCCgU and 133 CCC
0cos2sincossin 2 xxxx xx cossin 0cos2sin xx
only one solution.Answer ( C)
62. Since , the given matrix is singular
0142142223
xx
x
0142
110223
0142
2223
32
x
x
xxx
xRR
3,0033
0}2220413{
xxxxxxx
-
63 Trace of matrix is defined as 012222
1
xxan
iii
2,3 xHence (C) is the correct answer.
64. If AB=O then either of A and B are necessarily singular.Hence (C) is the correct answer.
65.
103122131
496121132
AB
.100010001
49612113201818043066
121863413621
AB
Hence (A) is the correct answer
66 TA transpose of A=
1000sincos0cossin
1sincoscossin
1000sincos0cossin
TA
Also, 1.A ASoAT
adjAAadjAA 1
1000sincos0cossin
1
adjAA .
11000sincos0cossin
1
A
-
67. Here ,
bacacbcba
A So,
bacacbcba
AT , interchanging rows and columns.
2
2
Abacacbcba
AAT
;2AAAT but IAAT (given)
Now, |I| = 22 1 AA
Now, bacacbcba
bacacbcba
A111
3211 RRRR
cbcacbabcbcba
bacacbcba
A
001
133
122
CCCCCC
}{ cabacbbccba bcabacabccbcba 222 2 abcabccbacba 222 abccba 3333
13333 abccba )1........(131 3332 cbaA
As a,b, c are positive, 133
333333
abccbacba
.3333 cba 13)1( 333 cba
4333 cba
-
71. Answer = (A)
73.The characteristic equation is 0442
331311
i.e., 08203
74. By Cayley- Hamilton theorem, 08203 IAA where 21
81
25 AIA
75. 21 81
25 AIA
4/14/14/14/34/14/5
2/313
2222622101284
81
100010001
25
-
ADDITIONAL QUESTIONS
72. If
cfgfbhgha
Then which of the following is are true ? where A,B,C,F,G,H are the
co factor & of a,b,c,f,g,h(A) aFBC 2 (B) bGCA 2
(C) cHAB 2 (D) None of these.
73 The value of
CABABCBCA
2sinsinsinsin2sinsinsinsin2sin
is
(A) 1 (B) 4 sin A. SinB . Sin C(C) 0 (D) None of these.
74. If
cfgfbhgha
& If A, B, C, F, G, H, are the co factors of a,b,c,f,g,h then which of the
following is (are) true ?(A) FAFGH (B) GBGHF(C) HCHFG (D) None of these
75. Let hgxfxexdxcxbxax 234567
thanxxxx
xxxxxxxxx
1221
21
22
22
22
(a) g=3 and h = -5 (b) g= -3 and h = -5 (c) g = -3 and h = -9 (d) none of these.
76.If ,111))()((
32
32
32
zyxxzxyyx
zzzyyyxxx
nnn
nnn
nnn
then n equals
(a) 1 (b) 2 (c) 3 (d) none o these
-
77 Let{ k .....,3,2,1 }be the set of third-order determinants that can be made with the distinct nonzero
real numbers 9,3,2,1 .....aaaa .Than
(a) k = 9! (b)
k
ii
10 (c) at least one 0i (d) none of these
78 Let ,21
)(
22
11
22
11
nn
nn
nn
nn
nn
nn
CCCPPP
nnnnf where the symbols have their usual meanings.
the )(nf is divisible by(a) 12 nn (b) (n + 1)! (c) n! (d) none of these.
79 if A+B+CiciAiB
iAiBiC
iBiCiA
i
eeeeeeeee
andzie2
2
2
sincos,
than
(a) Re(z) = 4 (b) Im(z) = 0 (c) Re(z) = - 4 (d) Im(z) = -1
COMPRIHENSIONLet A be a square matrix of order 2 ot 3 and I be the identity matrix of the same order then the
matrix A I is called charactristic matrix of the matrix A where is same complex no. The determi-nant of the characteristic matrix is called characteristic determinant of the matrix A which willof coursebe a polynomial of degree 3 in . The equation Det(A I ) = 0 is called characheristic equation ofthe matrix A and its roots(the values of ) are called characteristic roots or eigen values. It is alsoknown that every square matrix has its characteristic equation.
80. The eigen values of the matrix 211
432112
A are
(a) 2,1,1 (b) 2,3,-2 (c) -1,1,3 (d) none of these
81. which of the following matrices do not have eigen values as 1 and -1?
(a)
0110
(b)
00i
i (c)
1001
(d)
1001
82.If one of the eigen values of a square matrix A order 3 * 3 is zero then,(a) det A must be non -zero (b) det A must be zero (c) adj A must be a zore matrix (d) none of these
-
ANSWERS
72(A) (B)(C) 73 (C)74 (A)(B) (C) 75 (D) 76 (D) 77 (A) (B) 78(A) (C) 79 (B)(C)80 (C) 81(D) 82 (B)
-
72.
cfgfbhgha
if A,B,C,D,F,G,H are the cofactors of a,b.c,d,e,f,g,h then(i) ;,, 222 cHABbGCAaFBC(ii) hCHFGgBGHFfAFGH ,, .it will be sufficient to prove one of each set. Consider first
bGCA 2 .
Now 2GCA COGFIHGaA
CGGA
COGFIHGOA
cfgfbhgha
GCA )( 2
cCfFgGfcGfHgAfCbFhGbfGbHhAgChFaGhgGhHaA
2
bfo
obooh
Hence bGCA 2 .
73.
Consider the expression:
CxyBzxAyzCzByAxE sin2sin2sin22sin2sin2sin 222 We prove the result by showing that E can be expressed as the product of two factors linear in x,y,zNow in any triangle ABC
kc
Cb
Ba
A
sinsinsin
using this property and the identity cos.sin22sin we obtain
cxybzxayzCcxBbyAaxk
E coscoscos
2222
Again
BaAbcAcCabBcCba
coscoscoscoscoscos
zxAcCayzBcCbCcxBbyAaxk
E coscoscoscoscoscoscos2
222
xyBaAb )coscos(
-
)coscoscos)(( CzByAxczbyax Hece E can be written as the product of two factors linear in x,y,z. Thuis
02sinsinsin
sin2sinsinsinsin2sin
CABABCBCA
74Next consider . gBGFH
Now CFFBGH
FBGH
BGHF100
Hence CFFBGH
cfgfbhgha
BGHF100
)(
cCfFgGcFfBgHcfCbFhGfFbBhHfgChFaGgFHbaHg
2
0000
gcfg
Hence gBGHF
Similarly, fAFGH and hCHFG
75 (d) : By putting x = 0 an both sides of the equation we have
g = 9102210021 differentiaing both sides and then putting x =0, we get f= -5.
76.(d): Degree of L.H.S. = n+n +2+n+3 and that of R.H.S.= 2 1 n
77. The number of third-order determinant = the number of arrangements of nine different numbers in nineplaces =9!. Corresponding to each determinant made, there is a determinant obtained by interchangingtwo consecutive rows (or columns). so, the sum of this pair will be 0.
the sum of all the determinants 02!9.....000 timesto
-
78. 00.1
!1.1!.!11
111!2!1!
21)(
nnnnnn
nnnnnn
nf
(using 122233 , CCCCCC )
}.1{!!!11 2 nnnnnnn
79.
iCCAiCBi
BAiiBBCi
ABiACiiA
iCiBiA
eeeeeeeee
eeez
because
,1sincos)( iee iCBAi etc.
iCiBiA
iA
iB
eeee
e
002020
(using 322311 , RRRRRR )
.444}2{2 )()( iCBAiCAiiB eeee
80.
211432112
A
000000
I
211432112
IA
)3)(1)(1()det( IA
this the characteristic roots are -1,1&3.
81. (a) is not correct since its characteristic determinant
11
characteristic equation is 012 1,1 eigen values are 1 & -1we similarly note that matrices given in choice (b) and (b) and (c) have eigen values 1 and -1.
they are not correct. (d) has characteristic equation 2)1( = 0 eigen values are not 1 & -1 choice (d) is correct
-
82. Let A=
333
222
111
cbacbacba
A than
333
222
111
cbacbacba
IA
]))()[(()det( 23321 cbcbaIA )].([])[( 233213231 babaccccbnow one of the eigen values is zero, one root of should be zero constant term in the above polynomial is zero
0231321321321321321 bacaaaacbcabcbacba (collecting constant terms).But this value is value of determinant of A. det A = 0A should be non-singular..
-
SUBJECTIVE QUESTIONS 83. If A is non-singular matrix, then show that adj (adj A) = AA n 1
84. If A is a square matrix of order n, prone that 21)( nAadjAadj
85. If A and B are two square matrixs. such that BAAB 1 , then prove that 222)( BABA
86. Find the value of )( 1p interms of p where p is non-singular matrix and hence show that
PAQBPQadj )( 11 given taht 1& QPAB
87. If hgfcd
aB
cbb
aA 0
11,
11
01
zyx
Xa
Vhgf
U ,00,
2
and AX = U has infinitely many solutions, prove that if 0agf , then BX = V has no solution.
88. find the invrse of the matrix
011101110
S and show that 1SAS is a diagonal matrix
where
bacacbbaacbcabaccb
A21
89 If A & B are different matrices satisfying ABBABA 2233 & Find 22 BA 90 Prove that If A is nn matrix with real entries then get 02 nFA91 Show that if A & B are nn matrices with realentries and AB = nO then get
0222 BAI Pn for any positive integer P & Q92 Let A,B, C be nn real matrices that are pairwise communthetics nOCBA ,, Prove that
Get getCBA 333 (A + B+ C) 093 Let A,B be two & square matrices such that A+B = AB . Prove that AB =BA
94. Show that
2
sin2
sin2
sin2
sin2
sin2
sinsinsinsin23sin2sinsin3sin2sinsin3sin2sinsin
6
95. Find all the values of t for which the system of equations.02)13()1( tzytxt 0)3()24()1( ztytxt 0)1(3)13(2 ztytx
has a non-trivial solution.
96. Prove that
3
222
222
222
)(2 cbaabcbacc
bacbaacb
-
97 Show that
)(2sin
2sin0sinsincossin
cossincos
2
2
yxyxyxyxyxayxyxyx
98. Evaluate the determinant1111
23
23
23
23
daacaabaaaaa
99. solve
0333
333
222222
cxbxaxcxbxaxcxbxax
100. show that -(a+b+c) is one root of the equation
0
bxacacxbcbax
and solve the equation completely..
101. solve the equation ccacbbabxxax
233
233
233
= 0 where b, c are unequal.
102. Solve the equation 0
4321432143214321
xx
xx
ANSWERS
86
-
Solution:83. Let nIBadjBB )( Replacing B by adj AA
(adj A)(adj (adj A)) = nIadjA or nn IAadjAadjadjA 1))()(( or n
n IAAadjAadjadjAA 1))()((
or nn AIAadjAadjA 1))(( or AAadjAadjA n 1))(( or AAadjAadj n 2)(
84. adjAadjAadj )( 1)( adjA = 1111111 AAAAAAAA n }{ AkkAas n AA n 2
AAAas 111
85. BAAB 1 BABAAAAB 1 0 BAAB and
22222 ))(()( BABBAABABABABA Hence Proved.
86. 1 XXadjX 1111 PPPadj }1{ 11P
PPas 1111 adjPPP
Also 1111111 )( BPQBPQBPQadj 11111111 QBPPPBQ
= QPBPBQ 111 }1{ PQas PAQQadjBPQBBP )(1 {as adj B = A}A}
87. Consider[A:U]=
hcbgdbfa
11
01
applying 233 RRR
ghdcbgdbfa
11
01
AX =U has infinitely many solutions c= d and h = g ...........(i)
it is obvious that if (i) holds then, B = 0 BX = V has no unique solution.
Now consider [B:V] =
000
11 2
hgfcd
aa
apply 133 RafRR
=
afafh
afgf
cdaa00
11 2
apply 2331 R
afg
dRR
=
afafg
dc
afh
cdaa
00
0011 2
faafd ,0 and d are nonzero af is nonzero
Also,as (i) holds 0
afg
dc
afh BX = V has no solution if 0afd .
-
88. 2s and adj
111111111
21
111111111
1SS
also,
022202220
21
011101110
21
ccbbaa
bacacbbaacbcabaccb
SA
and
cb
a
cb
a
ccbbaa
SAS00
0000
400040004
41
111111111
022202220
411
which is a diagonal matrix.
89. We have 322322 ))(( BABBAABABA . since BA , this shows that 22 BA has a zero divisor. Hence it is not invertible, so its determinant is 0.
90. Write nnn iIAiIAIA 2 , where i is the imaginary unit. If n ......., 21 are the eigenvlues ofA, then the eigen values of niIA are iii n ......., 21 ,Hence
d e t )).......((),( 212 iiiiIA nn . S i m i l a r l y
det )).......((),( 212 iiiiIA nn
. Subce A has reak entries, its complex eigen values come in pairs of conjugate
numbers. By using the formulas aibaibiaibia 2)1())(( 22 and
aibaibiaibia 2)1())(( 22
, we see that det niIA and det niIA can be written as products of terms that are complex conjugate of each other. Hence the determinats themselves are complex conjugates of
each other, which implies that their product is nonnegative real number.
91. By the previous problem, we have det 02 pn AI and det 02 qn BI . from nAB 0 We ob
tain nqp BA 022 .thus )det()det( 222222 qpqpn
qpn BABAIBAI
)))(det(( 22 qnp
n BIAI
)det()det( 22 qnp
n BIAI
92. Let 1 be a third root of unit. Since A,B,C commute and nABC 0 , we can write, ABCCBACBA 3333333
)( 222 CABCABCBACBA ))(( 22 CBACBACBA ))(( 22 CBACBACBA
Hence CBACBA det)det( 333
)det()det()det( 222 CBACBACBA
)det()det()(det 222 CBACBACBA .0)det()(det 222 CBACBA
-
93. Since ,0nBAAB by adding nI to both sides and factoring we obtain nnn IBIAI . follows that AIn is invertible, and its inverse is BIn . Hence nnn IBIAI , which implies ,0nBAAB . Consequently, ABBABA .
94. )1cos4(sincossin2sin)1cos4(sincossin2sin)1cos4(sincossin2sin
2
2
2
3sin4sin3,3sin
)]cos1(43[sin 2 sin)1cos4( 2
Take sin , sin , sin common from 321, andRRR .
sin sin sin1cos4cos211cos4cos211cos4cos21
2
2
2
=8 sin sin sin
2
2
2
coscos1coscos1coscos1
95. For non-triival solution 0)1(3132
32412131
ttttt
ttt
303
3302131
tttt
ttt
)(
133
122
RRRRRR
0
3011102131
)3( 2
t
tttt
0)]1(2)1)(13()1)(1[()3( 2 tttt 0]2131[)3( 3 tttt 0 t or 3t96. Subtracting the second column from the third and the first from the second
222
22222
22
)(0)()(
0)(
cbacacbbacb
acb
))((0))(())((
0))(()(
2
2
2
cbacbaccbacbacbabacb
cbacbacb
cbaccbacbab
cbacbcba
0
0)()(
2
2
2
2
subtract from the first row the sum of the second and third rows. then
cbaccbacbabcbcbc
cba
0
2222)(
2
22
-
cbaccbacbab
cbcbccba
0)(2
2
22
Now add the second column on to the third and we have
cbac
cbabbcbc
cba
00)(2
2
22
maltiply the third column by c and add to the first.
cbabaccbab
bccba
0)(0
0)(2 22
Finding along the frist column.
)})(()({)(2 22 cbabaccbacbcba
)})(()({)(2 2 cbabacbabcbabc 3)(2 cbaabc
97. Subtracting the third column from the first,
yyxyxyxyxyxyx
2sin02sin2sin)sin()cos(0)cos()sin()cos(2
yyxyxyxyxyxyxyx
2sin0)sin()cos(2)sin()cos(0)cos()sin()cos(2
yyxyxyxyxyx
yx2sin0)sin(
)sin()cos(0)cos()sin(1
)cos(2
multiply the first row by sin(x-y) and subtract from the third, then
)sin()sin(2sin)sin()sin(0
)sin()cos(0)cos()sin(1
)cos(2yxyxyyxyx
yxyxyxyx
yx
)cos()sin()sin()sin(0)sin()cos(0
)cos()sin(1)cos(2
yxyxyxyxyxyx
yxyxyx
)cos()sin()sin()cos(
)sin()cos(2yxyxyxyx
yxyx
).(2sin)}(sin)(){cos(2sin 22 yxyxyxyx
-
98. Subtract the first row from the second, third and fourth rows and them expand along the fourth column. thus
adadabacacacababab
adadabacacacababab
aaa
2233
2233
2233
2233
2233
2233
23
0001
111
))()((22
22
22
dadadacacacabababa
adacab
Now subtract the first row from the second and third fourth rows and then expand along the fourth . Thus
001
))()((22
22
22
bdadadbdcaabacbcbababa
adacab
bddbabdbccbabc
adacab
)()()()(
))()((
= IdbaIcba
bdbcadacab
))()()()((
).)()()()()(( dcdbcbdacaba 99. For x where a,b,c, are unequal. adding the second row to the third.
)3(2)3(2)3(2
)()()(222222
333
222222
cxxbxxaxxcxbxaxcxbxax
222222
333
222222
333)()()(2cxbxax
cxbxaxcxbxax
x
adding the first row to the third,
222222
333
222222
)()()(4cxbxax
cxbxaxcxbxax
x
222222
2222223
222222
)333)(()333)(()(4accbax
aaccbxaxxbaaabbbxaxxbaaxcabaax
x
)()(
333333)())((422
2222223
22
acbaaxaabcbxaxxaabbbxaxxax
cabaaxcabax
-
00333333)())((4
22
2222223
22
axaabcbxaxxaabbbxaxxax
cabaaxcabax
2222223
333333))((8
aabcbxaxxaabbbxaxxacba
cabax
acbabccxbxaabbbxaxxbcba
cabax
22222
3
33333))((8
cbaxaabbbxaxxba
bccabax
33331
))()((8 2223
).3)()()((8 23 cabcabxbccabax
100. Adding the second and third rows to the first,
bxac
acxbcbaxcbaxcbax
bxac
acxbcbax
111
Hence x = - (a+b+c) is one solution of the equation. Now subtract the first column from the second and fromthe third. we have
cbxcacbabcxbcbax
001
))(( 222 cabcabbaxcbax
101. Subtract the first row from the second and third. then
xcxcaxxbxbax
xxax
2222
2233
233
11))((
22
22
233
xcxcxcxbxbxb
xxaxxcxb
subtract the second row from the third.
01))((
22
22
233
xcbxcxbcxbxbxb
xxaxxcxb
011))()(( 22
233
xcbxbxbxb
xxaxbcxcxb
Expanding along the third row,
)}())(){()()(( 32233 xbxxbaxbxxcbbcxcxb
= ).)()()(( 3 bcxabcxcxb Hence the solutions of 0 are cbx , or ./3 bca
-
102. Subtract the second row from the first and from the thirds and fourth. Then
xx
xxx
xx
0000432100
1010011043210011
3
x
x
Now subtract the first rowfromthe second. then
1010011043300011
3
x
x 101011433
3
xx
And the first column to the third and expand along the third row. thus
101111
7333
xxx = )10()73( 33 xxxx
Hence x= 0,0,0 or -10.
-
94. Statement 1 : The order of the matrix A is 35 and that of the matrix B is 54 , then the product BA is not possible
becauseStatement 2 : Number of columns of A number of rows of B.
(A) Statement -1 is true, statement=2 is true ; statement -2 is a correct explanation for statement -1 (B) Statement -1 is true, statement-2 is true; statement-2 is not a correct explanation for statement-1 (C) Statement -1 is true, statement- is false. (D) Statement-1 is false, statement -2 is true.
ComprehensionLet A be a square matrix of order 2 or 3 and I be the identity matrix of the same order, then the matrix IA is called characteristic matrix of the matrix A, where is some complex number. The determinant of the characteristic matrix is called characteristic determinant of the matrix A which will of course be a polynomial of degree 3 in The equation det 0)( IA is called characteristic equation of the matrix A and its roots arecalled characteristic roots or eigen values. It is known that every square matrix satisfied itscharacteristic equation.
95 If one of the eigen values of a square matrix A or order 33 is zero , then(A) det A must be non -zero (B) det A must be zero(C) adj A must be a zero matrix (D) none of these.
96 If 'A denotes transpose of matrix A, A A = 1 and det A = 1, Then det (A-I) must be equalto(A) 0 (B) -1(C) 1 (D) none of these.
97 If A=
100010001
420110001
I and dIcAAA 21 61
, then the value of c and d are
(A) - 6, - 11 (B) 6, 11(C) -6, 11 (D) 6, -11
98 .If
0333
333
222222
cxbxaxcxbxaxcxbxax
Then value of x are where a,b, c, are unequal
(A) 0 (B) 2
(C) 3 (D) cabcab
31
-
1 EXTRA QUESTION
cossinsincos
cossinsincos
cossinsincos
2A
A
2cos2sinsin2cos
cossincos.sinsin.coscos.sinsin.cossincos
2
2222
2
cossinsincos
2cos2sin2sin2cos23
AAA
coscossin.sinsin.coscos.sincos.sinsin.cossin.sincos.2cos
2222
22
coscossin.sinsin.coscos.sincos.sinsin.cossin.sincos.2cos
2222
22