Matrices Practice Questions - DPP for JEE

 Page 2


3. (b) (aI + bA)
2
 = a
2
I
2
 + b
2
A
2
 + 2ab AI
= a
2
I
2
 + b
2
 A
2
 + 2abA
But  ? (aI + bA)
2
 = a
2
I + 2abA.
4. (b) Hence  f(a) f(ß) = 
= 
= 
similarly f(a) f(ß) f(?) = 
=  as a + ß + ? = p
=   = – I
2
5. (a)
6. (a) (A – 2I) (A + I) = 0
? AA – A – 2I = 0 ( AI = A)
? 
7. (a) Given Q = PAP
T
? P
T
Q = AP
T
, ( Q PP
T
 = I)
? P
T
Q
2005
P = AP
T
Q
2004
 P = AP
T
Q
2003
PA
( Q Q = PAP
T
 ? QP = PA)
Page 3


3. (b) (aI + bA)
2
 = a
2
I
2
 + b
2
A
2
 + 2ab AI
= a
2
I
2
 + b
2
 A
2
 + 2abA
But  ? (aI + bA)
2
 = a
2
I + 2abA.
4. (b) Hence  f(a) f(ß) = 
= 
= 
similarly f(a) f(ß) f(?) = 
=  as a + ß + ? = p
=   = – I
2
5. (a)
6. (a) (A – 2I) (A + I) = 0
? AA – A – 2I = 0 ( AI = A)
? 
7. (a) Given Q = PAP
T
? P
T
Q = AP
T
, ( Q PP
T
 = I)
? P
T
Q
2005
P = AP
T
Q
2004
 P = AP
T
Q
2003
PA
( Q Q = PAP
T
 ? QP = PA)
= AP
T
 Q
2002
PA
2
 = AP
T
PA
2004
= AIA
2004
 =A
2005
 
8. (d)
Hence, AB = BA only when a = b
? There can be infinitely many B's for which AB = BA
9. (a) Here, A =  where t = tan 
Now, 
and 
= 
=  
Page 4


3. (b) (aI + bA)
2
 = a
2
I
2
 + b
2
A
2
 + 2ab AI
= a
2
I
2
 + b
2
 A
2
 + 2abA
But  ? (aI + bA)
2
 = a
2
I + 2abA.
4. (b) Hence  f(a) f(ß) = 
= 
= 
similarly f(a) f(ß) f(?) = 
=  as a + ß + ? = p
=   = – I
2
5. (a)
6. (a) (A – 2I) (A + I) = 0
? AA – A – 2I = 0 ( AI = A)
? 
7. (a) Given Q = PAP
T
? P
T
Q = AP
T
, ( Q PP
T
 = I)
? P
T
Q
2005
P = AP
T
Q
2004
 P = AP
T
Q
2003
PA
( Q Q = PAP
T
 ? QP = PA)
= AP
T
 Q
2002
PA
2
 = AP
T
PA
2004
= AIA
2004
 =A
2005
 
8. (d)
Hence, AB = BA only when a = b
? There can be infinitely many B's for which AB = BA
9. (a) Here, A =  where t = tan 
Now, 
and 
= 
=  
= 
= 
= 
= = 
Also, = 
=  = (I – A) 
10. (a) A
2
 = I
Now, (A – I)
3
 + (A + I)
3
 – 7A
= A
3
 – I
3
 – 3A
2
I + 3AI
2
 + A
3
 + I
3
 + 3A
2
I + 3AI
2
 – 7A
= 2A
3
 + 6AI
2
 – 7A = 2A
2
A + 6AI – 7A
= 2IA +6A – 7A = 2A + 6A – 7A = A       
11. (a) Since B is an idempotent matrix, ? B
2
 = B.
Page 5


3. (b) (aI + bA)
2
 = a
2
I
2
 + b
2
A
2
 + 2ab AI
= a
2
I
2
 + b
2
 A
2
 + 2abA
But  ? (aI + bA)
2
 = a
2
I + 2abA.
4. (b) Hence  f(a) f(ß) = 
= 
= 
similarly f(a) f(ß) f(?) = 
=  as a + ß + ? = p
=   = – I
2
5. (a)
6. (a) (A – 2I) (A + I) = 0
? AA – A – 2I = 0 ( AI = A)
? 
7. (a) Given Q = PAP
T
? P
T
Q = AP
T
, ( Q PP
T
 = I)
? P
T
Q
2005
P = AP
T
Q
2004
 P = AP
T
Q
2003
PA
( Q Q = PAP
T
 ? QP = PA)
= AP
T
 Q
2002
PA
2
 = AP
T
PA
2004
= AIA
2004
 =A
2005
 
8. (d)
Hence, AB = BA only when a = b
? There can be infinitely many B's for which AB = BA
9. (a) Here, A =  where t = tan 
Now, 
and 
= 
=  
= 
= 
= 
= = 
Also, = 
=  = (I – A) 
10. (a) A
2
 = I
Now, (A – I)
3
 + (A + I)
3
 – 7A
= A
3
 – I
3
 – 3A
2
I + 3AI
2
 + A
3
 + I
3
 + 3A
2
I + 3AI
2
 – 7A
= 2A
3
 + 6AI
2
 – 7A = 2A
2
A + 6AI – 7A
= 2IA +6A – 7A = 2A + 6A – 7A = A       
11. (a) Since B is an idempotent matrix, ? B
2
 = B.
Now, A
2
 = (I – B)
2
 = (I – B)(I – B)
?  A is idempotent.
12. (c)
13. (c)
? A(x). A(y) = A(z).
14. (d) We have   A = 
Now, A
2
 = A . A = =  = – I
where I =  is identity matrix
(A
2
)
8
 = (– I)
8
 = I
Hence, A
16
= I
15. (d) =  ;  
? a
2
 + ß? = 1
16. (a)