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12th Grade > Mathematics

MATRICES MCQs

Total Questions : 44 | Page 3 of 5 pages
Question 21. If A is a square matrix A+AT is symmetric matrix, then
A−AT=
  1.    Unit matrix
  2.    Symmetric matrix
  3.    Skew symmetric matrix
  4.    Zero matrix
 Discuss Question
Answer: Option C. -> Skew symmetric matrix
:
C
(A−AT)T=AT−(AT)T=AT−A[∵(AT)T=A]=−(A−AT)
So, A−AT ​is a skew symmetric matrix.
Question 22. If A=⎡⎢⎣123⎤⎥⎦ then AA' =  
  1.    14
  2.    ⎡⎢⎣133⎤⎥⎦
  3.    ⎡⎢⎣123246369⎤⎥⎦
  4.    None of these
 Discuss Question
Answer: Option C. -> ⎡⎢⎣123246369⎤⎥⎦
:
C
A' = [1 2 3] therefore AA' = =⎡⎢⎣123⎤⎥⎦[123]=⎡⎢⎣123246369⎤⎥⎦
Question 23. If A is involutary matrix, then which of the following is/are correct?
  1.    I + A is idempotent
  2.    I – A is idempotent
  3.    (I + A)(I – A) is singular
  4.    I+A3 is idempotent
 Discuss Question
Answer: Option C. -> (I + A)(I – A) is singular
:
C
A2=I
(I+A)(I−A)=I−A2+I−I=O
Question 24. Out of the following which is a skew- symmetric matrix
  1.    ⎡⎢⎣045−40−6−560⎤⎥⎦
  2.    ⎡⎢⎣145−41−6−561⎤⎥⎦
  3.    ⎡⎢⎣145−42−6−563⎤⎥⎦
  4.    ⎡⎢⎣i+145−4i−6−56i⎤⎥⎦
 Discuss Question
Answer: Option A. -> ⎡⎢⎣045−40−6−560⎤⎥⎦
:
A
In a skew-symmetrix matrix aij=−aji∀i,j=1,2,3forj=i,aii=−aji⇒ each aii=0.
Hence the matrix ⎡⎢⎣045−40−6−560⎤⎥⎦is skew-symmetric.
Question 25. If A=⎡⎢⎣123−23−1312⎤⎥⎦ and  I is a unit matrix of 3rd order, then
​(A2+9I) equals
  1.    2A
  2.    4A
  3.    6A
  4.    None of these
 Discuss Question
Answer: Option D. -> None of these
:
D
A=⎡⎢⎣123−23−1312⎤⎥⎦⇒A.A=A2=⎡⎢⎣6117−114−1171112⎤⎥⎦,
I=⎡⎢⎣100010001⎤⎥⎦, then, A2+9I=⎡⎢⎣15117−1113−1171121⎤⎥⎦.
Question 26. If A=[cosαsinα−sinαcosα], then A2=
  1.    [cos2αsin2αsin2αcos2α]
  2.    [cos2α−sin2αsin2αcos2α]
  3.    [cos2αsin2α−sin2αcos2α]
  4.    [−cos2αsin2α−sin2α−cos2α]
 Discuss Question
Answer: Option C. -> [cos2αsin2α−sin2αcos2α]
:
C
SinceA2=A.A=[cosαsinα−sinαcosα][cosαsinα−sinαcosα]=[cos2αsin2α−sin2αcos2α].
Question 27. If A and B are two square matrices which are skew symmetric then (AB)T equals to
  1.    AB
  2.    BA
  3.    -AB
  4.    -BA
 Discuss Question
Answer: Option B. -> BA
:
B
Required Expression,
(AB)T=BT.AT
= (-B) . (-A) (Since A and B are said to be skew symmetric)
= BA
∴ (b) is the right option.
Question 28. If A=⎡⎢⎣12101−13−11⎤⎥⎦, then
  1.    A3+3A2+A−9I3=0
  2.    A3−3A2+A+9I3=0
  3.    A3+3A2−A+9I3=0
  4.    A3−3A2−A+9I3=0
 Discuss Question
Answer: Option D. -> A3−3A2−A+9I3=0
:
D
⎡⎢⎣12101−13−11⎤⎥⎦A2=A.A=⎡⎢⎣12101−13−11⎤⎥⎦⎡⎢⎣12101−13−11⎤⎥⎦=⎡⎢⎣430−32−2645⎤⎥⎦A.A2=⎡⎢⎣12101−13−11⎤⎥⎦⎡⎢⎣430−9−2−721117⎤⎥⎦⇒A3−3A2−A+9I3=0
Question 29. If A is a 3×3 non-singular matrix such that AAT=ATAandB=A−1AT,then BBT is equal to
  1.    I + B
  2.     I
  3.    B−1
  4.    (B−1)T
 Discuss Question
Answer: Option B. ->  I
:
B
Given,AAT=ATAandB=A−1AT
We are asked for the value of the expression BBT.Since B is given in terms of A we will substitute for the values of BT and B.
BBT=(A−1AT)(A−1AT)T
=A−1ATA(A−1)T[∴(AB)T=BTAT]
=A−1AAT(A−1)T
=IAT(A−1)T[∴(A−1A)=I]
=(A−1A)T
=IT=I
Question 30. If A=[42−11] and I is the identity matrix of order 2, then (A - 2I)(A - 3I) =  
  1.    I
  2.    0
  3.    [1000]
  4.    [0001]
 Discuss Question
Answer: Option B. -> 0
:
B
(A−2I)(A−3I)=[22−1−1][12−1−2]=[0000]=0

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