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12th Grade > Mathematics

MATRICES MCQs

Total Questions : 44 | Page 4 of 5 pages
Question 31. If 3X + 2Y = I and 2X - Y = O, where I and O  are unit and null matrices of order 3 respectively, then
  1.    X=(17),Y=(27)
  2.    X=(27),Y=(17)
  3.    X=(17)I,Y=(27)I
  4.    X=(27)I,Y=(17)I
 Discuss Question
Answer: Option C. -> X=(17)I,Y=(27)I
:
C
3X+2Y=I2X−Y=0⇒3X+2Y=I4X−2Y=0⇒7X=IX=17I
(Solving simultaneously)
Therefore from (i), 2Y=I−37I=47I⇒Y=27I
Question 32. If A=⎡⎢⎣a000b000c⎤⎥⎦, then An=
  1.    ⎡⎢⎣na000nb000nc⎤⎥⎦
  2.    ⎡⎢⎣a000b000c⎤⎥⎦
  3.    ⎡⎢⎣an000bn000cn⎤⎥⎦
  4.    None of these
 Discuss Question
Answer: Option C. -> ⎡⎢⎣an000bn000cn⎤⎥⎦
:
C
Since A2=A.A=⎡⎢⎣a000b000c⎤⎥⎦⎡⎢⎣a000b000c⎤⎥⎦=⎡⎢⎣a2000b2000c2⎤⎥⎦
And A3=⎡⎢⎣a3000b3000c3⎤⎥⎦,....⇒An=An−1.A=⎡⎢⎣an−1000bn−1000cn−1⎤⎥⎦⎡⎢⎣a000b000c⎤⎥⎦=⎡⎢⎣an000bn000cn⎤⎥⎦.
Note: Students should remember this question as a formula.
Question 33. Let,
A=⎡⎢⎣46−13021−25⎤⎥⎦, B=⎡⎢⎣2401−12⎤⎥⎦
and C=[123]
The expression which is not defined is: 
  1.    B′B
  2.    CAB
  3.    A+B′
  4.    A2+A
 Discuss Question
Answer: Option C. -> A+B′
:
C
We can see from the options that if we take transpose of B,
B′ will be of 2 x 3 matrix which cannot be added to a 3 x 3 matrix, as for the addition the order should be the same.
Question 34. If A=[cosθsinθsinθ−cosθ],B=[10−11],C=ABAT,then ATCnA equals to(nϵZ+)
  1.    [−n110]
  2.    [1−n01]
  3.    [011−n]
  4.    [10−n1]
 Discuss Question
Answer: Option D. -> [10−n1]
:
D
A=[cosθsinθsinθ−cosθ]AAT=I(i)Now,C=ABAT⇒ATC=BAT(ii)NowATCnA=ATC.Cn−1A=BATCn−1A(from(ii))=BATC.Cn−2A=B2ATCn−2A=.......=Bn−1ATCA=Bn−1BATA=Bn=[10−n1]
Question 35. If A=⎡⎣1tanθ2−tanθ21⎤⎦ and AB = I, then B =
  1.    cos2θ2.A
  2.    cos2θ2.AT
  3.    cos2θ2.I
  4.    None of these
 Discuss Question
Answer: Option B. -> cos2θ2.AT
:
B
|A|=1+tan2θ2=sec2θ2AB=I⇒B−IA−1[1001]⎡⎢⎣1−tanθ2tanθ21⎤⎥⎦sec2θ2=cos2θ2.AT.
Question 36. 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 37. If A=[abba] and A2=[αββα], then
  1.    α=a2+b2,β=ab
  2.    α=a2+b2,β=2ab
  3.    α=a2+b2,β=a2−b2
  4.    α=2ab,β=a2+b2
 Discuss Question
Answer: Option B. -> α=a2+b2,β=2ab
:
B
A2=[αββα]=[abba][abba];α=α2+b2;β=2ab
Question 38. 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 39. If A and B are two non singular matrices and both are symmetric and commute each other then
  1.    Both A−1B and A−1B−1 are symmetric
  2.    A−1B is symmetric but A−1B−1 is not symmetric
  3.    A−1B−1  is symmetric but A−1B is not symmetric
  4.    Neither A−1B nor A−1B−1 are symmetric
 Discuss Question
Answer: Option A. -> Both A−1B and A−1B−1 are symmetric
:
A
AB =BA
Previous & past multiplying both sides by A−1.
A−1(AB)A−1=A−1(BA)A−1(A−1A)(BA−1)=A−1B(AA−1)⇒(BA−1)1=(A−1B)1=(A−1)1B1(reversallaws)=A−1B(asB=B1)(A−1)1=A−1⇒A−1B is symmetric
Similarly for A−1B−1.
Question 40. A=[aij]n×n and aij=i2−j2 then A is necessarily
  1.    a unit matrix
  2.    symmetric matrix
  3.    skew symmetric matrix
  4.    zero matrix
 Discuss Question
Answer: Option C. -> skew symmetric matrix
:
C
aji=j2−i2=−(i2−j2)=−aij

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