12th Grade > Mathematics
MATRICES MCQs
Total Questions : 44
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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.
:
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.
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.
:
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.
Answer: Option B. -> α=a2+b2,β=2ab
:
B
A2=[αββα]=[abba][abba];α=α2+b2;β=2ab
:
B
A2=[αββα]=[abba][abba];α=α2+b2;β=2ab
Answer: Option B. -> symmetric
:
B
We are given that the matrix A is skew symmetric which implies that,
AT=−A --------(1)
(A4)T=(A.A.A.A.)T=ATATATAT (By applying the property of transpose of a matrix)
⇒ (-A) (-A) (-A) (-A) (Using equation (1))
=(−1)4A4=A4
i.e,(A4)T=(A4)
A4 is symmetric.
:
B
We are given that the matrix A is skew symmetric which implies that,
AT=−A --------(1)
(A4)T=(A.A.A.A.)T=ATATATAT (By applying the property of transpose of a matrix)
⇒ (-A) (-A) (-A) (-A) (Using equation (1))
=(−1)4A4=A4
i.e,(A4)T=(A4)
A4 is symmetric.
Answer: Option B. -> [25600256]
:
B
A2=[0−220],A4=[−400−4]A8=[160016],A16=[25600256]
:
B
A2=[0−220],A4=[−400−4]A8=[160016],A16=[25600256]
Answer: Option C. -> Re(aij)
:
C
By taking complex conjugate of a matrix we reverse the sign of imaginary parts of all the elements in the original matrix. i.e., if the element in A is x + iy, then the corresponding element in A∗is x - iy.
So when A and A∗is added the imaginary parts cancel out and the sum becomes 2 times the real part of element in A.
i.e., since (aij) is general element in A, the general element in A+A∗becomes 2Re(aij).
∴General element in A+A∗2=Re(aij).
:
C
By taking complex conjugate of a matrix we reverse the sign of imaginary parts of all the elements in the original matrix. i.e., if the element in A is x + iy, then the corresponding element in A∗is x - iy.
So when A and A∗is added the imaginary parts cancel out and the sum becomes 2 times the real part of element in A.
i.e., since (aij) is general element in A, the general element in A+A∗becomes 2Re(aij).
∴General element in A+A∗2=Re(aij).
Answer: Option B. -> [25600256]
:
B
A2=[0−220],A4=[−400−4]A8=[160016],A16=[25600256]
:
B
A2=[0−220],A4=[−400−4]A8=[160016],A16=[25600256]
Answer: Option A. -> 3×4
:
A
A3×4⇒A′4×3Now A'B defined
⇒Bis 3×p
Again B3×pA′4×3 defined ⇒p=4
∴B is 3×4.
:
A
A3×4⇒A′4×3Now A'B defined
⇒Bis 3×p
Again B3×pA′4×3 defined ⇒p=4
∴B is 3×4.
Answer: Option C. -> A - I is non zero singular
:
C
A2=I⇒A2−I=0
⇒ (A+I)(A-I)=0
∴ either |A+I|=0 or
|A−I|=0
If |A−I|≠0, then (A+I)(A−I)=0⇒A+I=0 which is not so
∴|A−I| and A−I≠0.
:
C
A2=I⇒A2−I=0
⇒ (A+I)(A-I)=0
∴ either |A+I|=0 or
|A−I|=0
If |A−I|≠0, then (A+I)(A−I)=0⇒A+I=0 which is not so
∴|A−I| and A−I≠0.
Answer: Option C. -> skew symmetric matrix
:
C
aji=j2−i2=−(i2−j2)=−aij
:
C
aji=j2−i2=−(i2−j2)=−aij