Free Matrices 01 Practice Test - 12th Grade - Commerce 

Question 1

If A=⎡⎢⎣123⎤⎥⎦ then AA' =  

A. 14
B. ⎡⎢⎣133⎤⎥⎦
C. ⎡⎢⎣123246369⎤⎥⎦
D. None of these

SOLUTION

Solution : C

A' = [1 2 3] therefore AA' = =⎡⎢⎣123⎤⎥⎦[123]=⎡⎢⎣123246369⎤⎥⎦

Question 2

If A', B' are transpose matrices of the square matrices A, B respectively , then (AB)' is equal to

A. A'B'
B. B'A'
C. AB'
D. BA'

SOLUTION

Solution : B

It is a fundamental concept ,i.e., (AB)' = B'A'

Question 3

Out of the following which is a skew- symmetric matrix

A. ⎡⎢⎣045−40−6−560⎤⎥⎦
B. ⎡⎢⎣145−41−6−561⎤⎥⎦
C. ⎡⎢⎣145−42−6−563⎤⎥⎦
D. ⎡⎢⎣i+145−4i−6−56i⎤⎥⎦

SOLUTION

Solution : A

In a skew-symmetrix matrix aij=−aji∀i,j=1,2,3 for j=i,aii=−aji⇒ each aii=0.       
Hence the matrix ⎡⎢⎣045−40−6−560⎤⎥⎦ is skew-symmetric.

Question 4

If A is a square matrix A+AT is symmetric matrix, then
A−AT=

A. Unit matrix
B. Symmetric matrix
C. Skew symmetric matrix
D. Zero matrix

SOLUTION

Solution : C

(A−AT)T=AT−(AT)T=AT−A[∵(AT)T=A]=−(A−AT)
So, A−AT ​is a skew symmetric matrix.

Question 5

If matrix A=[aij]3×3,matrix B=[bij]3×3 where aij+aji=0 and bij−bji=0,then A4.B3 is

A. skew-symmetric matrix
B. singular
C. symmetric
D. zero matrix

SOLUTION

Solution : B

Since matrix  A  is  skew-symmetric,
∴|A|=0
∴|A4.B3|=0
 

Question 6

If A and B are symmetric matrices of same order and X= AB + BA and Y = AB – BA, then (XY)T is equal to

A. XY
B. YX
C. – YX
D. None of these

SOLUTION

Solution : C

X=AB+BA⇒XT=X
and Y=AB−BA⇒YT=−Y
Now,(XY)T=YT×XT=−YX

Question 7

If A is involutary matrix, then which of the following is/are correct?

A. I + A is idempotent
B. I – A is idempotent
C. (I + A)(I – A) is singular
D. I+A3 is idempotent

SOLUTION

Solution : C

A2=I
(I+A)(I−A)=I−A2+I−I=O

Question 8

If A and B are two matrices such that AB = B and BA = A, then

A. (A6−B5)3=A−B
B. (A5−B5)3=A3−B3
C. A−B is idempotent
D. A−B is nilpotent

SOLUTION

Solution : D

Since AB = B and BA = A
∴ A and B both are idempotent
(A−B)2=A2−AB−BA+B2=A−B−A+B=0
∴ A - B is nilpotent

Question 9

If A=[abba] and A2=[αββα], then

A. α=a2+b2,β=ab
B. α=a2+b2,β=2ab
C. α=a2+b2,β=a2−b2
D. α=2ab,β=a2+b2

SOLUTION

Solution : B

A2=[αββα]=[abba][abba];α=α2+b2;β=2ab

Question 10

The rank of the matrix ⎡⎢⎣−1252−4a−41−2a+1⎤⎥⎦ is

A. 1 if a = 6 
B. 2 if a = 1
C. 3 if a = 2
D. 1 if a = -6

SOLUTION

Solution : B and D

A=∣∣ ∣∣−1252−4a−41−2a+1∣∣ ∣∣=∣∣ ∣∣00a+600−a−61−2a+1∣∣ ∣∣
(Operating (R1→R1+R3andR2→R2−2R3)
=∣∣ ∣∣00000−a−61−2a+1∣∣ ∣∣(Operating R1→R1+R2)
When a=−6,A=∣∣ ∣∣0000001−2−5∣∣ ∣∣, ∴ρ(A)=1
Where ρ(A)number of non-zero rows
When a=6,A=∣∣ ∣∣00000−121−27∣∣ ∣∣,∴ρ(A)=2
When a=1,A=∣∣ ∣∣00000−71−22∣∣ ∣∣,∴ρ(A)=2
When a=2,A=∣∣ ∣∣00000−81−22∣∣ ∣∣,∴ρ(A)=2
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