# Free Objective Test 02 Practice Test - 11th and 12th

If A=[abba] and A2=[αββα], then

A. α=a2+b2,β=ab
B. α=a2+b2,β=2ab
C. α=a2+b2,β=a2b2
D. α=2ab,β=a2+b2

#### SOLUTION

Solution : B

A2=[αββα]=[abba][abba];α=α2+b2;β=2ab

If A=1tanθ2tanθ21 and AB = I, then B =

A. cos2θ2.A
B. cos2θ2.AT
C. cos2θ2.I
D. None of these

#### SOLUTION

Solution : B

|A|=1+tan2θ2=sec2θ2AB=IBIA1[1001]1tanθ2tanθ21sec2θ2=cos2θ2.AT.

Let,
A=461302125, B=240112
and C=[123]
The expression which is not defined is:

A. BB
B. CAB
C. A+B
D. A2+A

#### SOLUTION

Solution : 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.

If A is 3×4 matrix and B is a matrix such that A'B and BA' are both defined. Then B is of the type

A. 3×4
B. 3×3
C. 4×4
D. 4×3

#### SOLUTION

Solution : A

A3×4A4×3Now A'B defined
Bis 3×p
Again B3×pA4×3 defined p=4
B is 3×4.

If A=a000b000c, then An=

A. na000nb000nc
B. a000b000c
C. an000bn000cn
D. None of these

#### SOLUTION

Solution : C

Since A2=A.A=a000b000ca000b000c=a2000b2000c2
And A3=a3000b3000c3,....An=An1.A=an1000bn1000cn1a000b000c=an000bn000cn.
Note: Students should remember this question as a formula.

For each real number x such that 1<x<1,let A(x) be the matrix (1x)1[1xx1] and z=x+y1+xyThen,

A. A(z)=A(x)+A(y)
B. A(z)=A(x)+[A(y)]1
C. A(z)=A(x)A(y)
D. A(z)=A(x)A(y)

#### SOLUTION

Solution : C

A(z)=A(x+y1+xy)=[1+xy(1x)(1y)] 1(x+y1+xy)(x+y1+xy)1
A(x).A(y)=A(z)

If A=[cosθsinθsinθcosθ],B=[1011],C=ABAT,then ATCnA equals to(nϵZ+)

A. [n110]
B. [1n01]
C. [011n]
D. [10n1]

#### SOLUTION

Solution : D

A=[cosθsinθsinθcosθ]AAT=I          (i)Now,C=ABATATC=BAT       (ii)Now ATCnA=ATC.Cn1A=BATCn1A(from(ii))=BATC.Cn2A=B2ATCn2A=.......=Bn1ATCA=Bn1BATA=Bn=[10n1]

A=[aij]n×n and aij=i2j2 then A is necessarily

A. a unit matrix
B. symmetric matrix
C. skew symmetric matrix
D. zero matrix

#### SOLUTION

Solution : C

aji=j2i2=(i2j2)=aij

If A is a non-diagonal involutory matrix, then

A. A - I = 0
B. A + I = 0
C. A - I is non zero singular
D. none of these

#### SOLUTION

Solution : C

A2=IA2I=0
(A+I)(A-I)=0
either |A+I|=0 or
|AI|=0
If |AI|0, then (A+I)(AI)=0A+I=0 which is not so
|AI| and  AI0.

If A and B are two non singular matrices and both are symmetric and commute each other then

A. Both A1B and A1B1 are symmetric
B. A1B is symmetric but A1B1 is not symmetric
C. A1B1  is symmetric but A1B is not symmetric
D. Neither A1B nor A1B1 are symmetric

#### SOLUTION

Solution : A

AB =BA
Previous & past multiplying both sides by A1.
A1(AB)A1=A1(BA)A1(A1A)(BA1)=A1B(AA1)(BA1)1=(A1B)1=(A1)1B1(reversal laws)=A1B(as B=B1)(A1)1=A1A1B is symmetric
Similarly for A1B1.

If A is a skew-symmetric matrix of order 3, then the matrix A4 is

A. skew symmetric
B. symmetric
C. diagonal
D. none of those

#### SOLUTION

Solution : 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.

If A is a 3×3 non-singular matrix such that AAT=ATAandB=A1AT,then BBT is equal to

A. I + B
B.  I
C. B1
D. (B1)T

#### SOLUTION

Solution : B

Given, AAT=ATAandB=A1AT
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=(A1AT)(A1AT)T
=A1ATA(A1)T[(AB)T=BTAT]
=A1AAT(A1)T
=IAT(A1)T[(A1A)=I]
=(A1A)T
=IT=I

All the elements in a matrix A are complex numbers with imaginary parts not equal to zero. If A is the conjugate of the matrix A, aij is the general element of matrix A, then what is the general element of the matrix, A+A2.

A. 2lm(aij)
B. lm(aij)
C. Re(aij)
D. Re(aij)2

#### SOLUTION

Solution : 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+A2=Re(aij).

If A=[1111] then A16 =

A. [02562560]
B. [25600256]
C. [160016]
D. [016160]

#### SOLUTION

Solution : B

A2=[0220],A4=[4004]A8=[160016],A16=[25600256]

If A and B are two square matrices which are skew symmetric then (AB)T equals to

A. AB
B. BA
C. -AB
D. -BA

#### SOLUTION

Solution : 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.