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

### Question 1

If A=[abba] and A2=[αββα], then

#### SOLUTION

Solution :B

A2=[αββα]=[abba][abba];α=α2+b2;β=2ab

### Question 2

If A=⎡⎣1tanθ2−tanθ21⎤⎦ and AB = I, then B =

#### SOLUTION

Solution :B

|A|=1+tan2θ2=sec2θ2AB=I⇒B−IA−1[1001]⎡⎢⎣1−tanθ2tanθ21⎤⎥⎦sec2θ2=cos2θ2.AT.

### Question 3

Let,

A=⎡⎢⎣46−13021−25⎤⎥⎦, B=⎡⎢⎣2401−12⎤⎥⎦

and C=[123]

The expression which is not defined is:

#### 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.

### Question 4

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

#### SOLUTION

Solution :A

A3×4⇒A′4×3Now A'B defined

⇒Bis 3×p

Again B3×pA′4×3 defined ⇒p=4

∴B is 3×4.

### Question 5

If A=⎡⎢⎣a000b000c⎤⎥⎦, then An=

#### SOLUTION

Solution :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 6

For each real number x such that −1<x<1,let A(x) be the matrix (1−x)−1[1−x−x1] and z=x+y1+xyThen,

#### SOLUTION

Solution :C

A(z)=A(x+y1+xy)=[1+xy(1−x)(1−y)] ⎡⎢⎣1−(x+y1+xy)−(x+y1+xy)1⎤⎥⎦

∴A(x).A(y)=A(z)

### Question 7

If A=[cosθsinθsinθ−cosθ],B=[10−11],C=ABAT,then ATCnA equals to(nϵZ+)

#### SOLUTION

Solution :D

A=[cosθsinθsinθ−cosθ]AAT=I (i)Now,C=ABAT⇒ATC=BAT (ii)Now ATCnA=ATC.Cn−1A=BATCn−1A(from(ii))=BATC.Cn−2A=B2ATCn−2A=.......=Bn−1ATCA=Bn−1BATA=Bn=[10−n1]

### Question 8

A=[aij]n×n and aij=i2−j2 then A is necessarily

#### SOLUTION

Solution :C

aji=j2−i2=−(i2−j2)=−aij

### Question 9

If A is a non-diagonal involutory matrix, then

#### SOLUTION

Solution :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.

### Question 10

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

#### SOLUTION

Solution :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(reversal laws)=A−1B(as B=B1)(A−1)1=A−1⇒A−1B is symmetric

Similarly for A−1B−1.

### Question 11

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

#### 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.

### Question 12

If A is a 3×3 non-singular matrix such that AAT=ATAandB=A−1AT,then BBT is equal to

#### SOLUTION

Solution :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 13

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+A∗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+A∗2=Re(aij).

### Question 14

If A=[1−111] then A16 =

#### SOLUTION

Solution :B

A2=[0−220],A4=[−400−4]A8=[160016],A16=[25600256]

### Question 15

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

#### 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.