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.