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IIT JAM Mathematics Practice Test- 15 - Mathematics MCQ


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30 Questions MCQ Test - IIT JAM Mathematics Practice Test- 15

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IIT JAM Mathematics Practice Test- 15 - Question 1

If  and  where [ . ] be the greatest integer function, then the maximum value of the determinant 

Detailed Solution for IIT JAM Mathematics Practice Test- 15 - Question 1

Now for the maximum value of the determinant take

[x] =1 , [y] = 0 , [z] = 2 ,

then

IIT JAM Mathematics Practice Test- 15 - Question 2

If  and  then a is equal to

Detailed Solution for IIT JAM Mathematics Practice Test- 15 - Question 2

Apply R2 → R2 - R1

a ∈ any real no.

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IIT JAM Mathematics Practice Test- 15 - Question 3

if  , then the value of xn is ,

Detailed Solution for IIT JAM Mathematics Practice Test- 15 - Question 3

Now if we put n = 2 in the given options, then none of the matrix will equal to (*). So option (D) is true.

IIT JAM Mathematics Practice Test- 15 - Question 4

Let V be a 3 dimensional vector space with A and B its subspaces of dimensions 2 and 1 , respectively. If , then

Detailed Solution for IIT JAM Mathematics Practice Test- 15 - Question 4

We know

dim A + dim B = dim (A + B) + dim (A ∩ B ) .............. (1)

Here A ∩) B = { 0 } => dim (A ∩ B ) = 0

then by equation (1), dim (A + B ) = 3 => option (C) and (D) are incorrect.

Now (A + B) be a subspace of V and has same dimension as V.

=> V = A + B

IIT JAM Mathematics Practice Test- 15 - Question 5

The rank of the following ( n + 1 ) x ( n + 1 ) matrix , were a is real no.

Detailed Solution for IIT JAM Mathematics Practice Test- 15 - Question 5

Apply

Which is in echelon form and the number of non-zero rows is 1, hence rank is 1.

IIT JAM Mathematics Practice Test- 15 - Question 6

Let A, B be n x n matrices such that BA+ B2 = I - BA2, where I is the nxn identity matrix, which of the following is always true ?

Detailed Solution for IIT JAM Mathematics Practice Test- 15 - Question 6

Given

⇒ A + B + AB be inverse of B

⇒ B is invertible => B is non - singular.

IIT JAM Mathematics Practice Test- 15 - Question 7

The determinant of the n x n permutation matrix

(where [x] be greatest integer not exceeding x)

Detailed Solution for IIT JAM Mathematics Practice Test- 15 - Question 7

if n = 2 then

n= 3 then 

n= 4 then 

⇒ Option ( A ), (C) and (D) cannot be true. 

⇒ det of given matrix will be 

IIT JAM Mathematics Practice Test- 15 - Question 8

Which of the following sets is a basis for the subspace  of fhe vector o f all 2 x 2 real matrices over R ?

Detailed Solution for IIT JAM Mathematics Practice Test- 15 - Question 8

x + y = z

y + z = w

⇒ y = w - z

⇒ x = 2z- w

IIT JAM Mathematics Practice Test- 15 - Question 9

Let a, b, c and d are the eigenvalues of the matrix 

The a2 + b2 + c2 + d2 is equal to?

Detailed Solution for IIT JAM Mathematics Practice Test- 15 - Question 9

Let A be the given 4 * 4 matrix. So for the eigenvalues of A take 

IIT JAM Mathematics Practice Test- 15 - Question 10

Consider the following two statements

(I) If V be a vector space over F of dimension 5 and U and W are subspaces of V of dimension 3. Then 

(II) Let V be the set of all polynomials over Q of degree 2 together with the zero polynomial. Then V be a vector space over Q.
Then which of the above statement’s are correct?

Detailed Solution for IIT JAM Mathematics Practice Test- 15 - Question 10

So by dim U + dim W = dim(U + W) + dim(U ∩ W)

We have dim(U + W) = dim U + dim W = 3 + 3 = 6 

⇒ dim(U + W) = 6 ...(i)

Now we know U + W be a subspace of V — > dim(U + W) ≤ dim V = 5

which contradict (i) ⇒ Our supposition is wrong.

(II) Here given set is 

 and F = Q

(Given field).
So in particular case let p(x) = 1 + x + x2

and q(x) = 1 + x - x2

Then clearly p, q ∈ V

But p(x) + q(x) = 2 + 2x ∉ V

⇒ V not a subspace over Q.

IIT JAM Mathematics Practice Test- 15 - Question 11

Let  are two vector spaces over R, then which of the following subset’s is a subspace?

Detailed Solution for IIT JAM Mathematics Practice Test- 15 - Question 11

⇒ W1 not a subspace.

⇒ W2 not a subspace.

(C) Let P1, P2 ∈ Vthen 

⇒ W3 be a subspace.

(D) Let P1, P2 ∈ W4 then P1" + P1' = αx

and P2" + P2' = αx 

Now if α be a fixed real no. then 

for any 

But α can be any real no. then 

So we cannot say directly that whether W4 be a subspace or not.

IIT JAM Mathematics Practice Test- 15 - Question 12

Let A be a 3 x 3 matrix with eigen values 1 , - 1 , 0 . then the determinant of  

Detailed Solution for IIT JAM Mathematics Practice Test- 15 - Question 12

Here the characteristic equation is .

So by cayley's hamiiton theorem ,

Now since A(A2 - I) = 0

so

So |A100 + 1| = 8

IIT JAM Mathematics Practice Test- 15 - Question 13

Let V be the vector space of all real polynomials. Consider the subspace W spanned by t2 + t +2, t2 +2t +5, 5t2 +3t + 4, and 2t2 +2t+ 4. Then the dimension of W is,

Detailed Solution for IIT JAM Mathematics Practice Test- 15 - Question 13

Take

IIT JAM Mathematics Practice Test- 15 - Question 14

Let T be a Linear Transformation from  be a Linear Transform ation defined by  T (a ,b ,c ) = ( 3a , a - b , 2a + b + c ) , then T-1 (1,2,3) is

Detailed Solution for IIT JAM Mathematics Practice Test- 15 - Question 14

Here T is one-one and onto ⇒ unique T-1 exist

If T (a , b ,c) = ( p , q , r ) then T-1 (p ,q ,r ) =( a,b,c)

Now T (a , b, c) = ( p ,q , r ) 

⇒  (3a, a - b, 2a + b + c) = (p,q,r)

IIT JAM Mathematics Practice Test- 15 - Question 15

The linear Transformation , whose image is generated by {(1,2,0,-4), (2,0,-1,-3)} is,

Detailed Solution for IIT JAM Mathematics Practice Test- 15 - Question 15

Since {e1 = (1,0,0), e2 = (0,1,0), e3 = (0,0,1)} is a usual basis of 

Define

T (1,0,0) = (1,2,0,-4)

T (0,1,0) = (2,0,-1,-3)

T (0,0,1) = (0,0,0, 0)

Now, such a Linear Transformation exists and is unique 

T (x,y,z) = T (x (1,0,0) + y (0,1,0) + z (0,0,1))

= xT (1,0,0) +yT (0,1,0) +zT (0,0,1)

= x (1,2,0,-4) + y (2,0,-1 ,-3) + z (0,0,0,0)

= (x + 2y, 2x, -y, -4x - 3y)

IIT JAM Mathematics Practice Test- 15 - Question 16

Let  then  is

Detailed Solution for IIT JAM Mathematics Practice Test- 15 - Question 16

Given

so

Thus we have 


So in general

now

IIT JAM Mathematics Practice Test- 15 - Question 17

Let P4 denote the real vector space of all polynomials with real coefficients of degree at most 4. Consider the map T : P4 → P4 given by T[P(x)] = P"(x) + P'(x) + P(x). Then,

Detailed Solution for IIT JAM Mathematics Practice Test- 15 - Question 17

⇒ N(T) = {0}

⇒ T is one -one and onto.

IIT JAM Mathematics Practice Test- 15 - Question 18

Let T be a linear operator on R3 (R), defined by,

T(a , b, c) = ( 3a , a - b , 2a+ b + c ) , then (T2 — I )(T — 3I) is, 

(where I be a linear operator on R3)

Detailed Solution for IIT JAM Mathematics Practice Test- 15 - Question 18

We have (T — 3I )(a,b,c) = T(a,b,c)- 3I(a,b,c)

= (3a ,a - b ,2a + b + c) -(3a,3b,3c)

= (0,a - 4b,2a +b - 2c)

Where A= 0, B = a - 4b , C = 2a + b - 2c

Then (T2 - l)(A, B, C) = T2 (A, B, C) - l(A, B, C) ----- (2)

Now T (A, B, C) = (3A, A - B, 2A + B + C) 

T2 ( A , B ,C) = T(T ( A , B , C )

= T(3A, A - B, 2A+B + C)

put there values in (2 ), we have 

(T2-I)( A, B,C) =(0, 0, 0)

Then by 1, we have

(T2 — I)(T — 3I) (a , b , c ) = ( 0 , 0 , 0 )

IIT JAM Mathematics Practice Test- 15 - Question 19

Let  be a linear transform ation satisfying T3 + 3T2 = 4I, where I is  the identity transformation. Then the linear transformation S = T4 + 3T3 - 4I is

Detailed Solution for IIT JAM Mathematics Practice Test- 15 - Question 19

Given

  be a linear transformation,

,  is also a linear transformation, on 

Now S = T4 + 3T3 - 4I is also a linear transformation , on 

⇒ S = T (T3 + 3T2) - 4I = T (4I) - 4I = 4 (T - 1)

Now ker
(S) = ( V ∈ R4 | S(v) = 0)

⇒ 1 is the eigenvalue of T

⇒ ker (S) be the set of eigenvectors corresponding to the eigenvalue 1.

⇒ S is not one -one

⇒ S i s not on to

now clearly 1 is an eigenvalue of T

⇒ S is not invertible

IIT JAM Mathematics Practice Test- 15 - Question 20

Choose the correct option

Detailed Solution for IIT JAM Mathematics Practice Test- 15 - Question 20

(a) We know dim V = dim (Range T) + dim (KerT) 

Here range T = KerT 

⇒ dim (Range T ) = dim (kerT) 

⇒ dim V = 2 dim (Range T)

⇒ dim V is even

(b) T (1 , 0 ) = ( 0 , 0 ) 

T ( 0 , 1 ) = ( 1 , 0 )

=> dim ( Range T ) = 1

⇒ Thus basis for kerT is {(1,0)} 

⇒ dim ker T = 1

⇒ ker T = Range T.

(c) T (1 ,0 ) = ( 1 , 1 , 0 )

T (0,1) = (0 , 1 , 1 )

⇒ RanK T = 2 => nullity = 0

IIT JAM Mathematics Practice Test- 15 - Question 21

Let  Let M be the matrix whose columns are, V1+V2, 2V1 + V2, V1 - 2V2 in that order. Then which of the followings is not true for Matrix M ?

Detailed Solution for IIT JAM Mathematics Practice Test- 15 - Question 21

Given

det M = 2 + 12 + 3 - ( 2 + 12 + 3) = 0

⇒ M not invertible.
Characteristic equation of M is given as :

T ake Mx = 0 ...(i)

Then we know no. of L.l. solution of (i) = nullity of M

= no. of column's of M - Rank M = 3 - 2 = 1

Since all the ev’s of M are distinct.

⇒ M is diagonalizable.

IIT JAM Mathematics Practice Test- 15 - Question 22

Define  into itself by,  Then, matrix of  T-1 relative to the standard basis for  is

Detailed Solution for IIT JAM Mathematics Practice Test- 15 - Question 22

Here kerT = {0}

⇒ T is one - one and onto 

⇒ Unique T-1 exist 

IIT JAM Mathematics Practice Test- 15 - Question 23

Let A and B are two 3 * 3 real matrices. Let CA and CB denotes the span of the columns of A and B respectively and RA, RB denotes the span of the rows of A and B respectively.

where

Then the max {c, d}, where  and  is given by,

Detailed Solution for IIT JAM Mathematics Practice Test- 15 - Question 23

Given

CA = span {(1 , 0, 1), (1, 2, - 1 ) , ( 0 , 1 , 1 ) } 

CB = span{(1, 0, -1), (1, 1, 1), (-1, 0, 1)}

⇒ CA ∩ CB = span {(1 , 0, 1), (1, 2, - 1 ) , (0, 1 ,1 ) , (1, 0, - 1 ) , (1, 1, 1), ( - 1 , 0 , 1 ) } = span{V1,V2, V3, V4, V5, V6} (say) 

Clearly V4 = - V6​​​​

 

IIT JAM Mathematics Practice Test- 15 - Question 24

Let  be linear transformation such that ToS is the identity map on , Then

Detailed Solution for IIT JAM Mathematics Practice Test- 15 - Question 24

Let

⇒ Basis for SoT is {( 0,0,0,1)} ≠{(0,0,0,0)}

⇒ SoT neither one - one nor onto.

IIT JAM Mathematics Practice Test- 15 - Question 25

The matrix is?

Detailed Solution for IIT JAM Mathematics Practice Test- 15 - Question 25

If the identity matrix has other numbers in place of 1 it is said to be a scalar matrix. Identity matrix is also a type of scalar matrix.

IIT JAM Mathematics Practice Test- 15 - Question 26

The matrix  has one eigen value equal to 3. The sum of the other two eigen values is

Detailed Solution for IIT JAM Mathematics Practice Test- 15 - Question 26

∵ sum of the eigen values of matrix = Sum of diagonal values present in the matrix

Hence option (c) is correct.

IIT JAM Mathematics Practice Test- 15 - Question 27

Let M be a 7 x 7 matrix of Rank 4 with real entries, let  be a column vector. Then Rank of M + XX7 is at least

Detailed Solution for IIT JAM Mathematics Practice Test- 15 - Question 27

Given Rank

M = 4 we Know Rank XXT = 1

we know if A and B are two (nxn) matrices then we have Rank (A+B) ≤ Rank A + Rank - Max {c,d}

where


where CA CB: space those are generated by columns of A and B respectively and RA, RB : space those are generated by rows of A and B respectively 

Here we have

IIT JAM Mathematics Practice Test- 15 - Question 28

Let A be a 10x10 real matrix, whose elements are defined as (where w is cube root of unity), Then trace of A is ,

Detailed Solution for IIT JAM Mathematics Practice Test- 15 - Question 28

we have

IIT JAM Mathematics Practice Test- 15 - Question 29

let A = [ aij] be a 3x3 invertible matrix with real entries and B = [bij] be a matrix which is formed such that bij is the sum of all the elements except aij in the ith row. Answer the following:
If there exist a matrix X with Constant elements such that A X = B, then X is

Detailed Solution for IIT JAM Mathematics Practice Test- 15 - Question 29

Now ginen A  is invertiable ⇒ |A| ≠ 0 and A-1 exit and AX  = B

and 


IIT JAM Mathematics Practice Test- 15 - Question 30

Let M be a m X n (m < n) matrix with rank m. Then

Detailed Solution for IIT JAM Mathematics Practice Test- 15 - Question 30

Mx = b 

where M = m * n matrix 

r = rank M = m(given)

∴ Rank (M) = Rank[M, b] = m

Hence equations are consistent (n - r) i.e. n - m → linearly independent solution.

Now any linear combination of the (n - m) independent solution will also be a solution. Hence other are infinite number of solutions.

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