IIT JAM Mathematics Practice Test- 15 - Mathematics MCQ

Now for the maximum value of the determinant take

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

then

Apply R₂ → R₂ - R₁

a ∈ any real no.

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

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

Apply

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

Given

⇒ A + B + AB be inverse of B

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

if n = 2 then

n= 3 then 

n= 4 then 

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

⇒ det of given matrix will be 

x + y = z

y + z = w

⇒ y = w - z

⇒ x = 2z- w

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

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.

⇒ W₁ not a subspace.

⇒ W₂ not a subspace.

(C) Let P₁, P₂ ∈ V₃ then 

⇒ W₃ be a subspace.

(D) Let P_1, P₂ ∈ W₄ then P₁" + P_1' = 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.

Here the characteristic equation is .

So by cayley's hamiiton theorem ,

Now since A(A² - I) = 0

so

So |A₁₀₀ + 1| = 8

Take

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)

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)

Given

so

Thus we have 

So in general

now

⇒ N(T) = {0}

⇒ T is one -one and onto.

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 )

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

(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

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.

Here kerT = {0}

⇒ T is one - one and onto 

⇒ Unique T^-1 exist 

Given

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

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

⇒ C_A ∩ C_B = 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​​​​

​

Let

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

⇒ SoT neither one - one nor onto.

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.

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

Hence option (c) is correct.

Given Rank

M = 4 we Know Rank XX^T = 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 C_A C_B: space those are generated by columns of A and B respectively and R_A, R_B : space those are generated by rows of A and B respectively 

Here we have

we have

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

and 

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.