Test: Linear Algebra - 2
20 Questions MCQ Test Topic-wise Tests & Solved Examples for IIT JAM Mathematics | Test: Linear Algebra - 2
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If {v1, v2, v3} is a linearly independent set of vectors in a vector space over 
which one of the following sets is also linearly independent?
which one of the following sets is also linearly independent?Consdierthe subspace W = {(x1, x2,.... ,x10) 
xn = xn-1 + xn - 2 for 3 < n < 10) of the vector space 
the dimension of W is
xn = xn-1 + xn - 2 for 3 < n < 10) of the vector space the dimension of W isWhich one of the following is a subspace of the vector space 
If A and B are 3 × 3 real matrices such that rank (AB) = 1, then rank (BA) cannot be
Here A & B a re 3 × 2 real matrices such that rank (AB) = 1
So, |AB| = 0 ⇒ |A| |B| = 0 (∴ |AB| = |A| |B|)
⇒ either |A| or |B| should be zero
So, |BA| = |B||A| = 0
⇒ BA is singular
Hence rank (BA) cannot be 3. (Because BA is 3 × 3 matrix)
Which of the following sets of functions from 
is a vector space over 
Let T : R3 → R3 be the linear transformation define by T(x, y, z) = (x + y, + z, z + x) for all (x, y, z) ∈ 3. Then
Let T : R2 → R3 be the L.T. defined as
T (x, y, z) = (x + y, y + z, z + x) ∀ (x, y, z) ∈ R3
N (T ) = {( x, y, z)|T ( x, y, z) = 0}
= {(x, y, z)|x + y = 0,y + z = 0, z + x = 0}
= {(0, 0, 0)| x, y, z ∈R}
⇒ dim N (T) = 0
Rank (N) + Nullity T = dim R3 = 3
Rank (T) + 0 = 3 ⇒ Rank of T = 3
Here,
So, Answer is skew-Hermitian matrix.
Let V be a 3 dimensional vector space over the field 
of 3 elements. The number of distinct 1 dimensional subspaces of V is
Let M = {(a1,a2,a3): ai ∈ (1,2,3,4); a1 + a2 + a3 = 6} then the number of elements in M is
The dimension of the vector space of all symmetric matrices A = (aij) of order n x n (n > 2) with real entries, aii = 0 and trace zero is
The dimension of the space of n×n symmetric matrices with diagonal equal to zero is 
. Now, in your case the diagonal is not zero but the sum of its elements is zero, that means that you have n−1 elements which can vary. SO you get 
as expected.
Let n be a positive integer and let Hn be the space of all n x n matrices A = (aij) with entries in 
satisfying aij = ars whenever i + j = r + s (i, j, r, s = 1, 2,..., n) then the dimention of Hn, as a vector space over 
The dimension of the vector space of all symmetric matrices of order n x n (n > 2) with real entries and trace equal to zero is
Let {X,Y,Z) be a basis of 
Consider the following statements P and Q
P : {X + Y,Y + Z, X - Z) is a basis of 
Q : {X + Y + Z,X + 2Y - Z, X - 3Z} is a basis of 
Which of the above statements hold true?
Let M be the space of all 4 x 3 matrices with entries in the finite field of three elements. Then the number of matrices of Rank three In M is
Consider the subspace W = {[aij] : aij = 0, if i is even) of all 10 x 10 matrices(real) then the dimension of W is
over the field 
Then the dim V is
and let V = {(x, y, z) 
: |A| = 0}.Then the dimension of V equals
A basis of V = {(x, y, z, w) 
: x + y - z = 0, y + z + w = 0, 2x + y - 3z = 0)
The system of 3 equations in 4 unknowns is
1x + 1y +(-1)z +0w =0……..(1)
0x + 1y + 1z + 1w = 0……….(2)
2x + 1y +(-3)z +1w = 0……..(3). The row operation (3) - 2×(1) reduces (3) to
0x +(-1)y+(-1)z+(-1)w =0…….(4), which is just (-1)× (3). Hence the given system is equivalent to the pair of equations:
x = -y+z and y= -z -w. This pair is equivalent to x=(z+w)+z = 2z+w and y = -z -w, where z and w are completely arbitrary. Thus
[x,y,z,w] =[2z+w,-z-w,z,w] =
z[2,-1,1,0] + w[1,-1,0,1].
We had already observed that the rank of the echelon matrix equivalent to the given coefficient matrix is 2, and hence its nullity = 4–2 = 2, which is the same as the dimension of the solution space V. Thus a basis of V may be taken as
B={[2,-1,1,0], [1,-1,0,1]}.
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