Test: Linear Algebra - 2 - Mathematics MCQ

AX = B

Where A is 2 × 4 matrices and b ≠ 0

The order of A^T is 4 × 2

The order of A^TA is 4 × 4

Rank of (A) ≤ min (2, 4) = 2

Rank of (A^T) ≤ min (2, 4) = 2

Rank (A^TA) ≤ min (2, 2) = 2

As the matrix A^TA is of order 4 × 4, to have a unique solution the rank of A^TA should be 4.

Therefore, the unique solution of this equation is not possible.

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)

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.

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.

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]}.