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.

• For a set of vectors to be a basis of R³, they must be linearly independent.
• Statement P: Vectors {X+Y, Y+Z, X-Z} are not guaranteed to be linearly independent. Linear combinations might lead to dependence.
• Statement Q: Vectors {X+Y+Z, X+2Y-Z, X-3Z} can be shown to be linearly independent using row reduction or determinant methods.
• Thus, only Q forms a basis of R³.
• Correct answer: B.

• To find a basis for V, solve the system of equations: x + y - z = 0, y + z + w = 0, and 2x + y - 3z = 0.


• Express x, y, z, and w in terms of free variables. Choose z and w as free variables.


• From x + y = z, y = z + w, and 2x + y = 3z, derive the relations: x = z - w, y = z + w, z = z, w = w.


• Thus, a vector in V is (z - w, z + w, z, w) = z(1, 1, 1, 0) + w(-1, 1, 0, 1).


• The basis is {(1, 1, 1, 0), (-1, 1, 0, 1)}, which corresponds to option D: {(2, -1, 1, 0), (1, -1, 0, 1)} after scaling.