Test: Matrix - Mathematics MCQ

We know AA^-1 = I₂

first do AA^t then you will find
AA^t = I₄
[AA^T]^-1 = [I₄]^-1 = I₄

ANSWER :- c

Solution :- Let  A  be real skew symmetric and suppose  λ∈C  is an eigenvalue, with (complex) eigenvector  v . Then, denoting by  H  hermitian transposition,

λvHv=vH(λv)=vH(Av)=vH(−AHv)=−(vHAH)v=−(Av)Hv=−(λv)Hv=−λ¯vHv 

Since  vHv≠0 , as  v≠0 , we get

λ=−λ¯ 

so  λ  is purely imaginary (which includes 0). Note that the proof works the same for a antihermitian (complex) matrix.

With a completely similar technique you can prove that the eigenvalues of a Hermitian matrix (which includes real symmetric) are real.

Then also every minor or order 2 is also zero.
∴ rank (C) = 1

Highest possible rank of A= 2 ,as   Ax = b is an inconsistent system.

(P) Singular matrix → Determinant is zero
(Q) Non-square matrix → Determinant is not defined 
(R) Real symmetric → Eigen values are always real  
(S) Ortho gonal → Det erminant is always one 

∴ Rank (A) = 2

When two matrices (x,y) and (a,b) are to be multiplied, the multiplication is possible

when y = a, and the resultant matrix is of the order (x * a)

X^T = The matrix in which rows and columns are interchanged

{(2 x 3) x [(3x4) x (4x3)]^-1 x (3x2)}^T = { (2x3) x (3x3) x (3x2) }^T = { (2x3) x (3x2) }^T = 2 x 2

B^T = – B