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We know AA-1 = I2
We know
AAt = I4
[AAT]-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.
The rank of a 3 x 3 matrix C (=AB), found by multiplying a non-zero column matrix A of size 3 x 1 and a non-zero row matrix B of size 1 x 3, is
Then also every minor or order 2 is also zero.
∴ rank (C) = 1
A is a 3 x 4 real matrix and A x = b is an inconsistent system of equations. The highest possible rank of A is
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
[A] is a square matrix which is neither symmetric nor skew-symmetric and [A]T is its transpose. The sum and difference of these matrices are defined as [S] = [A] + [A]T and [D] = [A] -[A]T, respectively. Which of the following statements is True?
∴ Rank (A) = 2
Real matrices [A]3 x 1, [B]3 x 3, [C]3 x 5,[D]5 x 3, [E]5 x 5 and [F]5 x 1 are given. Matrices [B] and [E] are symmetric.
Following statements are made with respect to these matrices.
1. Matrix product [F]T [C]T [B] [C] [F] is a scalar.
2. Matrix product [D]T [F] [D] is always symmetric.
With reference to above statements, which of the following applies?
Consider the matrices X (4 × 3), Y (4 × 3) and P (2 × 3). The order or P (XTY)–1PT] T will be
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)
XT = 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
(PQ)-1P = Q-1P-1P = Q-1
BT = – B
27 docs|150 tests
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27 docs|150 tests
|