Show that a skew-symmetric matrix of odd order is aways singular.

1.	Show that a skew-symmetric matrix of odd order is aways singular.
Answer» Let, A be a skew-symmetric square matrix ofn×n, where n is odd, By general properties of determinants, det(A)=det(A^T)…(i) However, since A is a skew-symmetric matrix where aij=−aij(i,j are rows and column numbers ), ∴∴In case of skew-symmetric matrix, A^T=−ANowdet(−A)=det(A^T)But,det(−A)=(−1)^ndet(A) where n is no. of rows/columns in a square Matrix. ∴det(A^T)=(−1)^n∵n is odd,(−1)°n=−1∴det(A^T)=−det(A)...iiSubtractingequation(ii)from(i),∴2det(A)=det(A^T)−det(A^T)=0∴det(A)=0 Hence matrix A is singular.... By definition of singular matrix Hence proved.

Answer»

Let, A be a skew-symmetric square matrix ofn×n, where n is odd, By general properties of determinants,

det(A)=det(A^T)…(i)

However, since A is a skew-symmetric matrix where

aij=−aij(i,j are rows and column numbers ),

∴∴In case of skew-symmetric matrix,

A^T=−ANowdet(−A)=det(A^T)But,det(−A)=(−1)^ndet(A)

where n is no. of rows/columns in a square Matrix.

∴det(A^T)=(−1)^n∵n is odd,(−1)°n=−1∴det(A^T)=−det(A)...iiSubtractingequation(ii)from(i),∴2det(A)=det(A^T)−det(A^T)=0∴det(A)=0

Hence matrix A is singular.... By definition of singular matrix

Hence proved.

Discussion