Question

The determinant of a skew symmetric matrix of odd order is *​

Answer 1

Step-by-step explanation:

written 4.4 years ago by

gravatar for shaily.mishra30 shaily.mishra30 • 160

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

det(A)=det(AT)…(i)det(A)=det(AT)…(i)

However, since A is a skew-symmetric matrix where

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

∴∴ In case of skew-symmetric matrix,

AT=−ANowdet(−A)=det(AT)But,det(−A)=(−1)ndet(A)AT=−ANowdet(−A)=det(AT)But,det(−A)=(−1)ndet(A)

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

∴det(AT)=(−1)ndet(A)∵nisodd,(−1)n=−1∴det(AT)=−det(A)…(ii)Subtractingequation(ii)from(i),∴2det(A)=det(AT)−det(AT)=0∴det(A)=0∴det(AT)=(−1)ndet(A)∵nisodd,(−1)n=−1∴det(AT)=−det(A)…(ii)Subtractingequation(ii)from(i),∴2det(A)=det(AT)−det(AT)=0∴det(A)=0

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

Hence proved.