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The determinant of a skew symmetric matrix of odd order is * |
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Answer» 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. |
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