Evaluate det(−A) and det(−AT). Compare det(−A) and det(−AT) for:
(2.1) A = −4 2
3 −3 ;
(2.2) A = 3 1 −2
−5 3 −6
−1 0 −4
2.1)
det( A ) =
∴ det( -A ) =
Expanding along 1st row, we have
det(-A) = 4 * 3 - (- 3 ) * (- 2 )
det( -A ) = 6
Again, det( - AT) =
Expanding along 1st row, we have
det(-A) = 4 * 3 - (- 2 ) * (- 3 )
det(-AT) = 6
Hence, we see that det( -A ) = det(-AT) .
2.2)
det(A) =
So, det( - A ) =
Expanding along 1st row, we have
det(-A) = -3[ ( -3 ) 4 - 0 6] - ( -1 )[5 4 - 1 6] + 2[ 5 0 - 1 (-3)]
det(-A) = 56
det(- AT) =
Expanding along 1st row, we have
det(-A) = -3[ ( -3 ) 4 - 6 0] - 5 [ ( -1 ) 4 - 2 0] + 1[ ( -1 ) 6 - 2 ( -3 ) ]
det(- AT) = 56
Hence, we see that det( -A ) = det(-AT) .
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