Question #198537

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


Expert's answer

2.1)


det( A ) = ∣−423−3∣\begin{vmatrix} -4 & 2 \\ 3 & -3 \\ \end{vmatrix}



∴ det( -A ) = ∣4−2−33∣\begin{vmatrix} 4 & -2 \\ -3 & 3 \end{vmatrix}




Expanding along 1st row, we have


det(-A) = 4 * 3 - (- 3 ) * (- 2 )


det( -A ) = 6



Again, det( - AT) = ∣4−3−23∣\begin{vmatrix} 4 & -3 \\ -2 & 3 \\ \end{vmatrix}


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) = ∣31−2−53−6−10−4∣\begin{vmatrix} 3 & 1 & -2 \\ -5 & 3 & -6 \\ -1&0&-4\\ \end{vmatrix}


So, det( - A ) = ∣−3−125−36104∣\begin{vmatrix} -3 & -1 & 2 \\ 5 & -3 & 6 \\ 1&0&4\\ \end{vmatrix}


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) = ∣−351−1−30264∣\begin{vmatrix} -3 & 5 & 1 \\ -1 & -3 & 0 \\ 2&6&4\\ \end{vmatrix}


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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