Question #200602

Evaluate det(−A) and det(−A T ). Compare det(−A) and det(−A T ) for:

(2.1) A = -4 2

3 -3


(2.2) A = 3 1 -2

-5 3 -6

-1 0 -4


1
Expert's answer
2021-05-31T16:26:56-0400

The determinant of the transpose of a square matrix is equal to the determinant of the matrix, that is,


det(AT)=det(A)\det(A^T)=\det (A)

2.1


A=[4233]A=\begin{bmatrix} -4 & 2 \\ 3 & -3 \end{bmatrix}

A=[4233]-A=\begin{bmatrix} 4 & -2 \\ -3 & 3 \end{bmatrix}

det(A)=4233=4(3)(2)(3)=18\det (-A)=\begin{vmatrix} 4 & -2 \\ -3 & 3 \end{vmatrix}=4(3)-(-2)(-3)=18

AT=[4233]T=[4323]A^T=\begin{bmatrix} -4 & 2 \\ 3 & -3 \end{bmatrix}^T=\begin{bmatrix} -4 & 3 \\ 2 & -3 \end{bmatrix}

AT=[4323]-A^T=\begin{bmatrix} 4 & -3 \\ -2 & 3 \end{bmatrix}


det(AT)=4323=4(3)(2)(3)=18\det (-A^T)=\begin{vmatrix} 4 & -3 \\ -2 & 3 \end{vmatrix}=4(3)-(-2)(-3)=18


det(A)=det(AT)\det (-A)=\det (-A^T)

2.2



A=[312536104]A=\begin{bmatrix} 3 & 1 & -2 \\ -5 & 3 & -6 \\ -1 & 0 & -4 \end{bmatrix}

A=[312536104]-A=\begin{bmatrix} -3 & -1 & 2 \\ 5 & -3 & 6 \\ 1 & 0 & 4 \end{bmatrix}

det(A)=312536104\det (-A)=\begin{vmatrix} -3 & -1 & 2 \\ 5 & -3 & 6 \\ 1 & 0 & 4 \end{vmatrix}

=1123603256+43153=1\begin{vmatrix} -1& 2 \\ -3 & 6 \end{vmatrix}-0\begin{vmatrix} -3 & 2 \\ 5 & 6 \end{vmatrix}+4\begin{vmatrix} -3 & -1 \\ 5 & -3 \end{vmatrix}

=6+6+4(9+5)=56=-6+6+4(9+5)=56

AT=[312536104]T=[351130264]A^T=\begin{bmatrix} 3 & 1 & -2 \\ -5 & 3 & -6 \\ -1 & 0 & -4 \end{bmatrix}^T=\begin{bmatrix} 3 &- 5 & -1 \\ 1 & 3 & 0 \\ -2 & -6 & -4 \end{bmatrix}

AT=[351130264]-A^T=\begin{bmatrix} -3 & 5 & 1 \\ -1 & -3 & 0 \\ 2 & 6 & 4 \end{bmatrix}


det(AT)=351130264\det (-A^T)=\begin{vmatrix} -3 & 5 & 1 \\ -1 & -3 & 0 \\ 2 & 6 & 4 \end{vmatrix}

=1132603526+43513=1\begin{vmatrix} -1& -3 \\ 2 & 6 \end{vmatrix}-0\begin{vmatrix} -3 & 5 \\ 2 & 6 \end{vmatrix}+4\begin{vmatrix} -3 &5 \\ -1 & -3 \end{vmatrix}


=6+69+4(9+5)=56=-6+6-9+4(9+5)=56


det(A)=det(AT)\det (-A)=\det (-A^T)


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