# How can det(B)=-det(A) when this happens?

There's a property that says when you interchange two rows/columns from a matrix A, the resulting determinant B will have its determinant equal to the original one, but with its sign inversed: $\det(B)=-\det(A)$. I've been wanting to know how to explain it for this specific case (without using elementary matrix):

For the A matrix (the original matrix) I've chosen to do cofactor expasion on the first row: $$|A| =\ \ \ \begin{vmatrix} a_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\\ \end{vmatrix} = a_{11} \begin{vmatrix} a_{22}& a_{23}\\ a_{32}& a_{33}\\ \end{vmatrix} -a_{12} \begin{vmatrix} a_{21}& a_{23}\\ a_{31}& a_{33}\\ \end{vmatrix} +a_{13} \begin{vmatrix} a_{21}& a_{22}\\ a_{31}& a_{32}\\ \end{vmatrix}$$ Now, when I interchange the first row of the A matrix with its second row, I'll have this (for which I've chosen to do the row expansion on the second row, since it's the same row as the one I used for A): $$|B| =\ \ \ \begin{vmatrix} a_{21}& a_{22}& a_{23}\\ a_{11}& a_{12}& a_{13}\\ a_{31}& a_{32}& a_{33}\\ \end{vmatrix} = a_{11} \begin{vmatrix} a_{22}& a_{23}\\ a_{32}& a_{33}\\ \end{vmatrix} -a_{12} \begin{vmatrix} a_{21}& a_{23}\\ a_{31}& a_{33}\\ \end{vmatrix} +a_{13} \begin{vmatrix} a_{21}& a_{22}\\ a_{31}& a_{32}\\ \end{vmatrix}$$ So, it clearly sounds like |A| = |B| by looking at these, although i'm perfectly sure it's not. What am I missing here? =\

• HINT: remember the "checkerboard" pattern of pluses and minuses for row/column expansion for determinant computation. – afedder Jun 22 '14 at 20:49
• Just googled "checkerboard", it really helps to make sure you're understanding it right =P. Thanks o/ – user2743765 Jun 22 '14 at 20:53

You've done the cofactor expansion incorrectly. When you expand along the middle row for $B$, you start with $-$ first, not $+$.
If you begin with a $(-)$ sign in the row expansion for $\text{det}(B)$, you will end up with exactly opposite $\text{det}(A)$, seeing as each term will be the same, except that the first and third terms will have a minus sign in front of them, while the second term will be positive. This proves the desired claim for the $3 \times 3$ case.