Determinant and cross product of rows/columns

Question

I have been learning about determinants and, in particular, calculating determinants by adding multiples of one row/column to another. However on doing some problems, I see that you cannot simply add and subtract these anyhow.

Specifically:

• you cannot transform row $i$ by subtracting some multiple of row $j$, and then transform row $j$ by subtracting some multiple of row $i$ (as in, what $i$ was before you transformed it). The resulting matrix will have a different determinant to help original.

• secondly, you cannot add/subtract multiples of one row from itself without changing the determinant.

I know these points are probably very obvious and elementary, however they were not at all obvious to me from how I was taught (i.e. that you can just subtract multiples of rows/columns to obtain the triangular form.)

Although I cannot come up with a geometric/intuitive picture of why the above statements are true, I can motivate it in the $3\times 3$ case considering the determinant as the vector triple product:

$det (A)=a \cdot (b \times c) $

I can see from this that if I add a multiple of a vector to itself, the result is changed. I can only add multiples of different rows. Also, I can see that the result is also changed if I add a multiple of one row to another, and simultaneously add a multiple of the second row to the first (say change $b$ to $b+2a$, and $a$ to $a+0.5b$ etc)

However I do not know how to rationalise this for higher dimension matrices. Is there an equivalent cross /wedge product formulation for the determinant in higher dimensions that would illustrate this easily? The only formulation if the determinant I do know is involving the cofactor, or the levi civita tensor and cycling around the elements. But it is difficult to see any result in these forms, as you can with the cross product formula.

Answer 1

INTUITIVE PICTURE

If the rows $r_k$ of a matrix $A$ represent vectors, then $\det(A)$ can be interpreted as the oriented volume of the parallelepiped defined by said vectors. This being the case

$r_k \to \alpha r_k$

If you multiply one of the lengths by a factor $\alpha$, then the volume of the parallelepiped changes by the same factor

$r_k \to r_k + \alpha r_l $

Note that in general, if you two rows are the same, then the volume is zero, because the lengths of one of the sides of the parallelepiped is zero. So this operation leaves the volume unchanged

$r_k \leftrightarrow r_l $

This a bit more tricky, but the idea is that as you swap any two vectors, then the sign of the volume changes. But the absolute value remains the same

FORMAL PICTURE

Let $A$ a $n\times n$ square matrix, then

$$ \det(A) = \epsilon_{i_1\cdots i_n} a_{1i_1}\cdots a_{1 i_n} $$

if we call $A'$ the resulting matrix after applying on $A$ an elementary operation, and $r_k$ the $k$-th row of $A$, then

$r_k \to \alpha r_k$

\begin{eqnarray} \det(A') &=& \epsilon_{i_1\cdots i_n} a'_{1i_1}\cdots a'_{k i_k} \cdots a'_{n i_n} \\ &=& \epsilon_{i_1\cdots i_n} a_{1i_1}\cdots (\alpha a_{k i_k}) \cdots a_{n i_n} \\ &=& \alpha \epsilon_{i_1\cdots i_n} a_{1i_1}\cdots a_{k i_k} \cdots a_{n i_n} \\ &=& \alpha \det(A) \end{eqnarray}

$r_k \to r_k + \alpha r_l $

\begin{eqnarray} \det(A') &=& \epsilon_{i_1\cdots i_n} a'_{1i_1}\cdots a'_{k i_k} \cdots a'_{l i_l} \cdots a'_{n i_n} \\ &=& \epsilon_{i_1\cdots i_n} a_{1i_1}\cdots (a_{k i_k} + \alpha a_{l i_k}) \cdots a_{l i_l} \cdots a_{1 i_n} \\ &=& (\epsilon_{i_1\cdots i_n} a_{1i_1}\cdots a_{k i_k} \cdots a_{l i_l} \cdots a_{1 i_n}) + (\alpha \epsilon_{i_1\cdots i_n} a_{1i_1}\cdots a_{l i_k} \cdots a_{l i_l} \cdots a_{1 i_n}) \\ &=& (\epsilon_{i_1\cdots i_n} a_{1i_1}\cdots a_{k i_k} \cdots a_{l i_l} \cdots a_{1 i_n}) + 0 \\ &=& \det(A) \end{eqnarray}

$r_k \leftrightarrow r_l $

\begin{eqnarray} \det(A') &=& \epsilon_{i_1\cdots i_k \cdots i_l \cdots i_n} a'_{1i_1}\cdots a'_{k i_k} \cdots a'_{l i_l} \cdots a'_{n i_n} \\ &=& \epsilon_{i_1\cdots i_k \cdots i_l \cdots i_n} a_{1i_1}\cdots a_{li_k} \cdots a_{k i_l} \cdots a_{1 i_n} \\ &=& \epsilon_{i_1\cdots i_l \cdots i_k \cdots i_n} a_{1i_1}\cdots a_{li_l} \cdots a_{k i_k} \cdots a_{1 i_n} \\ &=& -\epsilon_{i_1\cdots i_k \cdots i_l \cdots i_n} a_{1i_1}\cdots a_{li_l} \cdots a_{l i_l} \cdots a_{1 i_n} \\ &=& -\det(A) \end{eqnarray}