3.3. THE MATHEMATICAL THEORY OF DETERMINANTS 93

3.3.3 A Symmetric Definition

With the above, it is possible to give a more symmetric description of the determinant fromwhich it will follow that det (A) = det

(AT).

Corollary 3.3.8 The following formula for det (A) is valid.

det (A) =1

n!·∑

(r1,··· ,rn)

∑(k1,··· ,kn)

sgn (r1, · · · , rn) sgn (k1, · · · , kn) ar1k1· · · arnkn

. (3.10)

And also det(AT)= det (A) where AT is the transpose of A. (Recall that

(AT)ij= Aji.)

Proof: From Proposition 3.3.6, if the ri are distinct,

det (A) =∑

(k1,··· ,kn)

sgn (r1, · · · , rn) sgn (k1, · · · , kn) ar1k1 · · · arnkn .

Summing over all ordered lists, (r1, · · · , rn) where the ri are distinct, (If the ri are notdistinct, sgn (r1, · · · , rn) = 0 and so there is no contribution to the sum.)

n! det (A) =∑

(r1,··· ,rn)

∑(k1,··· ,kn)

sgn (r1, · · · , rn) sgn (k1, · · · , kn) ar1k1 · · · arnkn .

This proves the corollary since the formula gives the same number for A as it does for AT .■

Corollary 3.3.9 If two rows or two columns in an n × n matrix A, are switched, thedeterminant of the resulting matrix equals (−1) times the determinant of the original matrix.If A is an n×n matrix in which two rows are equal or two columns are equal then det (A) = 0.Suppose the ith row of A equals (xa1 + yb1, · · · , xan + ybn). Then

det (A) = xdet (A1) + y det (A2)

where the ith row of A1 is (a1, · · · , an) and the ith row of A2 is (b1, · · · , bn) , all other rowsof A1 and A2 coinciding with those of A. In other words, det is a linear function of eachrow A. The same is true with the word “row” replaced with the word “column”.

Proof: By Proposition 3.3.6 when two rows are switched, the determinant of the re-sulting matrix is (−1) times the determinant of the original matrix. By Corollary 3.3.8 thesame holds for columns because the columns of the matrix equal the rows of the transposedmatrix. Thus if A1 is the matrix obtained from A by switching two columns,

det (A) = det(AT)= −det

(AT

1

)= −det (A1) .

If A has two equal columns or two equal rows, then switching them results in the samematrix. Therefore, det (A) = −det (A) and so det (A) = 0.

It remains to verify the last assertion.

det (A) ≡∑

(k1,··· ,kn)

sgn (k1, · · · , kn) a1k1 · · · (xarki + ybrki) · · · ankn

= x∑

(k1,··· ,kn)

sgn (k1, · · · , kn) a1k1· · · arki

· · · ankn

+y∑

(k1,··· ,kn)

sgn (k1, · · · , kn) a1k1 · · · brki · · · ankn ≡ xdet (A1) + y det (A2) .

The same is true of columns because det(AT)= det (A) and the rows of AT are the columns

of A. ■