Let us begin with the determinant of product of matrices. The proof of the property will be given here, after the rigorous definition of determinants.
Taking the absolute value, the property means that the ratio for the change of volume under the composition AB is the product of the ratios under the transformations A and B. Intuitively, this is quite obvious: enlarging the volume twice and then enlarging again three times is the same as enlarging six (two times three) times.
Taking the signs, we may consider various combinations. For example, if 2 by 2 matrices A and B preserve orientations (i.e., detA > 0, detB > 0), then
(u, v) clockwise ⇒ (Au, Bv) clockwise ⇒ (ABu, ABv) clockwise,
so that AB also preserves orientation (i.e., detAB > 0). We may also verify signs of other combinations in a similar way.
We also note that the product property implies
det(Ak) = (detA)k, for k > 0.
The next property indicates that, as far as the determinant is concerned, there is no difference between rows and columns. In particular, one may use row operations as well as column operations. The proof of the property will be given here, after the rigorous definition of determinants.
The next property is quite useful for computation. The proof of the property will be given here, after the rigorous definition of determinants.
Geometrically, the absolute value of the determinant represents the volumn of a high dimensional parallelogram with B-paralleogram as "base" and A-parallelogram (skewed by # term) as "height". The size of the base is |detB|, and the size of the height is |detA|. Therefore the size of the whole parallelogram is |detA detB|.
The triangle property can be generalized (by induction and transpose, for example). For square matrices A1, A2, ..., Ak, we have
|det[||A1||#||. .||#||] = detA1 detA2 ... detAk,|
|det[||A1||O||. .||O||] = detA1 detA2 ... detAk.|
In particular, for upper and lower triangular matrices, we have
|det[||a1||#||. .||#||] = a1a2...an,|
|det[||a1||0||. .||0||] = a1a2...an.|
Further specializing to the identity matrix, we have
detI = 1.
Geometrically, this simply means that the cube with unit side length has volume 1, and the standard basis has the positive orientation. Moreover, from AA-1 = I, we have detA detA-1 = 1, so that det(A-1) = (detA)-1. Combined with det(Ak) = (detA)k for k > 0, it is not hard to conclude that
det(Ak) = (detA)k for invertible A and all k.
|det[||1||0||2||0||] = -det[||1||2||0||0||] = det[||1||2||0||0||]|
|0||3||0||4||0||0||3||4||5||6||0||0||= det[||1||2||] det[||3||4||] = (-4)(-4) = 16.|
Example Let A and B be n by n matrices. Suppose detA = 7, detB = 2. Then
det(AB) = detA detB = 14,
det(ATAB2A-3B) = detAT detA (detB)2 detA-3 detB = (detA)-1 (detB)3 = 8/7.
Finally, we remark that
det(A + B) ≠ detA + detB.
However, the determinant is multilinear with respect to rows and columns.
In dA, the number d is multiplied to each row of A. By the second of this property, for an n by n matrix A, we have
det(dA) = dn detA ≠ d detA.