Calculus: Analytic Geometry and Calculus, with Vectors

2.5 Determinants and applications 89

of the third row is demonstrated by the expansion

2 3 -1 6

1 3 7 3 -1^6

-2 = 3

3 0 0 0 3 7 -2 +0

-1 5 4

1 -1 5 4

which reduces the problem of evaluating a determinant of order 4 to
the problem of evaluating a single determinant of order 3.
It should be known and remembered that a determinant is 0 if two
of its rows (or two of its columns) are identical. For determinants of
order 2, this is obvious because

a b

= ab-ab = 0 b- b=0

l =

Th t

a b

,

b

a a.

b

a

a

a

b

b

c

c =0,

a

b

d a

e b =0

d e f c f c

can be seen by expanding the first determinant in terms of elements of
the bottom row and by expanding the second in terms of elements of the
middle column. When the result has been established for determinants
of order 3, the same trick enables us to establish the result for determi-
nants of order 4, and so on. A simple modification of the above pro-
cedure shows that if two adjacent rows (or columns) of a determinant
are interchanged, then the value of the determinant is multiplied by -1,
that is, the sign of the determinant is changed. It is sometimes useful
to know the formulas

al a2 a3 ala2a3

(2.56) kbi kb2 kb3 = k bl b2 b3

Cl C2 ca C1 C2 Ca

al a2 a3 aia2a3 al a2 a3

(2.561) bi + ci b2 + C 2 ba + Ca = bib2 b3 Cl C2 C3

dl d2 d3 dld2 d3 dld2d3

and others like them. They can be proved by expanding the determi-

nants in terms of elements of the middle row. The results are particu-

larly useful when, for some constant k, we have cl = kd1, C2 = kd2, and

c3 = kdi. In this case the last determinant in the above formula is zero

and we obtain the formula

al a2 a3 al a2 as

(2.562) bl + kdl b2 + kd2 b3 + kdi = b1 b2 b3

dl d2 d3 dl d2 d3