Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 171
As in the permutation expansion derivation, whenever two of the indices in i 1 , . . . , i n are equal
then the determinant has two equal arguments, and evaluates to 0. So we need only consider the cases
where i 1 , . . . , i n form a permutation of the numbers 1, . . . , n. We thus have
∑
det(t(⃗s 1 ), . . . , t(⃗s n )) =
s φ(1),1 . . . s φ(n),n det(⃗t φ(1) , . . . ,⃗t φ(n) ).
permutations φ
Swap the columns in det(⃗t φ(1) , . . . ,⃗t φ(n) ) to get the matrix T back, which changes the sign by a factor
of sgn φ, and then factor out the determinant of T .
= ∑ s φ(1),1 . . . s φ(n),n det(⃗t 1 , . . . ,⃗t n ) · sgn φ = det(T ) ∑ s φ(1),1 . . . s φ(n),n · sgn φ.
φ
φ
As in the proof that the determinant of a matrix equals the determinant of its transpose, we commute
the s’s so they are listed by ascending row number instead of by ascending column number (and we
substitute sgn(φ −1 ) for sgn(φ)).
= det(T ) ∑ φ
s 1,φ −1 (1) . . . s n,φ −1 (n) · sgn φ −1 = det(T ) det(⃗s 1 , ⃗s 2 , . . . , ⃗s n )
Four.II.1.29
(a) An algebraic check is easy.
0 = xy 2 + x 2 y 3 + x 3 y − x 3 y 2 − xy 3 − x 2 y = x · (y 2 − y 3 ) + y · (x 3 − x 2 ) + x 2 y 3 − x 3 y 2
simplifies to the familiar form
y = x · (x 3 − x 2 )/(y 3 − y 2 ) + (x 2 y 3 − x 3 y 2 )/(y 3 − y 2 )
(the y 3 − y 2 = 0 case is easily handled).
For geometric insight, this picture shows that the box formed by the three vectors. Note that all
three vectors end in the z = 1 plane. Below the two vectors on the right is the line through (x 2 , y 2 )
and (x 3 , y 3 ).
( x
)
y
1
( x2
)
y 21
( x3
)
y 31
The box will have a nonzero volume unless the triangle formed by the ends of the three is degenerate.
That only happens (assuming that (x 2 , y 3 ) ≠ (x 3 , y 3 )) if (x, y) lies on the line through the other two.
(b) This is how the answer was given in the cited source. The altitude through (x 1 , y 1 ) of a triangle
with vertices (x 1 , y 1 ) (x 2 , y 2 ) and (x 3 , y 3 ) is found in the usual way from the normal form of the
above:
∣ ∣∣∣∣∣ x
1
1 x 2 x 3
√ y 1 y 2 y 3
(x2 − x 3 ) 2 + (y 2 − y 3 ) 2 1 1 1 ∣ .
Another step shows the area of the triangle to be
1
x 1 x 2 x 3
2
y 1 y 2 y 3
∣ 1 1 1 ∣ .
This exposition reveals the modus operandi more clearly than the usual proof of showing a collection
of terms to be identitical with the determinant.
(c) This is how the answer was given in the cited source. Let
x 1 x 2 x 3
D =
y 1 y 2 y 3
∣ 1 1 1 ∣
then the area of the triangle is (1/2)|D|. Now if the coordinates are all integers, then D is an integer.
Subsection Four.III.1: Laplace’s Expansion