,__>;
[a b c d e]
[ ]
[f g h i j]
[ ]
M5 := [k l m n o]
[ ]
[p q r s t]
[ ]
[u v w x y]
> Det(M0):=Det(M0); Det(M1):=Det(M1); Det(M2):=Det(M2); Det(M3):=Det(M3); Det(M4):=Det(M4); Det(M5):=Det(M5);
Det([]) := 1
Det([a]) := a
[a b]
Det([ ]) := a d - b c
[c d]
[a b c]
[ ]
Det([d e f]) := a e i - a f h + d h c - d b i + g b f - g e c
[ ]
[g h i]
[a b c d]
[ ]
[e f g h]
Det([ ]) :=
[i j k l]
[ ]
[m n o p]
a f k p - a f l o + a j o h - a j g p + a n g l
- a n k h - e b k p + e b l o - e j o d + e j c p - e n c l + e n k d
+ i b g p - i b o h + i f o d - i f c p + i n h c - i n g d - m b g l
+ m b k h - m f k d + m f c l - m j h c + m j g d
[a b c d e]
[ ]
[f g h i j]
[ ]
Det([k l m n o]) :=
[ ]
[p q r s t]
[ ]
[u v w x y]
-f b m s y + f b m t x - f b r x o + f b r n y
- f b w n t + f b w s o + f l c s y - f l c t x + f l r x e - f l r d y
+ f l w d t - f l w s e - f q c n y + f q c x o - f q m x e + f q m d y
- f q w o d + f q w n e + f v c n t - f v c s o + f v m s e - f v m d t
+ k v h d t - k v r d j + k v r e i + f v r o d - f v r n e + k b h s y
- k b h t x + k b r x j - k b r i y + k b w i t - k b w s j - k g c s y
+ k g c t x - k g r x e + k g r d y - k g w d t + k g w s e + k q c i y
- k q c x j + k q h x e - k q h d y + k q w d j - k q w e i - k v c i t
+ k v c s j - k v h s e + a g m s y - a g m t x + a g r x o - a g r n y
+ a g w n t - a g w s o - a l h s y + a l h t x - a l r x j + a l r i y
- a l w i t + a l w s j + a q h n y - a q h x o + a q m x j - a q m i y
+ a q w i o - a q w n j - a v h n t + a v h s o - a v m s j + a v m i t
- a v r i o + a v r n j - p b h n y + p b h x o - p b m x j + p b m i y
- p b w i o + p b w n j + p g c n y - p g c x o + p g m x e - p g m d y
+ p g w o d - p g w n e - p l c i y + p l c x j - p l h x e + p l h d y
- p l w d j + p l w e i + p v c i o - p v c n j + p v h n e - p v h o d
+ p v m d j - p v m e i + u b h n t - u b h s o + u b m s j - u b m i t
+ u b r i o - u b r n j - u g c n t + u g c s o - u g m s e + u g m d t
- u g r o d + u g r n e + u l c i t - u l c s j + u l h s e - u l h d t
+ u l r d j - u l r e i - u q c i o + u q c n j - u q h n e + u q h o d
- u q m d j + u q m e i
==================================================
== A trick for computing Det() of a 3x3 matrix ==
==================================================
(This trick ONLY works for 3x3; nothing smaller, nothing larger.)
Recall that
[a b c]
[ ]
Det([d e f]) := a e i - a f h + d h c - d b i + g b f - g e c .
[ ]
[g h i]
We repeat the first two columns, to get a 3x5 matrix:
[a b c a b]
[ ]
[d e f d e]
[ ]
[g h i g h]
Compute the SUM of the three diagonal products shown:
a b c a b
\ \ \
d e f d e
\ \ \
g h i g h
From that sum, SUBTRACT the sum of these three opposite-slope diagonals:
a b c a b
/ / /
d e f d e
/ / /
g h i g h
The result is Det of the given 3x3.
============================
== Computing a Wronkskian ==
============================
Example of linear-Dependence:
Consider the follow functions/expressions:
C := cosh(x), S := sinh(x),
E := exp(x), G := exp(-x).
[The "G" is for neGative.]
Notice that function-triple
(C, E, G)
is linearly-DEPENDENT, since C is an element of
Span(E,G),
seeing as
C = [E + G]/2.
Thus, we expect the Wronskian of list (C, E, G) to be the zero-function.
Computing, the Wronskian-matrix of the list is
[C E G]
M := [S E -G] .
[C E G]
Applying the trick yields 3x5 matrix
C E G C E
S E -G S E
C E G C E
Using its diagonals, then,
Det(M) = [ C*E*G + E*[-G]*C + G*S*E ]
- [ G*E*C + C*[-G]*E + E*S*G ].
The terms cancel pairwise, yielding Det=0.
Also note that since
E*G = 1,
we could first simplify to
Det(M) = [ C - C + S ]
- [ C - C + S ]
= 0.
================
Example of linear-INdependence:
As functions of x, define
Z := 1, (zeroth power)
F := x, (first power)
S := x^2, (second power)
T := x^3. (third power)
The Wronskian-matrix of (Z,F,S) is
[1 x S]
[0 1 2x] .
[0 0 2]
The Det() of this matrix, since the matrix is upper-triangular, is 1*1*2,
which is /not/ the zero-function. Thus we see that
function-triple (1, x, x^2)
is linearly-independent, as we expect.
Second example of linear-Independence:
Now consider function-triple (x, x^2, x^3). We expect that this list
is linearly-independent. Its Wronskian-matrix is
[x S T]
[1 2x 3S]
[0 2 6x]
Use the trick to compute its Wronskian-function. Is its Wronskian the zero-fnc?
==================================================
== Characteristic polynomial of a square matrix ==
==================================================
CharMatB := BDetOne - (x * eye(3));
[-x 1 1 ]
[ ]
CharMatB := [1 5 - x -1 ]
[ ]
[0 5 4 - x]
> CharPolyB := Det(CharMatB) ;
2 3
CharPolyB := -24 x + 9 x - x + 1
> M := <| > ;
[a b]
M := [ ]
[c d]
> CharMatM := M - (x * eye(2));
[a - x b ]
CharMatM := [ ]
[ c d - x]
> CharPolyM := Det(CharMatM) ;
2
CharPolyM := a d - a x - x d + x - b c
> quit;
END: "determinant-as-poly.txt"
__