9 determinant and cross product

Determinant and Cross Product
A matrix is a rectangular table with numbers
or formulas as entries.

or formulas as entries. The size of a matrix is r x c
where r is the number of rows and c is the number of
columns.

columns. Following are examples of matrices:
5 2 x 2 –1
A2X2 = 5 2 –3
4 –1 B2X3 = C3X3 = 6 –2 y2
4 –1 0
11 9 –4

5 2 x 2 –1
A2X2 = 5 2 –3
4 –1 B2X3 = C3X3 = 6 –2 y2
4 –1 0
11 9 –4
Matrices of sizes n x n, such as A and C above,
are called square matrices.

5 2 x 2 –1
A2X2 = 5 2 –3
4 –1 B2X3 = C3X3 = 6 –2 y2
4 –1 0
11 9 –4
are called square matrices. For each square matrix A,
we extract a number called the determinant of A.

5 2 x 2 –1
A2X2 = 5 2 –3
4 –1 B2X3 = C3X3 = 6 –2 y2
4 –1 0
11 9 –4
are called square matrices. For each square matrix A,
we extract a number called the determinant of A.
Given a 2x2 matrix A the determinant of A is
a b
det(A) = det = ad – bc. Hence det(A) = –13.
c d

The 2x2 determinant
det a b = ad – bc
c d
gives the signed area of the
parallelogram defined by the
row–vectors (a, b) and (c, d),

The 2x2 determinant
det a b = ad – bc
c d (c, d)
(a, b)

(a+c,
The 2x2 determinant b+d)
det a b = ad – bc
c d (c, d)
(a, b)

(a+c,
det a b = ad – bc
c d (c, d)
(a, b)
i.e. |det(A)| = the area of

(a+c,
det a b = ad – bc
c d (c, d)
(a, b)
To see this, we dissect the
rectangle that boxes–in the
parallelogram into regions as
shown

(a+c,
det a b = ad – bc
c d (c, d)
(a, b)
(a+c,
To see this, we dissect the b+d)
parallelogram into regions as (c, d)

shown (a, b)

(a+c,
det a b = ad – bc
c d (c, d)
(a, b)
(a+c,
To see this, we dissect the .
b+d)

shown (a, b)

.

(a+c,
det a b = ad – bc
c d (c, d)
(a, b)
(a+c,
b+d)

shown and note that area of (a, b)
.
= – 2( + + )
.

.

(a+c,
det a b = ad – bc
c d (c, d)
(a, b)
(a+c,
b+d)

shown and note that area of (a, b)
.
= – 2( + + )
.

.
= ad – bc (check this.)

The sign of the 2x2 determinant D of the matrix
D = det a b = ad – bc
c d
gives the following information concerning directions.

c d
i. If D > 0, then the parallelogram D = ad – bc > 0
is formed by sweeping u = < a, b> (c, d)
in a counterclockwise motion, (a, b)
i.e. to the left of u. u

c d
ii. If D < 0, then the parallelogram
is formed by sweeping u = < a, b> D = ad – bc < 0
in a clockwise motion, (a, b)
i.e. to the right of u.
u (c, d)

c d
ii. If D < 0, then the parallelogram
is formed by sweeping u = < a, b> D = ad – bc < 0
in a clockwise motion, (a, b)
i.e. to the right of u.
u (c, d)
iii. If D = 0, there is no parallelogram.
It’s deformed to a line.

Example A.
det 1 0 = +1, so from i = < 1, 0> to j = <0, 1> is a
0 1
counterclockwise sweep.

Example A.
0 1
det 0 1 = –1, so from j = <0, 1> to i = <1, 0> is a
1 0
is a clockwise sweep.

Example A.
0 1
1 0
a b
Hence we say the matrix
c d
preserves orientation if D > 0,

Example A.
0 1
1 0
a b
c d
preserves orientation if D > 0, i.e. facing in the direction
of <a, b>, the vector <c, d> is to the left, just as in our
R2 if we face the x–axis, the y–axis is to the left,

Example A.
0 1
1 0
a b
c d
reverses orientation if D < 0,

Example A.
0 1
1 0
a b
c d
reverses orientation if D < 0, i.e. facing in the
direction of <a, b>, <c, d> is to the right which is the
mirror image to our choice of R2.

Let us generalize 2x2 determinants and the associated
geometry to 3x3 determinants.

a b c
det d e f
g h i

a b c a b c a b
det d e f  d e f d e
g h i
g h i g h

a b c a b c a b
g h i
= aei g h i g h

+ +
a b c a b c a b
g h i
= aei + bfg g h i g h

+ + +
a b c a b c a b
g h i
= aei + bfg + cdh g h i g h

+ + + –
a b c a b c a b
g h i
= aei + bfg + cdh – (ceg i g h
g h

+ + + – – –
a b c a b c a b
g h i
= aei + bfg + cdh – (ceg + afhg bdi)
g h i + h

+ + + – – –
a b c a b c a b
g h i
g h i + h

1 2 –1 1 2 –1 1 2
Hence det 6 –2 2  6 –2 2 6 –2
3 1 –4 3 1 –4 3 1

+ + + – – –
a b c a b c a b
g h i
g h i + h
8 12 –6
1 2 –1 1 2 –1 1 2
Hence det 6 –2 2  6 –2 2 6 –2
3 1 –4 3 1 –4 3 1

+ + + – – –
a b c a b c a b
g h i
g h i + h
8 12 –6 6 2 –48
1 2 –1 1 2 –1 1 2
Hence det 6 –2 2  6 –2 2 6 –2
3 1 –4 3 1 –4 3 1

+ + + – – –
a b c a b c a b
g h i
g h i + h
8 12 –6 6 2 –48
1 2 –1 1 2 –1 1 2
Hence det 6 –2 2  6 –2 2 6 –2
3 1 –4 3 1 –4 3 1
= 8 + 12 – 6 – (6 + 2 – 48)
= 54

+ + + – – –
a b c a b c a b
g h i
g h i + h
8 12 –6 6 2 –48
1 2 –1 1 2 –1 1 2
Hence det 6 –2 2  6 –2 2 6 –2
3 1 –4 3 1 –4 3 1
= 8 + 12 – 6 – (6 + 2 – 48)
= 54
Following are some of the important geometric and
algebraic properties concerning determinants.

The geometric significance of the
3x3 determinants is the same as
the 2x2 determinants.

the 2x2 determinants. z +

<g, h, i>
a b c <d, e, f>
If det d e f = D, <a, b, c>
g h i y
D<0
then D is the signed volume of x
the parallelepiped, i.e. a squashed box,
formed by the row–vectors of the matrix.


<g, h, i>
a b c <d, e, f>
g h i y
D<0
i. If D > 0, then the row vectors
<a, b, c> → <d, e, f>→ <g, h, i> form a right handed
system as in the case i → j → k.


<g, h, i>
a b c <d, e, f>
g h i y
D<0
i. If D > 0, then the row vectors
<a, b, c> → <d, e, f>→ <g, h, i> form a right handed
system as in the case i → j → k.
ii. If D < 0, then <a, b, c> → <d, e, f>→ <g, h, i>
form a left handed system as in the case i → j → –k.

Here are some important properties of determinants.

x * *
1 0 0
i. det 0 y * = xyz, hence det 0 1 0 = 1
0 0 z
0 0 1

x * *
1 0 0
0 0 z
0 0 1
a b c
ii. Given that det d e f = D, then
g h i
ka kb kc a b c a b c
det d e f = det kd ke kf = det d e f = kD
g h i g h i kg kh ki
where k is a constant,

x * *
1 0 0
0 0 z
0 0 1
a b c
ii. Given that det d e f = D, then
g h i
ka kb kc a b c a b c
det d e f = det kd ke kf = det d e f = kD
g h i g h i kg kh ki
where k is a constant, i.e. stretching an edge of the
box by a factor k, then the volume of the box is
changed by the factor k.

3. The determinant changes sign if we swap two rows
in the matrix.

in the matrix. For example,
a b c d e f
if det d e f = D, then det a b c = –D.
g h i g h i

a b c d e f
g h i g h i
Geometrically, this says that if we switch the order of
any two vector, we change the orientation of the
system defined by the row vectors.

a b c d e f
g h i g h i
Geometrically, this says that if we switch the order of
any two vector, we change the orientation of the
system defined by the row vectors.
1 0 0 1 0 0
So det 0 1 0 = 1 and det 0 0 1 = –1
0 0 1 0 1 0
since we changed from a right handed system to a
left handed system following the row vectors.

In fact the sign of the determinant is the algebraic
definition of the right handed system vs. the left
handed system, i.e. a positive determinant means the
row vectors form a right handed system and a negative
determinant means they form a left handed system.

Cross Product

Cross Product
Given two 3D vectors u = <a, b, c>, v = <d, e, f>,
their cross products are defined below

Cross Product
i j k
u x v = det a b c
d e f

Cross Product
i j k i j k
u x v = det a b c and v x u = det d e f
d e f a b c

Cross Product
i j k i j k
u x v = det a b c and v x u = det d e f
d e f a b c
The cross product is only defined for two 3D vectors,
the product yield another 3D vector.

Example B. Find the cross product of
u = <1, 2, –1>, v = <2, –1, 3 >.

u = <1, 2, –1>, v = <2, –1, 3 >.

i j k
u x v = det 1 2 –1
2 –1 3

u = <1, 2, –1>, v = <2, –1, 3 >.

i j k i j k i j
u x v = det 1 2 –1  1 2 –1 1 2
2 –1 3 2 –1 3 2 –1

u = <1, 2, –1>, v = <2, –1, 3 >.
6i –2j –k 4k i 3j
i j k i j k i j
u x v = det 1 2 –1  1 2 –1 1 2
2 –1 3 2 –1 3 2 –1

u = <1, 2, –1>, v = <2, –1, 3 >.
6i –2j –k 4k i 3j
i j k i j k i j
u x v = det 1 2 –1  1 2 –1 1 2
2 –1 3 2 –1 3 2 –1
= 6i – 2j – k – (4k + i + 3j) or that u x v = 5i – 5j – 5k.

u = <1, 2, –1>, v = <2, –1, 3 >.
6i –2j –k 4k i 3j
i j k i j k i j
u x v = det 1 2 –1  1 2 –1 1 2
2 –1 3 2 –1 3 2 –1
* u x v is a orthogonal to both
u and v in the direction determined
by the right-hand rule.

u = <1, 2, –1>, v = <2, –1, 3 >.
6i –2j –k 4k i 3j
i j k i j k i j
u x v = det 1 2 –1  1 2 –1 1 2
2 –1 3 2 –1 3 2 –1
* u x v is a orthogonal to both v
u

u = <1, 2, –1>, v = <2, –1, 3 >.
6i –2j –k 4k i 3j
i j k i j k i j
u x v = det 1 2 –1  1 2 –1 1 2
2 –1 3 2 –1 3 2 –1
* u x v is a orthogonal to both uxv v
u

u = <1, 2, –1>, v = <2, –1, 3 >.
6i –2j –k 4k i 3j
i j k i j k i j
u x v = det 1 2 –1  1 2 –1 1 2
2 –1 3 2 –1 3 2 –1
u
* The length |u x v| is equal to
the area of the parallelogram
defined by u and v.

u = <1, 2, –1>, v = <2, –1, 3 >.
6i –2j –k 4k i 3j
i j k i j k i j
u x v = det 1 2 –1  1 2 –1 1 2
2 –1 3 2 –1 3 2 –1
u
* The length |u x v| is equal to u x v v
u
defined by u and v.

u = <1, 2, –1>, v = <2, –1, 3 >.
6i –2j –k 4k i 3j
i j k i j k i j
u x v = det 1 2 –1  1 2 –1 1 2
2 –1 3 2 –1 3 2 –1
u
* The length |u x v| is equal to u x v v
u
defined by u and v.
|u x v| = area of the

Cross Products of i, j, and k

Since the unit coordinate vectors i, j, and k are
mutually perpendicular, we have the following
relations.

relations.
i

+
k j

ixj=k
jxk=i
kxi=j
(Forward ijk)

relations.
i i

+ –
k j k j

ixj=k i x k = –j
jxk=i k x j = –i
kxi=j j x i = –k
(Forward ijk) (Backward kji)

Algebra of Cross Product

Let u, v and w be 3D vectors:
1. u x u = u x 0 = 0

1. u x u = u x 0 = 0
2. (Orthogonal Property)
u • (v x u) = 0, v • (v x u) = 0

1. u x u = u x 0 = 0
u • (v x u) = 0, v • (v x u) = 0
3. (Anti-commutative) u x v = – (v x u)

1. u x u = u x 0 = 0
u • (v x u) = 0, v • (v x u) = 0
4. If k is a scalar, (ku) x v = u x (kv) = k(u x v)

1. u x u = u x 0 = 0
u • (v x u) = 0, v • (v x u) = 0
5. (Distributive Law)
(u + v) x w = u x w + v x w
u x (v + w) = u x v + u x w

1. u x u = u x 0 = 0
u • (v x u) = 0, v • (v x u) = 0
(u + v) x w = u x w + v x w
u x (v + w) = u x v + u x w
u
6. (Triple Scalar Product) u•(v x w) = det v
w

1. u x u = u x 0 = 0
u • (v x u) = 0, v • (v x u) = 0
(u + v) x w = u x w + v x w
u x (v + w) = u x v + u x w
u
6. (Triple Scalar Product) u•(v x w) = det v
w
In particular |u•(v x w)| = volume of the parallelepiped
defined by u, v, and w.

Example C. Given that u = <2, 0, 0>, v = <0, 3, 0>,
and w = <0, 0, 4>. Find
a. u x (2u – 3w)

and w = <0, 0, 4>. Find
a. u x (2u – 3w) = u x (2u) + u x (–3w)

and w = <0, 0, 4>. Find
a. u x (2u – 3w) = u x (2u) + u x (–3w)
= 2(u x u) – 3(u x w)

and w = <0, 0, 4>. Find
a. u x (2u – 3w) = u x (2u) + u x (–3w)
= 2(u x u) – 3(u x w)
= 2*0 – 3<0, –8, 0> = <0, 24, 0>

and w = <0, 0, 4>. Find
a. u x (2u – 3w) = u x (2u) + u x (–3w)
= 2(u x u) – 3(u x w)
= 2*0 – 3<0, –8, 0> = <0, 24, 0>
b. u • (2v x 3w) =

and w = <0, 0, 4>. Find
a. u x (2u – 3w) = u x (2u) + u x (–3w)
= 2(u x u) – 3(u x w)
= 2*0 – 3<0, –8, 0> = <0, 24, 0>
b. u • (2v x 3w) = u • [6(v x w)]

and w = <0, 0, 4>. Find
a. u x (2u – 3w) = u x (2u) + u x (–3w)
= 2(u x u) – 3(u x w)
= 2*0 – 3<0, –8, 0> = <0, 24, 0>
b. u • (2v x 3w) = u • [6(v x w)]
= 6[u • (v x w)]

and w = <0, 0, 4>. Find
a. u x (2u – 3w) = u x (2u) + u x (–3w)
= 2(u x u) – 3(u x w)
= 2*0 – 3<0, –8, 0> = <0, 24, 0>
b. u • (2v x 3w) = u • [6(v x w)]
= 6[u • (v x w)]
u
= 6 *det v
w

and w = <0, 0, 4>. Find
a. u x (2u – 3w) = u x (2u) + u x (–3w)
= 2(u x u) – 3(u x w)
= 2*0 – 3<0, –8, 0> = <0, 24, 0>
b. u • (2v x 3w) = u • [6(v x w)]
= 6[u • (v x w)]
u
= 6 *det v = 6 * 24 = 144
w
Volume of the box
formed by u,v and w

I don’t have the textbook with me so please do:
HW. From the textbook do all odd on the topic of 2×2
and 3×3 determinants and cross product.

9 determinant and cross product

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