The document defines the determinant of a square matrix. For a 1x1 matrix with value k, the determinant is defined to be k. For a 2x2 matrix with values a, b, c, d, the determinant is defined as ad - bc. This definition is motivated geometrically as representing the signed area of the parallelogram formed by the vector points (a,b) and (c,d). It is also motivated algebraically in that a system of equations has a unique solution if and only if the determinant of the coefficient matrix is non-zero. Cramer's rule is presented for solving systems of linear equations.

1. Determinant

2. The determinant of an nxn square matrix A,
or det(A), is a number.
Determinant

3. The determinant of an nxn square matrix A,
or det(A), is a number.
We define the determinant of a 1x1 matrix k ,
or det k to be k. Hence det –3 = –3.
Determinant

4. The determinant of an nxn square matrix A,
or det(A), is a number.
We define the determinant of a 1x1 matrix k ,
or det k to be k. Hence det –3 = –3.
We define the determinant of the 2x2 matrix
or det
a b
c d
a b
c d
Determinant

5. The determinant of an nxn square matrix A,
or det(A), is a number.
We define the determinant of a 1x1 matrix k ,
or det k to be k. Hence det –3 = –3.
We define the determinant of the 2x2 matrix
or det = ad
a b
c d
a b
c d
Determinant

6. The determinant of an nxn square matrix A,
or det(A), is a number.
We define the determinant of a 1x1 matrix k ,
or det k to be k. Hence det –3 = –3.
We define the determinant of the 2x2 matrix
or det = ad – bc
a b
c d
a b
c d
Determinant

7. The determinant of an nxn square matrix A,
or det(A), is a number.
We define the determinant of a 1x1 matrix k ,
or det k to be k. Hence det –3 = –3.
We define the determinant of the 2x2 matrix
or det = ad – bc
a b
c d
a b
c d
Here are two motivations for this definition.
Determinant

8. The determinant of an nxn square matrix A,
or det(A), is a number.
We define the determinant of a 1x1 matrix k ,
or det k to be k. Hence det –3 = –3.
We define the determinant of the 2x2 matrix
or det = ad – bc
a b
c d
a b
c d
Here are two motivations for this definition.
I. (Geometric) Let (a, b) and (c, d)
be two points in the rectangular
coordinate system.
(a, b)(c, d)
Determinant

9. The determinant of an nxn square matrix A,
or det(A), is a number.
We define the determinant of a 1x1 matrix k ,
or det k to be k. Hence det –3 = –3.
We define the determinant of the 2x2 matrix
or det = ad – bc
a b
c d
a b
c d
Here are two motivations for this definition.
I. (Geometric) Let (a, b) and (c, d)
be two points in the rectangular
coordinate system.
Then det
a b
c d
= ad – bc
= “Signed” area of the parallelogram as shown.
(a, b)(c, d)
“Signed” area
= ad – bc
Determinant

10. Determinant
Let’s verify this in the simple cases where one of the
points is on the axis.

11. Determinant
Let’s verify this in the simple cases where one of the
points is on the axis.
For example, Iet the points be
(a, 0) and (c, d) as shown.
(a, 0)
(c, d)

12. Determinant
Let’s verify this in the simple cases where one of the
points is on the axis.
For example, Iet the points be
(a, 0) and (c, d) as shown.
then the area of the
parallelogram is
det
a 0
c d
(a, 0)
(c, d)
d
= ad

13. Determinant
Let’s verify this in the simple cases where one of the
points is on the axis.
For example, Iet the points be
(a, 0) and (c, d) as shown.
then the area of the
parallelogram is
det
a 0
c d
(a, 0)
(c, d)
d
If we switch the order of the two rows then
det
c d
= – ad
= ad
a 0

14. Determinant
Let’s verify this in the simple cases where one of the
points is on the axis.
For example, Iet the points be
(a, 0) and (c, d) as shown.
then the area of the
parallelogram is
det
a 0
c d
(a, 0)
(c, d)
d
If we switch the order of the two rows then
det
a 0
c d
= – ad
= ad
What’s the meaning of the “–” sign?
(besides that the rows are switched)

15. Determinant
Let’s verify this in the simple cases where one of the
points is on the axis.
For example, Iet the points be
(a, 0) and (c, d) as shown.
then the area of the
parallelogram is
det
a 0
c d
(a, 0)
(c, d)
d
If we switch the order of the two rows then
det
a 0
c d
= – ad
= ad
What’s the meaning of the “–” sign?
(besides that the rows are switched). Actually it tells
us the “relative position” of these two points.

16. Determinant
Given a point A(a, b) ≠ (0, 0), the dial with the tip at A
defines a direction.
A(a, b)A(a, b)

17. Determinant
Given a point A(a, b) ≠ (0, 0), the dial with the tip at A
defines a direction. Let B(c, d) be another point,
there are two possibilities, A(a, b)A(a, b)

18. Determinant
Given a point A(a, b) ≠ (0, 0), the dial with the tip at A
defines a direction. Let B(c, d) be another point,
there are two possibilities,
either B is to the left or
it’s to the right of dial A
as shown.
A(a, b)
B(c, d)
A(a, b)
B(c, d)

19. Determinant
Given a point A(a, b) ≠ (0, 0), the dial with the tip at A
defines a direction. Let B(c, d) be another point,
there are two possibilities,
either B is to the left or
it’s to the right of dial A
as shown.
The determinant
identifies which case it is.
A(a, b)
B(c, d)
A(a, b)
B(c, d)

20. Determinant
If det
a b
c d
is +, then B is to the left of A.
Given a point A(a, b) ≠ (0, 0), the dial with the tip at A
defines a direction. Let B(c, d) be another point,
there are two possibilities,
either B is to the left or
it’s to the right of dial A
as shown.
i.
The determinant
identifies which case it is. det
A
B
A(a, b)
B(c, d)
A(a, b)
B(c, d)
> 0
B is to the left of A.

21. Determinant
If det
a b
c d
is +, then B is to the left of A.
Given a point A(a, b) ≠ (0, 0), the dial with the tip at A
defines a direction. Let B(c, d) be another point,
there are two possibilities,
either B is to the left or
it’s to the right of dial A
as shown.
i.
ii.
The determinant
identifies which case it is. det
A
B
A(a, b)
B(c, d)
A(a, b)
B(c, d)
> 0 det
A
B < 0
B is to the left of A. B is to the right of A.
If det
a b
c d is –, then B is to the right of A.

22. Determinant
If det
a b
c d
is +, then B is to the left of A.
Given a point A(a, b) ≠ (0, 0), the dial with the tip at A
defines a direction. Let B(c, d) be another point,
there are two possibilities,
either B is to the left or
it’s to the right of dial A
as shown.
i.
ii.
The determinant
identifies which case it is. det
A
B
A(a, b)
B(c, d)
A(a, b)
B(c, d)
> 0 det
A
B < 0
B is to the left of A. B is to the right of A.
If det
a b
c d is –, then B is to the right of A.
For example det
1 0
0 1
> 0 and det
1 0
0 1
< 0.

23. It is in the above sense that we have the
signed-area-answer of det A.
Determinant

24. So given the picture to the right: (a, b)(c, d)
It is in the above sense that we have the
signed-area-answer of det A.
Determinant

25. So given the picture to the right:
det
a b
c d
= + Area of
(a, b)(c, d)
It is in the above sense that we have the
signed-area-answer of det A.
Determinant

26. So given the picture to the right:
det
a b
c d
= + Area of
(a, b)(c, d)
det
a b
c d
= – Area of
and
It is in the above sense that we have the
signed-area-answer of det A.
Determinant

27. So given the picture to the right:
det
a b
c d
= + Area of
(a, b)(c, d)
det
a b
c d
= – Area of
and
It is in the above sense that we have the
signed-area-answer of det A.
Example A. Find the area.
Determinant
(5,–3)
(1,4)

28. So given the picture to the right:
det
a b
c d
= + Area of
(a, b)(c, d)
det
a b
c d
= – Area of
and
It is in the above sense that we have the
signed-area-answer of det A.
Example A. Find the area.
Determinant
(5,–3)
(1,4)
det
5 –3
1 4
To find the area, we calculate

29. So given the picture to the right:
det
a b
c d
= + Area of
(a, b)(c, d)
det
a b
c d
= – Area of
and
It is in the above sense that we have the
signed-area-answer of det A.
Example A. Find the area.
Determinant
(5,–3)
(1,4)
det
5 –3
1 4
To find the area, we calculate
= 5(4) – 1(–3) = 23 = area

30. So given the picture to the right:
det
a b
c d
= + Area of
(a, b)(c, d)
det
a b
c d
= – Area of
and
It is in the above sense that we have the
signed-area-answer of det A.
Example A. Find the area.
Determinant
(5,–3)
(1,4)
det
5 –3
1 4
To find the area, we calculate
= 5(4) – 1(–3) = 23 = area
Note that if a, b, c and d are integers,
then the -area is an integer also.

31. Here is another motivation for the 2x2 determinants.
Determinant

32. Here is another motivation for the 2x2 determinants.
Determinant
Il. (Algebraic) The system of equations
det
a b
c d
≠ 0
ax + by = #
cx + dy = #
has exactly one solution if and only if

33. Here is another motivation for the 2x2 determinants.
Here are examples where we don’t have exactly
one solution.
Determinant
Il. (Algebraic) The system of equations
det
a b
c d
≠ 0
ax + by = #
cx + dy = #
has exactly one solution if and only if
{x + y = 2
2x + 2y = 4

34. Here is another motivation for the 2x2 determinants.
Here are examples where we don’t have exactly
one solution.
Duplicated information.
Infinitely many solutions.
Determinant
Il. (Algebraic) The system of equations
det
a b
c d
≠ 0
ax + by = #
cx + dy = #
has exactly one solution if and only if
{x + y = 2
2x + 2y = 4

35. Here is another motivation for the 2x2 determinants.
Here are examples where we don’t have exactly
one solution.
Duplicated information.
Infinitely many solutions.
Determinant
Il. (Algebraic) The system of equations
det
a b
c d
≠ 0
ax + by = #
cx + dy = #
has exactly one solution if and only if
{x + y = 2
2x + 2y = 4
{x + y = 2
2x + 2y = 5

36. Here is another motivation for the 2x2 determinants.
Here are examples where we don’t have exactly
one solution.
Duplicated information.
Infinitely many solutions.
Determinant
Il. (Algebraic) The system of equations
det
a b
c d
≠ 0
ax + by = #
cx + dy = #
has exactly one solution if and only if
{x + y = 2
2x + 2y = 4
{x + y = 2
2x + 2y = 5
Contradictory information.
No solutions.

37. Here is another motivation for the 2x2 determinants.
Here are examples where we don’t have exactly
one solution.
Duplicated information.
Infinitely many solutions.
Determinant
Il. (Algebraic) The system of equations
det
a b
c d
≠ 0
ax + by = #
cx + dy = #
has exactly one solution if and only if
{x + y = 2
2x + 2y = 4
{x + y = 2
2x + 2y = 5
Contradictory information.
No solutions.
Note that the coefficient matrix has det
1 1
2 2 = 0.

38. Here is the Cramer’s rule that actually gives
the only solution of the system
Determinant
if det
a b
c d
= D ≠ 0
ax + by = u
cx + dy = v

39. Here is the Cramer’s rule that actually gives
the only solution of the system
Determinant
if det
a b
c d
= D ≠ 0
ax + by = u
cx + dy = v
{x + y = 2
x + 4y = 5
Example B. Use the Cramer’s rule to solve for x & y.

40. Here is the Cramer’s rule that actually gives
the only solution of the system
Determinant
if det
a b
c d
= D ≠ 0
ax + by = u
cx + dy = v
{x + y = 2
x + 4y = 5
Example B. Use the Cramer’s rule to solve for x & y.
1 1
1 4
= 3 = D ≠ 0det

41. Here is the Cramer’s rule that actually gives
the only solution of the system
Determinant
if det
a b
c d
= D ≠ 0
ax + by = u
cx + dy = v
namely that x =
det
u b
v d
D
{x + y = 2
x + 4y = 5
Example B. Use the Cramer’s rule to solve for x & y.
1 1
1 4
= 3 = D ≠ 0det

42. Here is the Cramer’s rule that actually gives
the only solution of the system
Determinant
if det
a b
c d
= D ≠ 0
ax + by = u
cx + dy = v
namely that x =
det
u b
v d
D
{x + y = 2
x + 4y = 5
Example B. Use the Cramer’s rule to solve for x & y.
1 1
1 4
= 3 = D ≠ 0det
so that x =
det
2 1
5 4
3

43. Here is the Cramer’s rule that actually gives
the only solution of the system
Determinant
if det
a b
c d
= D ≠ 0
ax + by = u
cx + dy = v
namely that x =
det
u b
v d
D
{x + y = 2
x + 4y = 5
Example B. Use the Cramer’s rule to solve for x & y.
1 1
1 4
= 3 = D ≠ 0det
so that x =
det
2 1
5 4
3
= 1

44. Here is the Cramer’s rule that actually gives
the only solution of the system
Determinant
if det
a b
c d
= D ≠ 0
ax + by = u
cx + dy = v
namely that x =
det
u b
v d
D
y =
det
a u
c v
D
{x + y = 2
x + 4y = 5
Example B. Use the Cramer’s rule to solve for x & y.
1 1
1 4
= 3 = D ≠ 0det
so that x =
det
2 1
5 4
3
= 1

45. Here is the Cramer’s rule that actually gives
the only solution of the system
Determinant
if det
a b
c d
= D ≠ 0
ax + by = u
cx + dy = v
namely that x =
det
u b
v d
D
y =
det
a u
c v
D
{x + y = 2
x + 4y = 5
Example B. Use the Cramer’s rule to solve for x & y.
1 1
1 4
= 3 = D ≠ 0det
so that x =
det
2 1
5 4
3
y =
det
1 2
1 5
3
= 1 = 1

46. The determinant of a 3×3 matrix, may be defined
in the following expansion of 2×2 determinants:
Determinant
det
a b c
d e f
g h i

47. The determinant of a 3×3 matrix, may be defined
in the following expansion of 2×2 determinants:
Example C. Calculate.
Determinant
det
a b c
d e f
g h i
det
1 0 2
2 1 0
2 3 –1

48. det
e f
h i
= a*
The determinant of a 3×3 matrix, may be defined
in the following expansion of 2×2 determinants:
Determinant
det
a b c
d e f
g h i
det
1 0 2
2 1 0
2 3 –1
Example C. Calculate.

49. det
e f
h i
= a*
The determinant of a 3×3 matrix, may be defined
in the following expansion of 2×2 determinants:
Determinant
det
a b c
d e f
g h i
det
1 0
3 –1= 1*det
1 0 2
2 1 0
2 3 –1
Example C. Calculate.

50. det
e f
h i
= a*
The determinant of a 3×3 matrix, may be defined
in the following expansion of 2×2 determinants:
Determinant
det
a b c
d e f
g h i
det– b*
d f
g i
det
1 0
3 –1= 1*det
1 0 2
2 1 0
2 3 –1
Example C. Calculate.

51. det
e f
h i
= a*
The determinant of a 3×3 matrix, may be defined
in the following expansion of 2×2 determinants:
Determinant
det
a b c
d e f
g h i
det– b*
d f
g i
det
1 0
3 –1= 1*det
1 0 2
2 1 0
2 3 –1
Example C. Calculate.
det– 0*
2 0
2 –1

52. det
e f
h i
= a*
The determinant of a 3×3 matrix, may be defined
in the following expansion of 2×2 determinants:
Determinant
det
a b c
d e f
g h i
det– b* det+ c*
d f
g i
d e
det
1 0
3 –1= 1*det
1 0 2
2 1 0
2 3 –1
Example C. Calculate.
det– 0*
2 0
2 –1

53. det
e f
h i
= a*
The determinant of a 3×3 matrix, may be defined
in the following expansion of 2×2 determinants:
Example C. Calculate.
Determinant
det
a b c
d e f
g h i
det– b* det+ c*
d f
g i
d e
det
1 0
3 –1= 1*det
1 0 2
2 1 0
2 3 –1
det– 0* det+ 2*
2 0
2 –1
2 1

54. det
e f
h i
= a*
The determinant of a 3×3 matrix, may be defined
in the following expansion of 2×2 determinants:
Example C. Calculate.
Determinant
det
a b c
d e f
g h i
det– b* det+ c*
d f
g i
d e
det
1 0
3 –1= 1*det
1 0 2
2 1 0
2 3 –1
det– 0* det+ 2*
2 0
2 –1
2 1
= 1(–1 – 0) – 0 + 2(6 – 2)

55. det
e f
h i
= a*
The determinant of a 3×3 matrix, may be defined
in the following expansion of 2×2 determinants:
Example C. Calculate.
Determinant
det
a b c
d e f
g h i
det– b* det+ c*
d f
g i
d e
det
1 0
3 –1= 1*det
1 0 2
2 1 0
2 3 –1
det– 0* det+ 2*
2 0
2 –1
2 1
= 1(–1 – 0) – 0 + 2(6 – 2)
= –1 + 8
= 7

56. Don’t remember this!
det
e f
h i
= a*
The determinant of a 3×3 matrix, may be defined
in the following expansion of 2×2 determinants:
Example C. Calculate.
Determinant
det
a b c
d e f
g h i
det– b* det+ c*
= a(ei – hf) – b(ei – hf) + c(dh – ge)
d f
g i
d e
det
1 0
3 –1= 1*det
1 0 2
2 1 0
2 3 –1
det– 0* det+ 2*
2 0
2 –1
2 1
= 1(–1 – 0) – 0 + 2(6 – 2)
= –1 + 8
= 7

57. Here is the useful Butterfly method
for the determinant of a 3×3 matrix.
Determinant
a b c
d e f
g h i
det
1 0 2
2 1 0
2 3 –1
det
Example D. Calculate using the Butterfly method.

58. Here is the useful Butterfly method
for the determinant of a 3×3 matrix.
Determinant
a b c
d e f
g h i
det
1 0 2
2 1 0
2 3 –1
det
Example D. Calculate using the Butterfly method.
a b c
d e f
g h i
a b
d e
g h
Enlarge the matrix
by adding the
first two columns.

59. Here is the useful Butterfly method
for the determinant of a 3×3 matrix.
Determinant
a b c
d e f
g h i
det
1 0 2
2 1 0
2 3 –1
det
Example D. Calculate using the Butterfly method.
a b c
d e f
g h i
a b
d e
g h
1 0 2
2 1 0
2 3 –1
1 0
2 1
2 3
Enlarge the matrix
by adding the
first two columns.

60. Here is the useful Butterfly method
for the determinant of a 3×3 matrix.
Determinant
a b c
d e f
g h i
det
1 0 2
2 1 0
2 3 –1
det
Example D. Calculate using the Butterfly method.
a b c
d e f
g h i
a b
d e
g h
1 0 2
2 1 0
2 3 –1
1 0
2 1
2 3
Enlarge the matrix
by adding the
first two columns.
Add all these diagonal products

61. Here is the useful Butterfly method
for the determinant of a 3×3 matrix.
Determinant
a b c
d e f
g h i
det
1 0 2
2 1 0
2 3 –1
det
Example D. Calculate using the Butterfly method.
a b c
d e f
g h i
a b
d e
g h
1 0 2
2 1 0
2 3 –1
1 0
2 1
2 3
Enlarge the matrix
by adding the
first two columns.
Add all these diagonal products
Add –1 0 12

62. Here is the useful Butterfly method
for the determinant of a 3×3 matrix.
Determinant
a b c
d e f
g h i
det
1 0 2
2 1 0
2 3 –1
det
Example D. Calculate using the Butterfly method.
a b c
d e f
g h i
a b
d e
g h
1 0 2
2 1 0
2 3 –1
1 0
2 1
2 3
Enlarge the matrix
by adding the
first two columns.
Subtract all these diagonal products
Add all these diagonal products
Add –1 0 12

63. Here is the useful Butterfly method
for the determinant of a 3×3 matrix.
Determinant
a b c
d e f
g h i
det
1 0 2
2 1 0
2 3 –1
det
Example D. Calculate using the Butterfly method.
a b c
d e f
g h i
a b
d e
g h
1 0 2
2 1 0
2 3 –1
1 0
2 1
2 3
Enlarge the matrix
by adding the
first two columns.
Subtract all these diagonal products
Add all these diagonal products
Subtract 4 0 0
Add –1 0 12

64. Here is the useful Butterfly method
for the determinant of a 3×3 matrix.
Determinant
a b c
d e f
g h i
det
1 0 2
2 1 0
2 3 –1
det
Example D. Calculate using the Butterfly method.
a b c
d e f
g h i
a b
d e
g h
1 0 2
2 1 0
2 3 –1
1 0
2 1
2 3
Enlarge the matrix
by adding the
first two columns.
Subtract all these diagonal products
Add all these diagonal products
Subtract 4 0 0
Add –1 0 12
(–1 +12) – 4 = 7

65. The det A of a 3×3 matrix A similarly gives
the volume of a solid in 3D space.
Determinant

66. Determinant
(a, b, c)
(d, e, f)
(0, 0, 0)
(g, h, i)
a b c
d e f
g h i
det = ± the volume of

67. The det A of a 3×3 matrix A similarly gives
the volume of a solid in 3D space.
Specifically,
Determinant
(a, b, c)
(d, e, f)
(0, 0, 0)
(g, h, i)
a b c
d e f
g h i
det = ± the volume of

68. The det A of a 3×3 matrix A similarly gives
the volume of a solid in 3D space.
Specifically,
Again there the ± are the two “orientations” of the
objects defined by these coordinates.
Determinant
(a, b, c)
(d, e, f)
(0, 0, 0)
(g, h, i)
a b c
d e f
g h i
det = ± the volume of

69. The det A of a 3×3 matrix A similarly gives
the volume of a solid in 3D space.
Specifically,
Again there the ± are the two “orientations” of the
objects defined by these coordinates.
If we switch two of the coordinates in our labeling,
say x and y, we would get a left handed copy
of the solid.
Determinant
(a, b, c)
(d, e, f)
(0, 0, 0)
(g, h, i)
a b c
d e f
g h i
det = ± the volume of

70. The det A of a 3×3 matrix A similarly gives
the volume of a solid in 3D space.
Specifically,
Again there the ± are the two “orientations” of the
objects defined by these coordinates.
If we switch two of the coordinates in our labeling,
say x and y, we would get a left handed copy
of the solid. But if we switch the labeling again,
we would get back to the original right handed solid.
Determinant
(a, b, c)
(d, e, f)
(0, 0, 0)
(g, h, i)
a b c
d e f
g h i
det = ± the volume of

71. Likewise the 3x3 system
Determinant
≠ 0det
a b c
d e f
g h i
has a unique answer iff the
ax + by + cz = #
dx + ey + fz = #
gx + hy + iz = #

72. Likewise the 3x3 system
Determinant
There is a 3x3 (in general nxn) Cramer’s Rule that
gives the solutions for x, y, and z but it’s not useful
for computational purposes-too many steps!
≠ 0det
a b c
d e f
g h i
has a unique answer iff the
ax + by + cz = #
dx + ey + fz = #
gx + hy + iz = #

73. Likewise the 3x3 system
Determinant
ax + by + cz = #
There is a 3x3 (in general nxn) Cramer’s Rule that
gives the solutions for x, y, and z but it’s not useful
for computational purposes-too many steps!
dx + ey + fz = #
gx + hy + iz = #
≠ 0det
a b c
d e f
g h i
has a unique answer iff the
We will show a method of finding the determinants of
n x n matrices.

74. Likewise the 3x3 system
Determinant
ax + by + cz = #
There is a 3x3 (in general nxn) Cramer’s Rule that
gives the solutions for x, y, and z but it’s not useful
for computational purposes-too many steps!
dx + ey + fz = #
gx + hy + iz = #
≠ 0det
a b c
d e f
g h i
has a unique answer iff the
We will show a method of finding the determinants of
n x n matrices. The Cramer’s Rule establishes that:
an nxn linear system has a unique answer iff
the det(the coefficient matrix) ≠ 0.

75. Determinant
We need two points to define the
two edges of a parallelogram

76. Determinant
We need two points to define the
two edges of a parallelogram
and three points for the three edges
of a tilted box.

77. Determinant
We need two points to define the
two edges of a parallelogram
and three points for the three edges
of a tilted box. The coordinates of
these points form 2x2 and 3x3
square matrices and the determinants
of these matrices give the signed
areas and volumes respectively.

78. Determinant
We need two points to define the
two edges of a parallelogram
and three points for the three edges
of a tilted box. The coordinates of
these points form 2x2 and 3x3
square matrices and the determinants
of these matrices give the signed
areas and volumes respectively.
Likewise, we need n points to
determine an “n-dimensional box”
whose coordinates form an nxn matrix.

79. Determinant
We need two points to define the
two edges of a parallelogram
and three points for the three edges
of a tilted box. The coordinates of
these points form 2x2 and 3x3
square matrices and the determinants
of these matrices give the signed
areas and volumes respectively.
Likewise, we need n points to
determine an “n-dimensional box”
whose coordinates form an nxn matrix.
Below we give one method to find the determinants of
nxn matrices which is the signed “volumes” of these
“n-dimensional” boxes.

80. Determinant
Finding Determinants-Expansion by the First Row

81. Determinant
Finding Determinants-Expansion by the First Row
Let A be an nxn matrix as shown.
a11 a12 . . a1n
a21 a22 . . a2n
. . . aij .
. . . . .
an1 . . . ann
A =

82. Determinant
Finding Determinants-Expansion by the First Row
Let A be an nxn matrix as shown.
a11 a12 . . a1n
a21 a22 . . a2n
. . . aij .
. . . . .
an1 . . . ann
The submatrix Aij of A
is the matrix left after deleting
the i'th row and j’th column of A.A =

83. Determinant
Finding Determinants-Expansion by the First Row
Let A be an nxn matrix as shown.
a11 a12 . . a1n
a21 a22 . . a2n
. . . aij .
. . . . .
an1 . . . ann
The submatrix Aij of A
is the matrix left after deleting
the i'th row and j’th column of A.
a11 a12 . . a1n
a21 a22 . . a2n
. . . . .
. . . . .
an1 . . . ann
A =
Aij = aij

84. Determinant
Finding Determinants-Expansion by the First Row
Let A be an nxn matrix as shown.
a11 a12 . . a1n
a21 a22 . . a2n
. . . aij .
. . . . .
an1 . . . ann
The submatrix Aij of A
is the matrix left after deleting
the i'th row and j’th column of A.
For example, if
a11 a12 . . a1n
a21 a22 . . a2n
. . . . .
. . . . .
an1 . . . ann
A =
Aij =
A =
1 0 2
2 1 0
2 3 –1
, then
A23 =aij

85. Determinant
Finding Determinants-Expansion by the First Row
Let A be an nxn matrix as shown.
a11 a12 . . a1n
a21 a22 . . a2n
. . . aij .
. . . . .
an1 . . . ann
The submatrix Aij of A
is the matrix left after deleting
the i'th row and j’th column of A.
For example, if
a11 a12 . . a1n
a21 a22 . . a2n
. . . . .
. . . . .
an1 . . . ann
A =
Aij =
A =
1 0 2
2 1 0
2 3 –1
, then
A23 =
1 0 2
2 1 0
2 3 –1
delete the 2nd row
and the 3rd column
aij

86. Determinant
Finding Determinants-Expansion by the First Row
Let A be an nxn matrix as shown.
a11 a12 . . a1n
a21 a22 . . a2n
. . . aij .
. . . . .
an1 . . . ann
The submatrix Aij of A
is the matrix left after deleting
the i'th row and j’th column of A.
For example, if
a11 a12 . . a1n
a21 a22 . . a2n
. . . . .
. . . . .
an1 . . . ann
A =
Aij =
A =
1 0 2
2 1 0
2 3 –1
, then
A23 =
1 0 2
2 1 0
2 3 –1
delete the 2nd row
and the 3rd column
aij
=
1 0
2 3

87. Determinant
Finding Determinants-Expansion by the First Row

88. Determinant
Finding Determinants-Expansion by the First Row
Let A be an nxn matrix as shown.
a11 a12 . . a1n
a21 a22 . . a2n
. . . . .
. . . . .
an1 . . . ann
A =

89. Determinant
Finding Determinants-Expansion by the First Row
Let A be an nxn matrix as shown.
a11 a12 . . a1n
a21 a22 . . a2n
. . . . .
. . . . .
an1 . . . ann
The determinant of A or det(A)
may be calculated using the
determinants of its
submatrices Aij’s.
A =

90. Determinant
Finding Determinants-Expansion by the First Row
Let A be an nxn matrix as shown.
a11 a12 . . a1n
a21 a22 . . a2n
. . . . .
. . . . .
an1 . . . ann
The determinant of A or det(A)
may be calculated using the
determinants of its
submatrices Aij’s.
Here is the formula for det(A)
using the determinants of the
submatrices from the 1st row.
A =

91. Determinant
Finding Determinants-Expansion by the First Row
Let A be an nxn matrix as shown.
a11 a12 . . a1n
a21 a22 . . a2n
. . . . .
. . . . .
an1 . . . ann
The determinant of A or det(A)
may be calculated using the
determinants of its
submatrices Aij’s.
Here is the formula for det(A)
using the determinants of the
submatrices from the 1st row.
A =
det(A) ≡ a11det(A11)–a12det(A12) +a13det(A13) – ..(±)a1ndet(A1n)

92. Determinant
Finding Determinants-Expansion by the First Row
Let A be an nxn matrix as shown.
a11 a12 . . a1n
a21 a22 . . a2n
. . . . .
. . . . .
an1 . . . ann
The determinant of A or det(A)
may be calculated using the
determinants of its
submatrices Aij’s.
Here is the formula for det(A)
using the determinants of the
submatrices from the 1st row.
A =
det(A) ≡ a11det(A11)–a12det(A12) +a13det(A13) – ..(±)a1ndet(A1n)
Hence using this formula,
det
1 0 2
2 1 0
2 3 –1

93. Determinant
Finding Determinants-Expansion by the First Row
Let A be an nxn matrix as shown.
a11 a12 . . a1n
a21 a22 . . a2n
. . . . .
. . . . .
an1 . . . ann
The determinant of A or det(A)
may be calculated using the
determinants of its
submatrices Aij’s.
Here is the formula for det(A)
using the determinants of the
submatrices from the 1st row.
A =
det(A) ≡ a11det(A11)–a12det(A12) +a13det(A13) – ..(±)a1ndet(A1n)
Hence using this formula,
det
1 0 2
2 1 0
2 3 –1
= 1det – 0 + 2det1 0
3 –1
2 1
2 3

94. Determinant
Finding Determinants-Expansion by the First Row
Let A be an nxn matrix as shown.
a11 a12 . . a1n
a21 a22 . . a2n
. . . . .
. . . . .
an1 . . . ann
The determinant of A or det(A)
may be calculated using the
determinants of its
submatrices Aij’s.
Here is the formula for det(A)
using the determinants of the
submatrices from the 1st row.
A =
det(A) ≡ a11det(A11)–a12det(A12) +a13det(A13) – ..(±)a1ndet(A1n)
Hence using this formula,
det
1 0 2
2 1 0
2 3 –1
= 1det – 0 + 2det1 0
3 –1
2 1
2 3
= –1 + 8 = 7 (same answer as in eg. D)

95. Determinant
Example E. Find the following determinant using
the 1st row expansion.
2 0 −1 0
1 0 0 2
0 1 1 0
−1 2 0 1
det

96. Determinant
Example E. Find the following determinant using
the 1st row expansion.
2 0 −1 0
1 0 0 2
0 1 1 0
−1 2 0 1
det
= 2 det
0 0 2
1 1 0
2 0 1
2

97. Determinant
Example E. Find the following determinant using
the 1st row expansion.
2 0 −1 0
1 0 0 2
0 1 1 0
−1 2 0 1
det
= 2 det
0 0 2
1 1 0
2 0 1
– 0
0

98. Determinant
Example E. Find the following determinant using
the 1st row expansion.
2 0 −1 0
1 0 0 2
0 1 1 0
−1 2 0 1
det
= 2 det
0 0 2
1 1 0
2 0 1
– 0 + (−1)det
1 0 2
0 1 0
–1 2 1
−1

99. Determinant
Example E. Find the following determinant using
the 1st row expansion.
2 0 −1 0
1 0 0 2
0 1 1 0
−1 2 0 1
det
= 2 det
0 0 2
1 1 0
2 0 1
– 0 + (−1)det
1 0 2
0 1 0
–1 2 1
0
– 0

100. Determinant
Example E. Find the following determinant using
the 1st row expansion.
2 0 −1 0
1 0 0 2
0 1 1 0
−1 2 0 1
det
= 2 det
0 0 2
1 1 0
2 0 1
– 0 + (−1)det
1 0 2
0 1 0
–1 2 1
– 0
= −11