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Determinant
The determinant of an nxn square matrix A,
or det(A), is a number.
Determinant
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
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
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
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
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
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
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 deta b
c d = ad – bc
= “Signed” area of the parallelogram as shown.
(a, b) (c, d)
“Signed” area
= ad – bc
Determinant
DeterminantLet’s verify this in the simple cases where one of the
points is on the axis.
DeterminantLet’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)
DeterminantLet’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
deta 0
c d
(a, 0)
(c, d)
d
= ad
DeterminantLet’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
deta 0
c d
(a, 0)
(c, d)
d
If we switch the order of the two rows then
detc d
= – ad
= ad
a 0
DeterminantLet’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
deta 0
c d
(a, 0)
(c, d)
d
If we switch the order of the two rows then
deta 0
c d = – ad
= ad
What’s the meaning of the “–” sign?
(besides that the rows are switched)
DeterminantLet’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
deta 0
c d
(a, 0)
(c, d)
d
If we switch the order of the two rows then
deta 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.
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)
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)
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)
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)
Determinant
If deta 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. detAB
A(a, b)
B(c, d)
A(a, b)
B(c, d)
> 0
B is to the left of A.
Determinant
If deta 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. detAB
A(a, b)
B(c, d)
A(a, b)
B(c, d)
> 0 detAB < 0
B is to the left of A. B is to the right of A.
If deta b
c d is –, then B is to the right of A.
Determinant
If deta 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. detAB
A(a, b)
B(c, d)
A(a, b)
B(c, d)
> 0 detAB < 0
B is to the left of A. B is to the right of A.
If deta b
c d is –, then B is to the right of A.
For example det1 0
0 1 > 0 and det
1 0
0 1 < 0.
It is in the above sense that we have the
signed-area-answer of det A.
Determinant
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
So given the picture to the right:
deta 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
So given the picture to the right:
deta b
c d = + Area of
(a, b) (c, d)
deta b
c d = – Area of
and
It is in the above sense that we have the
signed-area-answer of det A.
Determinant
So given the picture to the right:
deta b
c d = + Area of
(a, b) (c, d)
deta 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)
So given the picture to the right:
deta b
c d = + Area of
(a, b) (c, d)
deta 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)
det5 –3
1 4
To find the area, we calculate
So given the picture to the right:
deta b
c d = + Area of
(a, b) (c, d)
deta 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)
det5 –3
1 4
To find the area, we calculate
= 5(4) – 1(–3) = 23 = area
So given the picture to the right:
deta b
c d = + Area of
(a, b) (c, d)
deta 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)
det5 –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.
Here is another motivation for the 2x2 determinants.
Determinant
Here is another motivation for the 2x2 determinants.
Determinant
Il. (Algebraic) The system of equations
deta b
c d ≠ 0
ax + by = #
cx + dy = #
has exactly one solution if and only if
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
deta b
c d ≠ 0
ax + by = #
cx + dy = #
has exactly one solution if and only if
{ x + y = 2
2x + 2y = 4
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
deta b
c d ≠ 0
ax + by = #
cx + dy = #
has exactly one solution if and only if
{ x + y = 2
2x + 2y = 4
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
deta 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
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
deta 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.
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
deta 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 det1 12 2 = 0.
Here is the Cramer’s rule that actually gives
the only solution of the system
Determinant
if deta b
c d = D ≠ 0
ax + by = u
cx + dy = v
Here is the Cramer’s rule that actually gives
the only solution of the system
Determinant
if deta 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.
Here is the Cramer’s rule that actually gives
the only solution of the system
Determinant
if deta 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
Here is the Cramer’s rule that actually gives
the only solution of the system
Determinant
if deta 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 ≠ 0 det
Here is the Cramer’s rule that actually gives
the only solution of the system
Determinant
if deta 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 ≠ 0 det
so that x =det
2 15 4
3
Here is the Cramer’s rule that actually gives
the only solution of the system
Determinant
if deta 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 ≠ 0 det
so that x =det
2 15 4
3= 1
Here is the Cramer’s rule that actually gives
the only solution of the system
Determinant
if deta b
c d = D ≠ 0
ax + by = u
cx + dy = v
namely that x =det
u b v d
Dy =
deta uc 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 ≠ 0 det
so that x =det
2 15 4
3= 1
Here is the Cramer’s rule that actually gives
the only solution of the system
Determinant
if deta b
c d = D ≠ 0
ax + by = u
cx + dy = v
namely that x =det
u b v d
Dy =
deta uc 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 ≠ 0 det
so that x =det
2 15 4
3y =
det1 21 5
3= 1 = 1
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
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
dete 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.
dete 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
det1 0
3 –1= 1*det
1 0 2
2 1 0
2 3 –1
Example C. Calculate.
dete 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
det1 0
3 –1= 1*det
1 0 2
2 1 0
2 3 –1
Example C. Calculate.
dete 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
det1 0
3 –1= 1*det
1 0 2
2 1 0
2 3 –1
Example C. Calculate.
det– 0*2 0
2 –1
dete 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
det1 0
3 –1= 1*det
1 0 2
2 1 0
2 3 –1
Example C. Calculate.
det– 0*2 0
2 –1
dete 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
det1 0
3 –1= 1*det
1 0 2
2 1 0
2 3 –1
det– 0* det+ 2*2 0
2 –1
2 1
dete 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
det1 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)
dete 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
det1 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
Don’t remember this!
dete 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
det1 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
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.
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.
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.
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
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
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
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
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
The det A of a 3×3 matrix A similarly gives
the volume of a solid in 3D space.
Determinant
Determinant
(a, b, c)
(d, e, f)
(0, 0, 0)
(g, h, i)
a b c
d e f
g h idet = ± the volume of
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 idet = ± the volume of
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 idet = ± the volume of
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 idet = ± the volume of
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 idet = ± the volume of
Likewise the 3x3 system
Determinant
≠ 0det
a b c
d e f
g h ihas a unique answer iff the
ax + by + cz = #
dx + ey + fz = #
gx + hy + iz = #
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 ihas a unique answer iff the
ax + by + cz = #
dx + ey + fz = #
gx + hy + iz = #
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 ihas a unique answer iff the
We will show a method of finding the determinants of
n x n matrices.
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 ihas 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.
DeterminantWe need two points to define the
two edges of a parallelogram
DeterminantWe need two points to define the
two edges of a parallelogram
and three points for the three edges
of a tilted box.
DeterminantWe 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.
DeterminantWe 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.
DeterminantWe 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.
DeterminantFinding Determinants-Expansion by the First Row
DeterminantFinding Determinants-Expansion by the First Row
Let A be an nxn matrix as shown.
a11 a12 . . a1n
a21 a22 . . a2n
. . . aij .
. . . . .
an1 . . . ann
A =
DeterminantFinding 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 =
DeterminantFinding 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
DeterminantFinding 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
DeterminantFinding 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
DeterminantFinding 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
DeterminantFinding Determinants-Expansion by the First Row
DeterminantFinding Determinants-Expansion by the First Row
Let A be an nxn matrix as shown.
a11 a12 . . a1n
a21 a22 . . a2n
. . . . .
. . . . .
an1 . . . ann
A =
DeterminantFinding 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 =
DeterminantFinding 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 =
DeterminantFinding 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)
DeterminantFinding 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,
det1 0 2
2 1 0
2 3 –1
DeterminantFinding 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,
det1 0 2
2 1 0
2 3 –1
= 1det – 0 + 2det 1 03 –1
2 12 3
DeterminantFinding 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,
det1 0 2
2 1 0
2 3 –1
= 1det – 0 + 2det 1 03 –1
2 12 3
= –1 + 8 = 7 (same answer as in eg. D)
DeterminantExample E. Find the following determinant using
the 1st row expansion.
2 0 −1 01 0 0 20 1 1 0−1 2 0 1
det
DeterminantExample E. Find the following determinant using
the 1st row expansion.
2 0 −1 01 0 0 20 1 1 0−1 2 0 1
det
= 2 det
0 0 2
1 1 0
2 0 1
2
DeterminantExample E. Find the following determinant using
the 1st row expansion.
2 0 −1 01 0 0 20 1 1 0−1 2 0 1
det
= 2 det
0 0 2
1 1 0
2 0 1
– 0
0
DeterminantExample E. Find the following determinant using
the 1st row expansion.
2 0 −1 01 0 0 20 1 1 0−1 2 0 1
det
= 2 det
0 0 2
1 1 0
2 0 1
– 0 + (−1)det1 0 2
0 1 0
–1 2 1
−1
DeterminantExample E. Find the following determinant using
the 1st row expansion.
2 0 −1 01 0 0 20 1 1 0−1 2 0 1
det
= 2 det
0 0 2
1 1 0
2 0 1
– 0 + (−1)det1 0 2
0 1 0
–1 2 1
0
– 0
DeterminantExample E. Find the following determinant using
the 1st row expansion.
2 0 −1 01 0 0 20 1 1 0−1 2 0 1
det
= 2 det
0 0 2
1 1 0
2 0 1
– 0 + (−1)det1 0 2
0 1 0
–1 2 1
– 0
= −11