Determinants - Ch. 2.1, 2.2, 2.3, 2.4, 2.5

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Section Three: Determinants

Textbook: Ch. 2.1, 2.2, 2.3, 2.4, 2.5GOALS OF THIS CHAPTER

- introduce further the matrix determinant

- calculate determinants using the weave method

- calculate determinants using cofactor expansions

- calculate determinants using elementary row operations

- see properties of the matrix determinant

- application to geometry: areas of parallelograms andtriangles

- application to linear systems: solving systems using

Cramers Rule

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WHAT IS A MATRIX DETERMINANT?

Sometimes it is nice to know if a matrix isinvertible BEFORE we do all the work toactually find the inverse.

It turns out that there is a special numberassociated to a square matrix that tells us if thatmatrix is invertible!

We call this special number the determinant ofthe matrix A, or write it as det(A).

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Recall the 2x2 matrix determinant:

How do we actually get this number?

A =a b

c d

det(A) = ad - bc

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On our list of Non-singular Equivalences, wenoted that a matrix was invertible if its RREF

was an identity matrix.

Lets try to reduce this matrix to RREF and see what happens:

A =

a b

c d

Ex. 1 Where the 2x2 Determinant Comes From

R1 R1 / a

(here we are assuming that a is not zero)

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1 b/a

c d


A =


R2 R2 c*R1

1 b/a

0 d (bc)/aA =

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1 b/a

0 d (bc)/aA =

So, in order for A to be an identity matrix, wewould need the number d (bc)/a to not be zero:

d (bc)/a 0

ad bc 0

det(A)

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We have just shown the most important propertyof the determinant:

IF det(A) = 0, THEN THE MATRIX A IS NOTINVERTIBLE.

IF det(A) 0, THEN THE MATRIX A ISINVERTIBLE.

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THE WEAVE METHOD

The weave method gives us an easy way toremember the determinant of a 2x2 or 3x3matrix. We cannot use this method for higherorder matrices.

A =a b

c d

det(A) =

Ex. 2 The 2x2 Weave Method

- bc+ ad

+-

Down and to theright, we add.

Down and to the left,we subtract.

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THE WEAVE METHOD

A =

a b c

d e f

g h i

det(A) =


- ceg - afh - bdi+ aei + bfg + cdh

+-

To make it work, you have toadd the first two columns ofthe matrix on the right side.

Down and to the left,we subtract.

a b

d e

g h

Down and to theright, we add.+ +-

-

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THE WEAVE METHOD

A =

1 0 3

2 -8 7

5 2 1

det(A) =


- 3*(-8)*5 - 1*7*2 - 0*2*11*(-8)*1 + 0*7*5 + 3*2*2

+-

1 0

2 -8

5 2

+ +--

det(A) = -8 + 0 + 12 + 120 - 14 - 0

det(A) = 110

REMARK: Since thedeterminant is not

zero, it would bepossible to find theinverse of this matrix.

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COFACTOR EXPANSION

How do we deal with determinants of higherorder matrices? Wouldnt it be nice if we

could write these formulas in a simpler way?

To answer these questions, we will use the cofactors of matrixentries. Recall that the cofactor of a matrix entry a(i,j) isdefined to be:

Aij = (-1)i+jMij

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a11 a12

a21 a22

Lets revisit the 2x2 determinant:

A =

det(A) =

Ex. 4 The 2x2 Cofactor Expansion

- a12a21+ a11a22

+-Lets try to rewrite

this in terms ofcofactors.

COFACTOR EXPANSION

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a11 a12

a21 a22



A =

det(A) = a11a22 a12a21det(A) = a11M11 a12a21

We can replace a22with the determinantof a minor!

COFACTOR EXPANSION

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a11 a12

a21 a22A =

det(A) = a11a22 a12a21det(A) = a11M11 a12M12

We can do the samething for a21.

COFACTOR EXPANSION

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det(A) = a11A11 + a12A12

a11 a12

a21 a22A =

det(A) = a11a22 a12a21det(A) = a11M11 a12M12

Finally, we get rid ofthe negative signs byusing the cofactor.This is why thecofactor has a (-1) inthe formula!

COFACTOR EXPANSION

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det(A) = a11A11

a11 a12

a21 a22A =

We call this a cofactor expansionalong the first row. Pretend you aredrawing an arrow and whateverentry you hit, multiply that entry byits corresponding cofactor.

+ a12A12

COFACTOR EXPANSION

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Next, we will show that a cofactor expansionalong the first row of a 3x3 matrix gives the3x3 weave formula.


COFACTOR EXPANSION

A =

a b c

d e f

g h i

det(A) = a*A11 + b*A12 + c*A13

det(A) = a(-1)2M11 + b(-1)3M12 + c(-1)

4M13

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a b c

d e f

g h i


COFACTOR EXPANSION

A =

det(A) = a*A11 + b*A12 + c*A13

det(A) = a(-1)2M11 + b(-1)3M12 + c(-1)

4M13

det(A) = aM11 bM12 + cM13

det(A) = a(ei - fh) bM12 + cM13

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COFACTOR EXPANSION

A =

a b c

d e f

g h i

det(A) = a*A11 + b*A12 + c*A13

det(A) = a(ei - fh) b(di - fg) + cM13

det(A) = a(-1)2M11 + b(-1)3M12 + c(-1)

4M13


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COFACTOR EXPANSION

A =

a b c

d e f

g h i

det(A) = a*A11 + b*A12 + c*A13

det(A) = a(ei - fh) b(di - fg) + c(dh - eg)

det(A) = a(-1)2M11 + b(-1)3M12 + c(-1)

4M13


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COFACTOR EXPANSION

A =

a b c

d e f

g h i

det(A) = a*A11 + b*A12 + c*A13

det(A) = a(ei - fh) b(di - fg) + c(dh - eg)

det(A) = aei - afh - bdi + bfg + cdh - cegThis is theformula we

saw before!

det(A) = a(-1)2M11 + b(-1)3M12 + c(-1)

4M13


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COFACTOR EXPANSION

So this gives us a way to find determinants of4x4 or higher matrices! We expand using

cofactors (since we cant use the weaveformula).

Before I give some examples, I will state a theorem that mightsave you some time.

Thm. 6 If A is a square matrix, then det(A) can be expressedas a cofactor expansion along any row or any columnof thematrix.

In the next few examples, we will see just howawesome this theorem is.

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COFACTOR EXPANSION

Ex. 7 Illustration of Thm. 6

- done in class

Ex. 9 Speedy 4x4 Expansion- done in class

Ex. 8 4x4 Cofactor Expansion

- done in class

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ELEMENTARY ROW OPERATIONS

Recall the three types of elementary matrices:

TYPE I multiplying a row by a non-zero number

TYPE II adding a multiple of one row to another

TYPE III switching two rows

What we will do next is to calculate the determinantsof these types of elementary matrices, so we can seehow elementary row operations affect determinantcalculations.

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1 0 0

0 1 0

0 0 2

TYPE I multiplying a row by a non-zero number

E1 =


Elementary matrix obtained from operation R3 2R3

det(E1) = 2

So the determinant of aType I elementary matrix isequal to the non-zeronumber you use!

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1 0 -3

0 1 0

0 0 1

TYPE II adding a multiple of one row to another

E2 =


Elementary matrix obtained from operation R1 R1 3R3

det(E2) = 1

So the determinant of aType II elementary matrixis equal to one!

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1 0 0

0 0 1

0 1 0

TYPE III switching two rows

E3 =


Elementary matrix obtained from operation R2 R3

det(E3) = -1

So the determinant of aType III elementary matrixis equal to negative one!

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Fact:

For any square matrix A and any elementary matrix E

det(EA) = det(E) det(A).

This means if we want to use an elementary rowoperation on A to make our determinant calculationeasier, we can balance this action by dividing by thedeterminant of our elementary matrix.

det(EA)

det(E)= det(A)

performing anE.R.O. on a

determinant

balancing bydividing

keeps det(A) in

tact

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Ex. 11 Algebraic Row Operation Method- done in class

Ex. 10 5x5 Elementary Row Operation Method

- done in class


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PROPERTIES OF THE DETERMINANT

Thm. 12 Properties of the Determinant- proof done in class

(1) A is singular if and only if det(A) = 0

(2) det(AB) = det(A)det(B)

(3) det(At) = det(A)

(4) If A is upper or lower triangular, thendet(A) is the product of the elements onthe main diagonal.

(5) det(cB) = cndet(B) (c is a scalar number)

(6) If A is invertible det(A-1) = 1 / det(A)

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PROVING THE EQUIVALENT STATEMENTS

Now we have enough tools to prove the equivalent

statements. We also add two more items to the list.

(1)The nxn matrix A is invertible.(2) det(A) 0.(3) The system Ax=b has a unique solution for

every b.(4) The only solution of the homogeneous system

Ax=0 is the trivial solution x=0.(5) A is row equivalent to I (ie. there is not a full

row of zeros in the RREF of A).

(6) A can be written as a product of elementarymatrices

Proof done in class

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Consider a triangle in the first quadrant:

FINDING AREAS USING DETERMINANTS

(x1,y1)

(x2,y2)

(x3,y3)

To calculate the area, we will calculate thearea of three rectangles and three triangles.

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(x1,y1)

(x2,y2)

(x3,y3)

The area of the first rectangle can becalculated by:

l*w = (x2 - x1)y1

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(x1,y1)

(x2,y2)

(x3,y3)

The area of the first triangle can becalculated by:

b*h/2 = (x2 - x1)(y2 - y1)*1/2

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(x1,y1)

(x2,y2)

(x3,y3)

The total area of this shape is:

l*w + b*h/2 = (x2 - x1)y1 + (x2 - x1)(y2 - y1)*1/2

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(x1,y1)

(x2,y2)

(x3,y3)

The area of the second rectangle can becalculated by:

l*w = (x3 x2)y3

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(x1,y1)

(x2,y2)

(x3,y3)

The area of the second triangle can becalculated by:

b*h/2 = (x3 x2)(y2 - y3)*1/2

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(x1,y1)

(x2,y2)

(x3,y3)

The total area of this shape:

l*w + b*h/2 = (x3 x2)y3 + (x3 x2)(y2 - y3)*1/2

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(x1,y1)

(x2,y2)

(x3,y3)

The area of the final rectangle can becalculated by:

l*w = (x3 x1)y3

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(x1,y1)

(x2,y2)

(x3,y3)

The area of the final triangle can becalculated by:

b*h/2 = (x3 x1)(y1 - y3)*1/2

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(x1,y1)

(x2,y2)

(x3,y3)

The area of the final shape is:

l*w + b*h/2 = (x3 - x1)y3 + (x3 x1)(y1 - y3)*1/2

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So the area of the triangle is:


(x1,y1)

(x2,y2)

(x3,y3)

[(x2 - x1)y1 + (x2 - x1)(y2 - y1)*1/2]

+ [(x3 x2)y3 + (x3 x2)(y2 - y3)*1/2]

- [(x3 - x1)y3 + (x3 x1)(y1 - y3)*1/2]

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If we expand, we get:


1/2*x2y1 1/2*x1y2 + 1/2*x3y2 1/2*x2y3 1/2*x3y1 + 1/2*x1y3

It turns out that this expression above can be written as

x1 y1 1

x2 y2 1

x3 y3 1

det

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Sometimes the points will be not in the firstquadrant or the points will be labeled in adifferent order. This has an effect ofchanging the determinant by (-1), so we justtake the absolute value:


x1 y1 1

x2 y2 1

x3 y3 1

Area of triangle = det

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Since a parallelogram is just two triangles puttogether, we can use the previous formula forareas of parallelograms too! All we require arethree vertices of the parallelogram (thefourth one is just extra information).


x1 y1 1

x2 y2 1

x3 y3 1

Area of parallelogram = det

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Ex. 13 Area of Triangle and Parallelogram Using Determinants- done in class


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CRAMERS RULE

We can use determinants to solve equally determinedlinear systems in a much quicker way that usingGauss-Jordan elimination. The next method we see iscalled Cramers Rule.

Thm. 14 Let A be an nxn square matrix that is non-singular. Let bbe an nx1 column matrix. Let Ai be the matrix obtained by replacingthe ith column of A with b. If x is the unique solution of Ax=b, then

xi = det(Ai) , as i ranges from 1 to n.

det(A)

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CRAMERS RULE

Proof: Since A is non-singular, we know the inverse ofA exists. We can use it to solve the linear system:

Ax = bx = A-1b

If we rewrite the inverse of A using the adjoint formula, we obtain:

x = (1/det(A) * adj(A))b

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CRAMERS RULE

Write the formula using general matrices:

x = (1/det(A) * adj(A))b

A11 A21 An1

A12 A22 An2

A1n A2n Ann

x1

x2

xn

b1

b2

bn

= 1

det(A)

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CRAMERS RULE

Lets calculate one of the entries in x:

A11 A21 An1

A12 A22 An2

A1n A2n Ann

x1

x2

xn

b1

b2

bn

=1

det(A)

x1 = 1/det(A) (A11b1 + A21b2+ + An1bn)

x1 = det(A1)det(A)

This is a cofactorexpansion down the firstrow of A1!

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CRAMERS RULE

This works for all the other entries in the x-vector, so we have proved that Cramers Ruleworks.

Ex. 15 Cramers Rule for a 3x3 Linear System- done in class

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GABRIEL CRAMER (1704-1752)

- Swiss mathematician born in Geneva

- finished his education at eighteen and wasco-chair of a mathematics department attwenty!

- travelled a lot; he met Euler (that guy who has e named after him),Halley (that guy who has a comet named after him), Johann and DanielBernoulli (many things named after the Bernoullis)

- after Johann Bernoulli died, Cramer was asked to edit all ofBernoullis works and publish them

- Cramers Rule appears in Introduction lanalyse des lignes courbesalgbriques

- falling from his carriage while travelling and the over-work from

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