Week 4

2.6 Matrix Transformations (Revisited)

Composition of Transformations

Suppose TA : Rn Rk and TB : Rk Rm. Then if ~x Rn, we can compute TA(~x) Rk and thenwe can compute TB(TA(~x)) Rm or TB TA(~x).Now, if A is the standard matrix of TA and B is the standard matrix of TB, we have TA(~x) = A~xand TB(~w) = B~w.What is the standard matrix for TB TA?

(TB TA)(~x) = TB(TA(~x))= TB(A~x)

= (BA)~x

Thus, BA is the standard matrix for TB TA.

Example 5

Find the standard matrix for a reflection in the x-axis followed by a rotation through an angle ofpi

2

in R2 and use it to find the image of[38

].

All the transformations we have looked at have been matrix transformations since they can be writtenin the form T (~x) = A~x. They have also been linear.

Definition:

A transformation T : Rn Rm is linear ifT (~u+ ~v) = T (~u) + T (~v), for all ~u,~v Rn

T (a~u) = aT (~u), where a is a scalar, and ~u is any vector in Rn

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We say that T preserves addition and scalar multiplication.The combination of these two operations is called a linear combination.

T (a1~u1 + a2~u2 + . . .+ ak~uk) = a1T (~u1) + a2T (~u2) + . . .+ akT (~uk)

Example 6

If T : R2 R2 is a linear transformation where T[11

]=

[23]and T

[12]=

[51

], find T

[43

].

Theorem 1:

Every matrix transformation T : Rn Rm is linear.

Proof:

Let A be the standard matrix for T .Then T (~x) = A~x.We need to check that T preserves scalar multiplication and addition.T (~u+ ~v) = A(~u+ ~v)

= A~u+ A~v= T (~u) + T (~v)

T (a~u) = Aa~u= aA(~u)= aT (~u)

Theorem 2:

Every linear transformation is a matrix transformation.

Proof:

We define the standard basis of Rn to be the columns ~e1, ~e2, . . . , ~en of In.Note: Any vector in Rn can be written as a linear combination of ~e1, ~e2, . . . , ~en.

2

ie. ~u =

u1u2u3...un

=100...0

u1 +010...0

u2 +001...0

u3 + . . .+000...1

un = u1~e1 + u2~e2 + u3~e3 + . . .+ un~en.Let T : Rn Rm be a linear transformation.

T (~u) = T (u1~e1 + u2~e2 + . . .+ un~en)

= u1T (~e1) + u2T (~e2) + . . .+ unT (~en)

=[T (~e1) T (~e2) T (~en)

]u1u2...un

= A~u

Since A is the standard matrix for T , we have shown it is a matrix transformation.

Note: This proof gives us another way to find the standard matrix A by taking T (~ei) as the ith

column of A.

3.1 Determinants

In general a determinant is a function that takes a square matrix as input and outputs a real number.

For a 2 2 matrix A =[a bc d

], the determinant is defined to be ad bc. This number is significant

since it determines whether or not a matrix is invertible or not. If the determinant is 0, the matrixis not invertible. The determinant also has some geometric significance in R2 and R3 when we aredealing with 2 2 or 3 3 matrices. It also arises in a new formula for A1, a new method forsolving linear systems and is useful in applications where we want to predict the future behaviour ofa system. (Section 3.5)

Evaluating a 3 3 Determinant

If A =

a11 a12 a13a21 a22 a23a31 a32 a33

, we definedet(A) = a11a22a33 + a12a23a31 + a13a21a32 a13a22a31 a11a23a32 a12a21a33 (1)

We can rewrite this expression by factoring out the entries along the first row from each pair of terms:

det(A) = a11(a22a33 a23a32) a12(a21a33 a23a31) + a13(a21a32 a22a31)We can now think of the determinant of a 33 matrix in terms of three determinants of 22 matrices.

det(A) = a11

a22 a23a32 a33 a12 a21 a23a31 a33

+ a13 a21 a22a31 a32

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Example 1

Find det(A) and det(AT ), where A =

1 2 30 1 52 6 1

.

Note: det(AT ) = det(A)

Definition:

If A is an n n matrix, let Aij denote the (n 1) (n 1) sub-matrix that remains after the ithrow and jth column are deleted from A. Then the (i, j)-cofactor cij(A) is

cij(A) = (1)i+j det(Aij)

Example 2

Find c23(A) in the example above.

The sign (1)i+j in the cofactor can be determined quickly using a checkerboard diagram.+ + + + + + + + ...

......

...

Returning to our computation of a 3 3 determinant, we can now express it in terms of the

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cofactors of A. This is called the Cofactor Expansion of A along the first row of A.

det(A) = a11

a22 a23a32 a33 a12 a21 a23a31 a33

+ a13 a21 a22a31 a32

= a11c11(A) + a12c12(A) + a13c13(A)

In general, we can use cofactor expansion along any row or column to compute the determinant

Definition:

The cofactor expansion of an n n matrix A along the ith row of A is

det(A) = ai1ci1(A) + ai2ci2(A) + . . .+ aincin(A)

and the cofactor expansion of A along the jth column of A is

det(A) = a1jc1j(A) + a2jc2j(A) + . . .+ anjcnj(A)

Example 3:

If A =

3 1 02 4 35 4 2

, evaluate |A| by a cofactor expansion along the second column of A.

Is there an easier choice of row/column that we could expand along?Our strategy will be to expand along the row or column that contains the most zeros.If there are no zeros in our matrix, we will row-reduce the matrix in order to introduce some zeros.We need to see how EROs affect the value of the determinant.

Theorem 1:

Let A be an n n matrix.1. If A has a row of zeros, then |A| = 0.

5

2. If A has two proportional rows, then |A| = 0.3. If two rows of A are interchanged, the resulting matrix has determinant |A|.4. If a row of A is multiplied by a non-zero scalar k, the determinant of the resulting matrix is

k|A|.5. If a scalar multiple of one row of A is added to another, the determinant of the resulting matrix

is |A|.Note: Because |A| = |AT |, we can replace the word row by column in the list of operations abovesince transpose simply interchanges the rows and columns.

Definition:

A matrix is called lower triangular if it has zeros above the main diagonal and is called uppertriangular if it has zeros below the main diagonal. A matrix that is either upper or lower triangularis called triangular. A diagonal matrix has zeros above and below the main diagonal.

Theorem 2:

The determinant of a triangular matrix is the product of the entries along the main diagonal.

Examples:

4. If A =

1 0 0 32 7 0 60 6 3 07 3 1 5

, find |A|.

5. Find det(B), where B =

0 1 53 6 92 6 1

.

6. Find

1 2 3 5 72 0 1 5 64 7 3 9 43 1 2 2 3

5 1 3 7 9

.

7. If

a b cp q rx y z

= 6, finda x b y c z3x 3y 3zp q r

.8. Find values of x for which

1 x xx 1 xx x 1

= 0.

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3.2 Determinants and Matrix Inverses

Theorem 1:

Suppose A is an n n matrix and k is a scalar. Thendet(kA) = kn det(A)

since there are n rows and we can pull a common factor of k out of each row.

Theorem 2:

If A and B are square matrices of the same size, then

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

Theorem 3:

A square matrix A is invertible iff det(A) 6= 0.

Proof:

Let R be the RREF of A.Then there is a sequence of EROs that reduce A to R.Hence, there exist elementary matrices E1, E2, . . . Ek such that Ek E2E1A = R.() Assume det(A) 6= 0.Taking the determinant of both sides of the equation above, we have

det(Ek E2E1A) = det(R)det(Ek) det(E2) det(E1) det(A) = det(R)

Since the determinant of an elementary matrix is either 1, -1 or k 6= 0, then the left side of theequation is non-zero.Hence the right side of the equation is non-zero and so det(R) 6= 0.Therefore, R does not contain a row of zeros.Thus, R = I and A is invertible.() Assume A is invertible.Then R = I.Then there is a sequence of EROs that reduce A to I.Hence, there exist elementary matrices E1, E2, . . . Ek such that Ek E2E1A = I.Taking the determinant of both sides of the equation above, we have

det(Ek E2E1A) = det(I)det(Ek) det(E2) det(E1) det(A) = I

Thus, det(A) 6= 0.

Theorem 4:

If A is invertible, then det(A1) =1

det(A).

7

Proof:

Since AA1 = I, then det(AA1) = det(I) or det(A) det(A1) = 1, and since det(A) 6= 0, we havedet(A1) =

1

det(A).

Examples:

1. If det(A) = 2 and det(B) = 5, find det(A3B1ATB2).

2. (a) If A is a 2 2 matrix where |A| = 4, find det(3A), det(3A2), and det[(3A)2].(b) Is A invertible?

Definition:

The adjugate of an n n matrix is the transpose of the matrix of cofactors, or

adj(A) = [cij(A)]T

where [cij(A)] is the matrix whose (i, j) entry is the (i, j) cofactor of A.

Example 4:

If A =

1 3 20 1 52 6 7

, find adj(A).

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Theorem 5:

If A is an n n matrix, then A(adj(A)) = det(A)I.Furthermore, if A is invertible, then A1 =

1

det(A)(adj(A)).

Proof:

Example 5

Calculate the inverse of A, where A is the matrix in the Example 4.

Another use of determinants is in solving systems of equations. So far, we have seen two methods:augmented matrices and coefficient matrices (if A is invertible).

Cramers Rule

If A is an invertible n n matrix, then the solution to A~x = ~b is

x1 =|A1||A| , x2 =

|A2||A| , . . . , xn =

|An||A|

9

where Aj is the matrix obtained by replacing the entries in the jth column of A by the entries in the

matrix ~b =

b1b2...bn

.

Proof:

Let A be an invertible n n matrix where A~x = ~b.Then ~x = A1~b =

adj(A)

det(A)~b or

x1x2...xj...xn

=

1

|A|

c11(A) c21(A) cn1(A)c12(A) c22(A) cn2(A)

......

...c1j(A) c2j(A) cnj(A)

......

...c1n(A) c2n(A) cnn(A)

b1b2...bn

Then by matrix multiplication, xj =1

|A|(c1j(A)b1 + c2j(A)b2 + . . .+ cnj(A)bn).

Now consider the matrix Aj =

a11 a12 b1 a1na21 a22 b2 a2n...

......

...an1 an2 bn ann

By a cofactor expansion along columnj, we have

|Aj| = b1c1j(A) + b2c2j(A) + . . .+ bncnj(A)

Therefore|Aj||A| =

b1c1j(A) + b2c2j(A) + . . .+ bncnj(A)

|A| = xj.

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Cramers Rule is useful because it allows us to solve for a single component of ~x without computingthe rest.

Example 6

Use Cramers Rule to solve for x1 in the following system of equations:

x1 + 2x3 = 6

3x1 + 4x2 + 6x3 = 30x1 2x2 + 3x3 = 8

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Polynomial Interpolation

Suppose we are trying to describe the relationship between two variables x and y in terms of somepolynomial. We have a collection of experimental data points (x1, y1) (x2, y2), (x3, y3). We would liketo fit this data to a polynomial of the form ax2 + bx+ c. The unknowns here are the constants a, band c. By substituting each of our data points, we obtain a system of 3 equations in 3 unknowns,which we can then solve for.

Example:

Find a polynomial p(x) = ax2 + bx+ c that fits the following data points: (5, 3), (10, 5), (15, 6).

Solution:

By substituting the data points in for x and y, we obtain the following system:

a(5)2 + b(5) + c = 3

a(10)2 + b(10) + c = 5

a(15)2 + b(15) + c = 6

We could solve this using the matrix equationabc

= 25 5 1100 10 1225 15 1

1 356

=

125 5 1100 10 1225 15 1

adj

25 5 1100 10 1225 15 1

356

=

150710

0

The polynomial is 1

50x2 +

7

10x.

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Explanations and Proofs of Concepts Introduced this Week

Where does the determinant expression come from?

We look for conditions on the entries so that a 3 3 matrix A is invertible. We do so by trying toreduce A to I leaving any necessary divisions until the very end. When we finally have to divide toobtain a leading one, we will end up with the expression for the determinant and require it not toequal 0.Note: We cannot have a column of zeros if A is to be invertible, so we will assume at least one of theentries in column 1 is non-zero. Assuming a11 6= 0, we attempt to reduce A to I by performing thefollowing EROs on A:

1. a11R2

2. a11R3

3. R2 a21R14. R3 a31R15. (a11a22 a12a21)R36. R3 (a11a32 a31a12)R2,

we obtain

A =

a11 a12 a13a21 a22 a23a31 a32 a33

a11 a12 a13a11a21 a11a22 a11a23a11a31 a11a32 a11a33

a11 a12 a130 a11a22 a21a12 a11a23 a21a13

0 a11a32 a31a12 a11a33 a31a13

a11 a12 a130 a11a22 a21a12 a11a23 a21a13

0 (a11a22 a21a12)(a11a32 a31a12) (a11a22 a21a12)(a11a33 a31a13)

a11 a12 a130 a11a22 a21a12 a11a23 a21a13

0 0 (a11a22 a21a12)(a11a33 a31a13) (a11a32 a12a31)(a11a23 a21a13)

We want each column to contain a leading one. We require the first non-zero entry in each row tobe non-zero.The entry is row 3, column 3 simplifies toa11(a11a22a33 + a12a23a31 + a13a21a32 a13a22a31 a11a23a32 a12a21a33).The bracketed expression is what we define as the determinant of a 3 3 matrix.We require this number to be non-zero in order for A to be invertible.(Note that we also require a11a22 a21a12 to be non-zero in row 2.)

Cofactor expansion using cofactors and entries from different rows equals 0

Let A =

a11 a12 a13a21 a22 a23a31 a32 a33

.We know |A| = a11c11 + a12c12 + a13c13.

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Consider a11c31 + a12c32 + a13c33 where we have used entries from row 1, but cofactors from row 3.

We construct a new matrix A =

a11 a12 a13a21 a22 a23a11 a12 a13

where row 3 is simply a copy of row 1.Clearly |A| = 0 since it has 2 identical rows.Let c31, c

32 and c

33 be the cofactors of the entries in the 3rd row of A

.Now since the first two rows of A and A are the same, and the computations of c31, c32, c33, c31, c

32,

and c33 involve only these entries in the first two rows, then c31 = c31, c32 = c

32, and c33 = c

33.

Then by evaluating |A| by a cofactor expansion along the 3rd row of A, we have

|A| = a11c31 + a12c32 + a13c33= a11c31 + a12c32 + a13c33

But |A| = 0 so a11c31 + a12c32 + a13c33 = 0.

Illustration of why ERO Ri + kRj does not change a 3 3 determinant

Let A =

a11 a12 a13a21 a22 a23a31 a32 a33

. If we perform the ERO R1 + kR2, we have the following determinant:a11 + ka21 a12 + ka22 a13 + ka23

a21 a22 a23a31 a32 a33

= (a11 + ka21)c11 + (a12 + ka22)c12 + (a13 + ka23)c13= a11c11 + a12c12 + a13c13 + k(a21c11 + a22c12 + a23c13)

= det(A) + k

a21 a22 a23a21 a22 a23a31 a32 a33

= det(A) + k(0)

= det(A)

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