Lecture 5Matrix inverses, augmented matrices, matrix equations

Recall that the inverse of a square matrix A is a matrix A1

such that AA1 = I.

Example 5.1

A =

2

6

4

1 0 1

2 1 32 2 1

3

7

5

) A1 =

How do we find A1?

The sequence of row operations that reduces A to I is the samesequence of operations that turns I to A1.

I.5.1

Finding the inverse of a matrix

1. Construct the augmented matrix [A|I], where I is the nnidentity matrix.

2. Using row operations, reduce A to reduced row echelonform to give [C|D]. (Use Gauss-Jordan elimination)

3. If C is the identity matrix, then D = A1.

Otherwise A is singular and A1 does not exist.

I.5.2

Example 5.2 Find the inverse of A =

2

6

4

1 2 1

1 1 10 1 3

3

7

5

. Start

with the augmented matrix.2

6

4

1 2 1

1 1 10 1 3

1 0 0

0 1 0

0 0 1

3

7

5

I.5.3

We say that a system of equations is homogeneous if the righthand side is zero: Ax = 0.

To solve the equation

Ax = b

we can find the reduced row echelon form of the augmentedmatrix. If this is already done for the matrix Aequivalentlyif we have solved the associated homogeneous systemweneed only remember the row operations and apply them to b.

If A is invertible we can use its inverse A1.

Ax = b)

The product A1b is the same as remembering the row opera-tions and applying them to b.

I.5.4

Example 5.3 Solve the system

x+2y + z = 2

x y + z = 1y +3z = 2

I.5.5

Example 5.4 Solve the associated homogeneous system

x+2y + z = 0

x y + z = 0y +3z = 0

I.5.6

Theorem 5.1 For a matrix A, the following are equivalent:

1. A is invertible.

2. The reduced row echelon form of A is I.

3. Ax = 0 has only the solution x = 0.

4. Ax = b has a unique solution.

I.5.7

Example 5.5 Solve

x+2y + z = 0

x y + z = 0y +2z = 0

After this class, read section 2.1 of Anton and Rorres.You can attempt questions 18 to 20 in problem sheet 1of your workbook.

Lecture 6Determinants, cofactors

A matrix M is invertible if there exists a matrix M1 such thatMM1 = I.

We have given an algorithm to find M1 if it exists.

Is there a simpler way to tell if M is invertible before we applythe algorithm?

Yes. The determinant of a matrix gives us this answer.

Example 6.1

M =

"

1 2

2 1

#

I.6.1

Let A = [aij] be an n n matrix.

The determinant of A is denoted by detA, det[aij] or |aij|.

for the 1 1 matrix A = [a11

], detA =

for A ="

a bc d

#

, detA =

I.6.2

Example 6.2 Calculate det I2

and

1 42 3

.

Recall the 2 2 inverse

A1 =

When does a 2 2 matrix not have an inverse?

I.6.3

Area of parallelograms. A parallelogram is defined by two vec-tors v

1

and v2

.

The area of the parallelogram is the determinant of the 2 2

matrix" #

.

Example 6.3 (a) v1

= (1,0), v2

= (1,1)

(b) v1

= (1,0), v2

= (1,0)

I.6.4

The determinant det(A) of a square matrix A is defined by:

1. det(A) is a real number

2. for the 1 1 matrix A = [a11

], detA = a11

3. det(AB) = det(A)det(B)

4. A ="

B *O C

#

) det(A) = det(B)det(C) (B,C square)

Theorem 6.1Properties 1-4 uniquely determine the determinant function.

An nn determinant is an n-dimensional volume of a parallel-ogram.

I.6.5

Submatrices, minors and cofactors.

Consider the n n matrix

A =

2

6

6

6

6

6

6

4

a11

a12

a13

. . . a1n

a21

a22

a23

. . . a2n

a31

a32

a33

. . . a3n... . . . ...

an,1 an,2 an3 . . . an,n

3

7

7

7

7

7

7

5

The i, j submatrix of A, denoted by Aij, is the (n1)(n1)matrix obtained by deleting the th row and th column from A.

The i, j minor of A is detAij.

The i, j cofactor of A, denoted by Cij, is (1)i+jdetAij.

I.6.6

Example 6.4 For the matrix A =

2

6

4

1 2 1

1 1 10 1 3

3

7

5

A23

=

the 2,3 minor of A is

the 2,3 cofactor of A, or C23

=

I.6.7

Theorem 6.2 The determinant of an n n matrix A can becomputed by multiplying the entries in any row (or column) bytheir cofactors and adding the resulting products.

Expanding along a row, for 1 i n

detA = ai1Ci1 + ai2Ci2 + . . .+ ainCin

Expanding along a column, for 1 j n

detA = a1jC1j + a2jC2j + . . .+ anjCnj

Check: What is

a bc d

?

I.6.8

The sign on the cofactor (1)i+j is found by starting in the up-per left corner and writing ++ . . . along each row/column.

[(1)i+j] =

2

6

6

6

4

+ + . . . + . . .+ + . . .... ... ... . . .

3

7

7

7

5

Start with + in top left and alternate left-to-right and top-to-bottom.

Example 6.5 Calculate det

2

6

4

1 2 1

1 1 10 1 3

3

7

5

.

I.6.9

Example 6.6 Calculate

1 2 0 13 1 2 0

1 0 1 0

2 2 1 2

.

I.6.10

Properties of determinants.

Let A and B be n n matrices.

det(AT ) = det(A).

if A has a row of zeros, then detA = 0.

det(A+B) 6= det(A) + det(B).

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

if A is invertible then det(A1) = 1detA.

Q. if A is not invertible does this imply detA = 0?

I.6.11

Example 6.7 Verify the first and fifth point for A ="

1 02 3

#

Another technique for 3 3 determinants.

det

2

6

4

a11

a12

a13

a21

a22

a23

a31

a32

a33

3

7

5

=

There is an easy way to remember this!

I.6.12

Example 6.8 Calculate

1 2 1

1 1 10 1 3

.

After this class, read section 2.2 of Anton and Rorres.You can attempt questions 21 to 27 in problem sheet 1of your workbook.

I.6.13

Lecture 7Determinants by row operations, properties

A matrix is said to be upper triangular if all the elements belowthe main diagonal are zero.

Example 7.1 Let A =

2

6

6

6

4

2 10 92 1170 3 28 310 0 1 270 0 0 2

3

7

7

7

5

.

det(A) =

Theorem 7.1 If A is an nn triangular matrix (upper triangular,lower triangular or diagonal) then detA is the product of theentries on the main diagonal of A.

I.7.1

Aim: manipulate a matrix to triangular form to make the deter-minant calculation easier.

Strategy: reduce A to row echelon form and keep track of howthe determinant changes with each row operation.

Row operations

Recall that an n n matrix is an elementary matrix if it can beobtained from In by performing a single elementary row opera-tion.

Pre-multiplying by an elementary matrix is equivalent to doingan elementary row operation.

I.7.2

For example. Starting with I3

Interchange rows 2 and 3 : E =

2

6

4

1 0 0

0 0 1

0 1 0

3

7

5

, detE =

Multiply Row 1 by 2 : F =

2

6

4

2 0 0

0 1 0

0 0 1

3

7

5

, detF =

Row 3 + 3 Row 1! Row 3: G =

2

6

4

1 0 0

0 1 0

3 0 1

3

7

5

, detG =

Using elementary matrices we can check the effects of row op-erations on the determinant of a matrix.

First calculate the determinant of any elementary matrix.

I.7.3

1. if E multiplies a row by c then det(E) =

2. if F adds a multiple of a row to a row then det(F ) =

3. if G swaps two rows then det(G) =

So if we perform any of these row operations on A,

det(EA) =

det(FA) =

det(GA) =

I.7.4

The effect of row operations on determinants

Interchanging two rows multiplies the value of the determinantby 1.

If all the elements of one row of a determinant are multipliedby a constant, the value of the determinant is multiplied by thatconstant.

Replacing a row by itself plus a multiple of another row does notchange the value of the determinant.

These rules apply to operations on columns as well!

I.7.5

Example 7.2 Calculate det

2

6

4

1 2 1

1 1 10 1 3

3

7

5

.

I.7.6

Example 7.3 Calculate

1 2 0 13 1 2 0

1 0 1 0

2 2 1 2

.

I.7.7

More properties of determinants

If A is an n n matrix,

det(A) = ndetA.

If A is a triangular matrix, then its determinant is the productof the elements on the main diagonal.

If A has a row of zeros, then detA = 0.

If A has two proportional rows (or columns) then detA = 0.

If A has an inverse then detA 6= 0.

If A has no inverse then detA = 0.

I.7.8

Example 7.4 If A ="

1 2

1 1

#

then det(3A) =

Example 7.5 Find the determinant of C =

2

6

4

2 2 4

0 3 4

1 3 1

3

7

5

.

After this class, read sections 3.1 and 3.2 of Anton andRorres. You can attempt all questions in problem sheet1 of your workbook.

Lecture 8Vectors in R3, dot product and cross product.

Geometrically, a vector is a directed line segment. Algebraically,a vector is an ordered n-tuple of components.

Some of the notation used (often interchangeably) includes:

u is in n-space

u 2 Rn

u = (u1

, u2

, u3

, . . . , un)

I.8.1

Vectors in R3

Let i = (1,0,0), j = (0,1,0) and k = (0,0,1).

Then any vector in 3-space can be written as

u = (u1

, u2

, u3

) =

The vector u is the position vector of P (relative to O).

I.8.2

The length, magnitude or norm of the vector u in n-space is

||u|| =q

u21

+ u22

+ u23

+ . . .+ u2n

(Proof uses Pythagoras Theorem)

Example 8.1 Let u = 2i j+2k.

||u|| =

A vector with length or norm equal to 1 is called a unit vector

Example 8.2 Find a unit vector parallel to u.

I.8.3

Distance To find the distance between two vectors u and v,or d(u,v), consider u and v to be position vectors and find thedistance between their endpoints. i.e.

d(u,v) = ||vu|| =q

(v1

u1

)

2

+ (v2

u2

)

2

+ . . .+ (vn un)2

Example 8.3 Find the distance between the points P (1,3,1)and Q(2,1,1)

I.8.4

Two vectors are equal if they have the same number of compo-nents and their corresponding components are equal just as formatrices, since vectors are examples of matrices.

Similarly, scalar multiplication and vector addition are definedas for matrices.

Scalar Multiplication If u is in n-space, and 2 R, then

u = (u1

,u2

,u3

, . . .un)

Vector Addition If u and v are from Rn, then

u+ v = (u1

+ v1

, u2

+ v2

, . . . , un + vn)

i.e. you add two vectors by adding corresponding components.

Geometrically, what happens when you add two vectors?

I.8.5

Vector properties

u+ v = v+ u (commutative)

(u+ v) +w = u+ (v+w) (associative)

(u+ v) = u+ v (distributive)

||ku|| = |k| ||u||

Example 8.4 If u 2 R3, show that ||ku|| = |k| ||u||.

I.8.6

Dot Product

The dot product or scalar product or Euclidean inner product isdefined geometrically by

u v = ||u||||v|| cos

where , 0 is the angle between u and v.

If u = (u1

, u2

, u3

, . . . , un) and v = (v1

, v2

, v3

, . . . , vn), thenthe dot product of u and v is defined algebraically by

u v = u1

v1

+ u2

v2

+ u3

v3

+ . . .+ unvn

Example - angle between vectors

If u = 2i+ j k and v = i+2j+ k calculate

(a) u v

(b) the angle between u and v. I.8.7

Properties of dot products

u v is a scalar!!!!!!!

u v = v u

u (v+w) = u v+ u w

u u = ||u||2

u v = (u) v = u (v) where 2 R.

After this class, read sections 3.3 and 3.5 of Anton andRorres. You can attempt questions 1 to 4 in problemsheet 2 of your workbook.

I.8.8