MAS151: Civil EngineeringMathematics

Dr James Cranchmas-engineering@sheffield.ac.uk

Friday 25th October 2019, 12noonHicks Building LT1

Course matters

Your comments

We are interested to know your opinions about this course:

usethe discussion board and the end of semester questionnaire.

If you have comments about your class tutor, please mentionthem by name.

We are interested to know your opinions about this course: usethe discussion board and the end of semester questionnaire.

If you have comments about your class tutor, please mentionthem by name.

We are interested to know your opinions about this course: usethe discussion board and the end of semester questionnaire.

If you have comments about your class tutor, please mentionthem by name.

Reading week

Week 7 (November 11–15) is a reading week.

There will be no classes or new videos released in readingweek: the videos released late in week 6 will be due in early inweek 8.

You should use the time to revise or catch up with thematerial so far, e.g by working on exercises.

Week 7 (November 11–15) is a reading week.

There will be no classes or new videos released in readingweek: the videos released late in week 6 will be due in early inweek 8.

You should use the time to revise or catch up with thematerial so far, e.g by working on exercises.

Week 7 (November 11–15) is a reading week.

There will be no classes or new videos released in readingweek: the videos released late in week 6 will be due in early inweek 8.

You should use the time to revise or catch up with thematerial so far, e.g by working on exercises.

Matrices

Later in this course (Semester 2) we will spend a good amountof time studying matrices.

However, they are so fundamentalto engineering mathematics that they may have alreadyappeared elsewhere in your course or could come up before weget to them. To help you to get comfortable in their use, wewill cover some of the basics today.

Later in this course (Semester 2) we will spend a good amountof time studying matrices. However, they are so fundamentalto engineering mathematics that they may have alreadyappeared elsewhere in your course or could come up before weget to them.

To help you to get comfortable in their use, wewill cover some of the basics today.

Later in this course (Semester 2) we will spend a good amountof time studying matrices. However, they are so fundamentalto engineering mathematics that they may have alreadyappeared elsewhere in your course or could come up before weget to them. To help you to get comfortable in their use, wewill cover some of the basics today.

Why matrices?

Matrices to model systems

A certain city consists of an urban area and suburbs.

Eachyear 5% of those living in the urban area move to the suburbsand 3% of those living in the suburbs move to the urban area.If there are initially 600, 000 people in the urban area and400, 000 in the suburbs, how many are in each 25 years later?

Matrices to model systems

A certain city consists of an urban area and suburbs. Eachyear 5% of those living in the urban area move to the suburbs

and 3% of those living in the suburbs move to the urban area.If there are initially 600, 000 people in the urban area and400, 000 in the suburbs, how many are in each 25 years later?

Matrices to model systems

A certain city consists of an urban area and suburbs. Eachyear 5% of those living in the urban area move to the suburbsand 3% of those living in the suburbs move to the urban area.

If there are initially 600, 000 people in the urban area and400, 000 in the suburbs, how many are in each 25 years later?

Matrices to model systems

A certain city consists of an urban area and suburbs. Eachyear 5% of those living in the urban area move to the suburbsand 3% of those living in the suburbs move to the urban area.If there are initially 600, 000 people in the urban area and400, 000 in the suburbs, how many are in each 25 years later?

Definitions

Let m and n be positive integers.

Then an m× n matrix A isan array of real numbers, with m rows and n columns; that is

A =

a11 a12 . . . a1na21 a22 . . . a2n

......

. . ....

am1 am2 . . . amn

We sometimes write A = (aij) for the above matrix.

Let m and n be positive integers. Then an m× n matrix A isan array of real numbers, with m rows and n columns;

that is

A =

a11 a12 . . . a1na21 a22 . . . a2n

......

. . ....

am1 am2 . . . amn

We sometimes write A = (aij) for the above matrix.

Let m and n be positive integers. Then an m× n matrix A isan array of real numbers, with m rows and n columns; that is

A =

a11 a12 . . . a1na21 a22 . . . a2n

......

. . ....

am1 am2 . . . amn

We sometimes write A = (aij) for the above matrix.

Let m and n be positive integers. Then an m× n matrix A isan array of real numbers, with m rows and n columns; that is

A =

a11 a12 . . . a1na21 a22 . . . a2n

......

. . ....

am1 am2 . . . amn

We sometimes write A = (aij) for the above matrix.

Let m and n be positive integers. Then an m× n matrix A isan array of real numbers, with m rows and n columns; that is

A =

a11 a12 . . . a1na21 a22 . . . a2n

......

. . ....

am1 am2 . . . amn

We sometimes write A = (aij) for the above matrix.

For example,

(3 1 20 2 1

)is a 2× 3 matrix, and

2√2 7

0 1

3 4−√2

7 7

is a 4× 2 matrix.

For example,(3 1 20 2 1

)

is a 2× 3 matrix, and

2√2 7

0 1

3 4−√2

7 7

is a 4× 2 matrix.

For example,(3 1 20 2 1

)is a 2× 3 matrix,

and

2√2 7

0 1

3 4−√2

7 7

is a 4× 2 matrix.

For example,(3 1 20 2 1

)is a 2× 3 matrix, and

2√2 7

0 1

3 4−√2

7 7

is a 4× 2 matrix.

For example,(3 1 20 2 1

)is a 2× 3 matrix, and

2√2 7

0 1

3 4−√2

7 7

is a 4× 2 matrix.

For example,(3 1 20 2 1

)is a 2× 3 matrix, and

2√2 7

0 1

3 4−√2

7 7

is a 4× 2 matrix.

Identity matrices

Let n be a positive integer.

Then the n× n matrix In given by

In =

1 0 . . . 00 1 . . . 0...

.... . .

...0 0 . . . 1

is called the identity matrix of size n.

Identity matrices

Let n be a positive integer. Then the n× n matrix In given by

In =

1 0 . . . 00 1 . . . 0...

.... . .

...0 0 . . . 1

is called the identity matrix of size n.

Identity matrices

Let n be a positive integer. Then the n× n matrix In given by

In =

1 0 . . . 00 1 . . . 0...

.... . .

...0 0 . . . 1

is called the identity matrix of size n.

Identity matrices

Let n be a positive integer. Then the n× n matrix In given by

In =

1 0 . . . 00 1 . . . 0...

.... . .

...0 0 . . . 1

is called the identity matrix of size n.

For example,

I2 =

(1 00 1

)and I3 =

1 0 00 1 00 0 1

.

The identity matrix In is always square. That is, it has thesame number of rows and columns.

For example,

I2 =

(1 00 1

)

and I3 =

1 0 00 1 00 0 1

.

The identity matrix In is always square. That is, it has thesame number of rows and columns.

For example,

I2 =

(1 00 1

)and I3 =

1 0 00 1 00 0 1

.

The identity matrix In is always square. That is, it has thesame number of rows and columns.

For example,

I2 =

(1 00 1

)and I3 =

1 0 00 1 00 0 1

.

The identity matrix In is always square.

That is, it has thesame number of rows and columns.

For example,

I2 =

(1 00 1

)and I3 =

1 0 00 1 00 0 1

.

The identity matrix In is always square. That is, it has thesame number of rows and columns.

Matrix operations

To use matrices we have to define some basic mathematicaloperations.

Two basic operations are matrix addition andmatrix multiplication.

To use matrices we have to define some basic mathematicaloperations. Two basic operations are matrix addition

andmatrix multiplication.

To use matrices we have to define some basic mathematicaloperations. Two basic operations are matrix addition andmatrix multiplication.

Matrix addition

Let A = (aij) and B = (bij) both be m× n matrices.

Thenwe define the sum of A and B by

A+B =

a11 + b11 a12 + b12 . . . a1n + b1na21 + b21 a22 + b22 . . . a2n + b2n

......

. . ....

am1 + bm1 am2 + bm2 . . . amn + bmn

.

In other words, to add two matrices of the same dimensionssimply add their entries componentwise.

Matrix addition

Let A = (aij) and B = (bij) both be m× n matrices. Thenwe define the sum of A and B by

A+B =

a11 + b11 a12 + b12 . . . a1n + b1na21 + b21 a22 + b22 . . . a2n + b2n

......

. . ....

am1 + bm1 am2 + bm2 . . . amn + bmn

.

In other words, to add two matrices of the same dimensionssimply add their entries componentwise.

Matrix addition

Let A = (aij) and B = (bij) both be m× n matrices. Thenwe define the sum of A and B by

A+B =

a11 + b11 a12 + b12 . . . a1n + b1na21 + b21 a22 + b22 . . . a2n + b2n

......

. . ....

am1 + bm1 am2 + bm2 . . . amn + bmn

.

In other words, to add two matrices of the same dimensionssimply add their entries componentwise.

Matrix addition

Let A = (aij) and B = (bij) both be m× n matrices. Thenwe define the sum of A and B by

A+B =

a11 + b11 a12 + b12 . . . a1n + b1na21 + b21 a22 + b22 . . . a2n + b2n

......

. . ....

am1 + bm1 am2 + bm2 . . . amn + bmn

.

In other words, to add two matrices of the same dimensions

simply add their entries componentwise.

Matrix addition

Let A = (aij) and B = (bij) both be m× n matrices. Thenwe define the sum of A and B by

A+B =

a11 + b11 a12 + b12 . . . a1n + b1na21 + b21 a22 + b22 . . . a2n + b2n

......

. . ....

am1 + bm1 am2 + bm2 . . . amn + bmn

.

In other words, to add two matrices of the same dimensionssimply add their entries componentwise.

For example,(1 0 00 1 0

)+

(2 0 34 2 0

)=

(3 0 34 3 0

).

For example,(1 0 00 1 0

)+

(2 0 34 2 0

)=

(3 0 34 3 0

).

Warning!

It is not possible to add two matrices if their dimensions aredifferent, so take care!

Matrix multiplication

A big reason why matrices are so useful comes down to therule for how they multiply.

Matrices can only be multiplied when the dimensions matchup in the right way.

The thing to remember is that thenumber of columns of the first matrix must be the same as thenumber of rows of the second one.

That is, if A is p× q and B is q × r, then we can find theirproduct. The result, AB, is a p× r matrix.

A big reason why matrices are so useful comes down to therule for how they multiply.

Matrices can only be multiplied when the dimensions matchup in the right way. The thing to remember is that thenumber of columns of the first matrix must be the same as thenumber of rows of the second one.

That is, if A is p× q and B is q × r, then we can find theirproduct. The result, AB, is a p× r matrix.

A big reason why matrices are so useful comes down to therule for how they multiply.

Matrices can only be multiplied when the dimensions matchup in the right way. The thing to remember is that thenumber of columns of the first matrix must be the same as thenumber of rows of the second one.

That is, if A is p× q and B is q × r,

then we can find theirproduct. The result, AB, is a p× r matrix.

A big reason why matrices are so useful comes down to therule for how they multiply.

Matrices can only be multiplied when the dimensions matchup in the right way. The thing to remember is that thenumber of columns of the first matrix must be the same as thenumber of rows of the second one.

That is, if A is p× q and B is q × r, then we can find theirproduct.

The result, AB, is a p× r matrix.

A big reason why matrices are so useful comes down to therule for how they multiply.

Matrices can only be multiplied when the dimensions matchup in the right way. The thing to remember is that thenumber of columns of the first matrix must be the same as thenumber of rows of the second one.

That is, if A is p× q and B is q × r, then we can find theirproduct. The result, AB, is a p× r matrix.

We illustrate the procedure with an example.

Let

A =

(1 2 30 1 1

)and B =

2 03 42 0

To find AB, we take each row from A (starting from the top)and ‘multiply it’ by each column from B (starting from theleft) in the following way:

AB =

(1.2 + 2.3 + 3.2 1.0 + 2.4 + 3.00.2 + 1.3 + 1.2 0.0 + 1.4 + 1.0

)=

(14 85 4

).

We illustrate the procedure with an example. Let

A =

(1 2 30 1 1

)and B =

2 03 42 0

To find AB, we take each row from A (starting from the top)and ‘multiply it’ by each column from B (starting from theleft) in the following way:

AB =

(1.2 + 2.3 + 3.2 1.0 + 2.4 + 3.00.2 + 1.3 + 1.2 0.0 + 1.4 + 1.0

)=

(14 85 4

).

We illustrate the procedure with an example. Let

A =

(1 2 30 1 1

)and B =

2 03 42 0

To find AB, we take each row from A

(starting from the top)and ‘multiply it’ by each column from B (starting from theleft) in the following way:

AB =

(1.2 + 2.3 + 3.2 1.0 + 2.4 + 3.00.2 + 1.3 + 1.2 0.0 + 1.4 + 1.0

)=

(14 85 4

).

We illustrate the procedure with an example. Let

A =

(1 2 30 1 1

)and B =

2 03 42 0

To find AB, we take each row from A (starting from the top)

and ‘multiply it’ by each column from B (starting from theleft) in the following way:

AB =

(1.2 + 2.3 + 3.2 1.0 + 2.4 + 3.00.2 + 1.3 + 1.2 0.0 + 1.4 + 1.0

)=

(14 85 4

).

We illustrate the procedure with an example. Let

A =

(1 2 30 1 1

)and B =

2 03 42 0

To find AB, we take each row from A (starting from the top)and ‘multiply it’ by each column from B

(starting from theleft) in the following way:

AB =

(1.2 + 2.3 + 3.2 1.0 + 2.4 + 3.00.2 + 1.3 + 1.2 0.0 + 1.4 + 1.0

)=

(14 85 4

).

We illustrate the procedure with an example. Let

A =

(1 2 30 1 1

)and B =

2 03 42 0

To find AB, we take each row from A (starting from the top)and ‘multiply it’ by each column from B (starting from theleft)

in the following way:

AB =

(1.2 + 2.3 + 3.2 1.0 + 2.4 + 3.00.2 + 1.3 + 1.2 0.0 + 1.4 + 1.0

)=

(14 85 4

).

We illustrate the procedure with an example. Let

A =

(1 2 30 1 1

)and B =

2 03 42 0

To find AB, we take each row from A (starting from the top)and ‘multiply it’ by each column from B (starting from theleft) in the following way:

AB =

(1.2 + 2.3 + 3.2 1.0 + 2.4 + 3.00.2 + 1.3 + 1.2 0.0 + 1.4 + 1.0

)=

(14 85 4

).

We illustrate the procedure with an example. Let

A =

(1 2 30 1 1

)and B =

2 03 42 0

To find AB, we take each row from A (starting from the top)and ‘multiply it’ by each column from B (starting from theleft) in the following way:

AB =

(1.2 + 2.3 + 3.2 1.0 + 2.4 + 3.00.2 + 1.3 + 1.2 0.0 + 1.4 + 1.0

)=

(14 85 4

).

We illustrate the procedure with an example. Let

A =

(1 2 30 1 1

)and B =

2 03 42 0

To find AB, we take each row from A (starting from the top)and ‘multiply it’ by each column from B (starting from theleft) in the following way:

AB =

(1.2 + 2.3 + 3.2

1.0 + 2.4 + 3.00.2 + 1.3 + 1.2 0.0 + 1.4 + 1.0

)=

(14 85 4

).

We illustrate the procedure with an example. Let

A =

(1 2 30 1 1

)and B =

2 03 42 0

To find AB, we take each row from A (starting from the top)and ‘multiply it’ by each column from B (starting from theleft) in the following way:

AB =

(1.2 + 2.3 + 3.2 1.0 + 2.4 + 3.0

0.2 + 1.3 + 1.2 0.0 + 1.4 + 1.0

)=

(14 85 4

).

We illustrate the procedure with an example. Let

A =

(1 2 30 1 1

)and B =

2 03 42 0

To find AB, we take each row from A (starting from the top)and ‘multiply it’ by each column from B (starting from theleft) in the following way:

AB =

(1.2 + 2.3 + 3.2 1.0 + 2.4 + 3.00.2 + 1.3 + 1.2

0.0 + 1.4 + 1.0

)=

(14 85 4

).

We illustrate the procedure with an example. Let

A =

(1 2 30 1 1

)and B =

2 03 42 0

To find AB, we take each row from A (starting from the top)and ‘multiply it’ by each column from B (starting from theleft) in the following way:

AB =

(1.2 + 2.3 + 3.2 1.0 + 2.4 + 3.00.2 + 1.3 + 1.2 0.0 + 1.4 + 1.0

)=

(14 85 4

).

We illustrate the procedure with an example. Let

A =

(1 2 30 1 1

)and B =

2 03 42 0

To find AB, we take each row from A (starting from the top)and ‘multiply it’ by each column from B (starting from theleft) in the following way:

AB =

(1.2 + 2.3 + 3.2 1.0 + 2.4 + 3.00.2 + 1.3 + 1.2 0.0 + 1.4 + 1.0

)=

(14 85 4

).

In the previous example, A is 2× 3 and B is 3× 2

and theresult is 2× 2. Of course, BA will not be the same matrix, asthe result will be 3× 3.

In the previous example, A is 2× 3 and B is 3× 2 and theresult is 2× 2.

Of course, BA will not be the same matrix, asthe result will be 3× 3.

In the previous example, A is 2× 3 and B is 3× 2 and theresult is 2× 2. Of course, BA will not be the same matrix,

asthe result will be 3× 3.

In the previous example, A is 2× 3 and B is 3× 2 and theresult is 2× 2. Of course, BA will not be the same matrix, asthe result will be 3× 3.

Column vectors

One case that occurs frequently is when the second matrix is acolumn vector (i.e. an n× 1 matrix) of a suitable length.

Forexample,

(1 3 12 0 −1

) 322

=

(1.3 + 3.2 + 1.2

2.3 + 0.2 + (−1).2

)=

(114

).

The result will always be a column vector, although not ingeneral with the same length.

Column vectors

One case that occurs frequently is when the second matrix is acolumn vector (i.e. an n× 1 matrix) of a suitable length. Forexample,

(1 3 12 0 −1

) 322

=

(1.3 + 3.2 + 1.2

2.3 + 0.2 + (−1).2

)=

(114

).

The result will always be a column vector, although not ingeneral with the same length.

Column vectors

One case that occurs frequently is when the second matrix is acolumn vector (i.e. an n× 1 matrix) of a suitable length. Forexample,

(1 3 12 0 −1

) 322

=

(1.3 + 3.2 + 1.2

2.3 + 0.2 + (−1).2

)=

(114

).

The result will always be a column vector, although not ingeneral with the same length.

Column vectors

One case that occurs frequently is when the second matrix is acolumn vector (i.e. an n× 1 matrix) of a suitable length. Forexample,

(1 3 12 0 −1

) 322

=

(1.3 + 3.2 + 1.2

2.3 + 0.2 + (−1).2

)=

(114

).

The result will always be a column vector, although not ingeneral with the same length.

Column vectors

One case that occurs frequently is when the second matrix is acolumn vector (i.e. an n× 1 matrix) of a suitable length. Forexample,

(1 3 12 0 −1

) 322

=

(1.3 + 3.2 + 1.2

2.3 + 0.2 + (−1).2

)=

(114

).

The result will always be a column vector, although not ingeneral with the same length.

Column vectors

One case that occurs frequently is when the second matrix is acolumn vector (i.e. an n× 1 matrix) of a suitable length. Forexample,

(1 3 12 0 −1

) 322

=

(1.3 + 3.2 + 1.2

2.3 + 0.2 + (−1).2

)=

(114

).

The result will always be a column vector,

although not ingeneral with the same length.

Column vectors

One case that occurs frequently is when the second matrix is acolumn vector (i.e. an n× 1 matrix) of a suitable length. Forexample,

(1 3 12 0 −1

) 322

=

(1.3 + 3.2 + 1.2

2.3 + 0.2 + (−1).2

)=

(114

).

The result will always be a column vector, although not ingeneral with the same length.

Another thing to notice is that multiplication by the identitymatrix

(of the correct size) will leave the other matrixunchanged. For example,(

1 00 1

)(1 3 12 0 −1

)=

(1.1 + 0.2 1.3 + 0.0 1.1 + 0.(−1)0.1 + 1.2 0.3 + 1.0 0.1 + 1.(−1)

)=

(1 3 12 0 −1

).

Another thing to notice is that multiplication by the identitymatrix (of the correct size)

will leave the other matrixunchanged. For example,(

1 00 1

)(1 3 12 0 −1

)=

(1.1 + 0.2 1.3 + 0.0 1.1 + 0.(−1)0.1 + 1.2 0.3 + 1.0 0.1 + 1.(−1)

)=

(1 3 12 0 −1

).

Another thing to notice is that multiplication by the identitymatrix (of the correct size) will leave the other matrixunchanged.

For example,(1 00 1

)(1 3 12 0 −1

)=

(1.1 + 0.2 1.3 + 0.0 1.1 + 0.(−1)0.1 + 1.2 0.3 + 1.0 0.1 + 1.(−1)

)=

(1 3 12 0 −1

).

Another thing to notice is that multiplication by the identitymatrix (of the correct size) will leave the other matrixunchanged. For example,

(1 00 1

)(1 3 12 0 −1

)=

(1.1 + 0.2 1.3 + 0.0 1.1 + 0.(−1)0.1 + 1.2 0.3 + 1.0 0.1 + 1.(−1)

)=

(1 3 12 0 −1

).

Another thing to notice is that multiplication by the identitymatrix (of the correct size) will leave the other matrixunchanged. For example,(

1 00 1

)(1 3 12 0 −1

)=

(1.1 + 0.2 1.3 + 0.0 1.1 + 0.(−1)0.1 + 1.2 0.3 + 1.0 0.1 + 1.(−1)

)

=

(1 3 12 0 −1

).

Another thing to notice is that multiplication by the identitymatrix (of the correct size) will leave the other matrixunchanged. For example,(

1 00 1

)(1 3 12 0 −1

)=

(1.1 + 0.2 1.3 + 0.0 1.1 + 0.(−1)0.1 + 1.2 0.3 + 1.0 0.1 + 1.(−1)

)=

(1 3 12 0 −1

).

Activity. Working in groups of two or three, find a matrix Asuch that

A

(xurban

xsuburban

)=

(0.95xurban + 0.03xsuburban

0.05xurban + 0.97xsuburban

).

A =

(0.95 0.030.05 0.97

).

In the example at the beginning of the lecture,

A

(600, 000400, 000

)will give the number of people in the urban and suburbanareas after one year. Multiplying by A repeatedly means thepopulations after 25 years will be given by

A25

(600, 000400, 000

).

A =

(0.95 0.030.05 0.97

).

In the example at the beginning of the lecture,

A

(600, 000400, 000

)will give the number of people in the urban and suburbanareas after one year.

Multiplying by A repeatedly means thepopulations after 25 years will be given by

A25

(600, 000400, 000

).

A =

(0.95 0.030.05 0.97

).

In the example at the beginning of the lecture,

A

(600, 000400, 000

)will give the number of people in the urban and suburbanareas after one year. Multiplying by A repeatedly means thepopulations after 25 years will be given by

A25

(600, 000400, 000

).

And finally. . .

Reminders:

• email address mas-engineering@sheffield.ac.uk

• website http://engmaths.group.shef.ac.uk/mas151

(also accessible through MOLE).