math 1300: section 4-5 inverse of a square matrix

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Identity Matrix for MultiplicationInverse of a Square MatrixApplication: Cryptography

Math 1300 Finite MathematicsSection 4.5 Inverse of a Square Matrix

Jason Aubrey

Department of MathematicsUniversity of Missouri

Jason Aubrey Math 1300 Finite Mathematics

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Definition (Identity Matrix for Multiplication)An n × n matrix with the properties that

every element on the principal diagonal is a 1, andevery other element is 0

is called the n × n identity matrix.


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every element on the principal diagonal is a 1, and

every other element is 0is called the n × n identity matrix.


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every element on the principal diagonal is a 1, andevery other element is 0

is called the n × n identity matrix.


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For example,

I2 =

[1 00 1

]is the 2× 2 identity matrix.

I3 =

1 0 00 1 00 0 1

is the 3× 3 identity matrix.The reason In is called ’the n× n identity matrix’ is because

AIn = AInB = B

whenever those products are defined.


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For example,

I2 =

[1 00 1


I3 =

1 0 00 1 00 0 1

is the 3× 3 identity matrix.

The reason In is called ’the n× n identity matrix’ is because

AIn = AInB = B



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For example,

I2 =

[1 00 1


I3 =

1 0 00 1 00 0 1

is the 3× 3 identity matrix.The reason In is called ’the n× n identity matrix’ is because

AIn = AInB = B



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Example:[2 −11 3

] [1 00 1

]

=

[2(1)− 1(0) 2(0)− 1(1)1(1) + 3(0) 1(0) + 3(1)

]=

[2 − 11 3

][1 00 1

] [2 −11 3

]=

[2(1) + 1(0) 1(−1) + 3(0)0(2) + 1(1) 0(−1) + 1(3)

]=

[2 − 11 3

]


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Example:[2 −11 3

] [1 00 1

]=

[

2(1)− 1(0) 2(0)− 1(1)1(1) + 3(0) 1(0) + 3(1)

]=

[

2 − 11 3

]

[1 00 1

] [2 −11 3

]=

[2(1) + 1(0) 1(−1) + 3(0)0(2) + 1(1) 0(−1) + 1(3)

]=

[2 − 11 3

]


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Example:[2 −11 3

] [1 00 1

]=

[2(1)− 1(0)

2(0)− 1(1)1(1) + 3(0) 1(0) + 3(1)

]=

[

2 − 11 3

]

[1 00 1

] [2 −11 3

]=

[2(1) + 1(0) 1(−1) + 3(0)0(2) + 1(1) 0(−1) + 1(3)

]=

[2 − 11 3

]


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Example:[2 −11 3

] [1 00 1

]=

[2(1)− 1(0)

2(0)− 1(1)1(1) + 3(0) 1(0) + 3(1)

]=

[2

− 11 3

]

[1 00 1

] [2 −11 3

]=

[2(1) + 1(0) 1(−1) + 3(0)0(2) + 1(1) 0(−1) + 1(3)

]=

[2 − 11 3

]


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Example:[2 −11 3

] [1 00 1

]=

[2(1)− 1(0) 2(0)− 1(1)

1(1) + 3(0) 1(0) + 3(1)

]=

[2

− 11 3

]

[1 00 1

] [2 −11 3

]=

[2(1) + 1(0) 1(−1) + 3(0)0(2) + 1(1) 0(−1) + 1(3)

]=

[2 − 11 3

]


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Example:[2 −11 3

] [1 00 1

]=

[2(1)− 1(0) 2(0)− 1(1)

1(1) + 3(0) 1(0) + 3(1)

]=

[2 − 1

1 3

]

[1 00 1

] [2 −11 3

]=

[2(1) + 1(0) 1(−1) + 3(0)0(2) + 1(1) 0(−1) + 1(3)

]=

[2 − 11 3

]


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Example:[2 −11 3

] [1 00 1

]=

[2(1)− 1(0) 2(0)− 1(1)1(1) + 3(0)

1(0) + 3(1)

]=

[2 − 1

1 3

]

[1 00 1

] [2 −11 3

]=

[2(1) + 1(0) 1(−1) + 3(0)0(2) + 1(1) 0(−1) + 1(3)

]=

[2 − 11 3

]


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Example:[2 −11 3

] [1 00 1

]=

[2(1)− 1(0) 2(0)− 1(1)1(1) + 3(0)

1(0) + 3(1)

]=

[2 − 11

3

]

[1 00 1

] [2 −11 3

]=

[2(1) + 1(0) 1(−1) + 3(0)0(2) + 1(1) 0(−1) + 1(3)

]=

[2 − 11 3

]


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Example:[2 −11 3

] [1 00 1

]=

[2(1)− 1(0) 2(0)− 1(1)1(1) + 3(0) 1(0) + 3(1)

]=

[2 − 11

3

]

[1 00 1

] [2 −11 3

]=

[2(1) + 1(0) 1(−1) + 3(0)0(2) + 1(1) 0(−1) + 1(3)

]=

[2 − 11 3

]


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Example:[2 −11 3

] [1 00 1

]=

[2(1)− 1(0) 2(0)− 1(1)1(1) + 3(0) 1(0) + 3(1)

]=

[2 − 11 3

]

[1 00 1

] [2 −11 3

]=

[2(1) + 1(0) 1(−1) + 3(0)0(2) + 1(1) 0(−1) + 1(3)

]=

[2 − 11 3

]


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Example:[2 −11 3

] [1 00 1

]=

[2(1)− 1(0) 2(0)− 1(1)1(1) + 3(0) 1(0) + 3(1)

]=

[2 − 11 3

][1 00 1

] [2 −11 3

]=

[

2(1) + 1(0) 1(−1) + 3(0)0(2) + 1(1) 0(−1) + 1(3)

]=

[

2 − 11 3

]


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Example:[2 −11 3

] [1 00 1

]=

[2(1)− 1(0) 2(0)− 1(1)1(1) + 3(0) 1(0) + 3(1)

]=

[2 − 11 3

][1 00 1

] [2 −11 3

]=

[2(1) + 1(0)

1(−1) + 3(0)0(2) + 1(1) 0(−1) + 1(3)

]=

[

2 − 11 3

]


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Example:[2 −11 3

] [1 00 1

]=

[2(1)− 1(0) 2(0)− 1(1)1(1) + 3(0) 1(0) + 3(1)

]=

[2 − 11 3

][1 00 1

] [2 −11 3

]=

[2(1) + 1(0)

1(−1) + 3(0)0(2) + 1(1) 0(−1) + 1(3)

]=

[2

− 11 3

]


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Example:[2 −11 3

] [1 00 1

]=

[2(1)− 1(0) 2(0)− 1(1)1(1) + 3(0) 1(0) + 3(1)

]=

[2 − 11 3

][1 00 1

] [2 −11 3

]=

[2(1) + 1(0) 1(−1) + 3(0)

0(2) + 1(1) 0(−1) + 1(3)

]=

[2

− 11 3

]


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Example:[2 −11 3

] [1 00 1

]=

[2(1)− 1(0) 2(0)− 1(1)1(1) + 3(0) 1(0) + 3(1)

]=

[2 − 11 3

][1 00 1

] [2 −11 3

]=

[2(1) + 1(0) 1(−1) + 3(0)

0(2) + 1(1) 0(−1) + 1(3)

]=

[2 − 1

1 3

]


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Example:[2 −11 3

] [1 00 1

]=

[2(1)− 1(0) 2(0)− 1(1)1(1) + 3(0) 1(0) + 3(1)

]=

[2 − 11 3

][1 00 1

] [2 −11 3

]=

[2(1) + 1(0) 1(−1) + 3(0)0(2) + 1(1)

0(−1) + 1(3)

]=

[2 − 1

1 3

]


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Example:[2 −11 3

] [1 00 1

]=

[2(1)− 1(0) 2(0)− 1(1)1(1) + 3(0) 1(0) + 3(1)

]=

[2 − 11 3

][1 00 1

] [2 −11 3

]=

[2(1) + 1(0) 1(−1) + 3(0)0(2) + 1(1)

0(−1) + 1(3)

]=

[2 − 11

3

]


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Example:[2 −11 3

] [1 00 1

]=

[2(1)− 1(0) 2(0)− 1(1)1(1) + 3(0) 1(0) + 3(1)

]=

[2 − 11 3

][1 00 1

] [2 −11 3

]=

[2(1) + 1(0) 1(−1) + 3(0)0(2) + 1(1) 0(−1) + 1(3)

]=

[2 − 11

3

]


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Example:[2 −11 3

] [1 00 1

]=

[2(1)− 1(0) 2(0)− 1(1)1(1) + 3(0) 1(0) + 3(1)

]=

[2 − 11 3

][1 00 1

] [2 −11 3

]=

[2(1) + 1(0) 1(−1) + 3(0)0(2) + 1(1) 0(−1) + 1(3)

]=

[2 − 11 3

]


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Example:1 0 00 1 00 0 1

2 0 2−1 1 −31 0 3

=

1(2) + 0(−1) + 0(1) 1(0) + 0(1) + 0(0) 1(2) + 0(−3) + 0(3)0(2) + 1(−1) + 0(1) 0(0) + 1(1) + 0(0) 0(2) + 1(−3) + 0(3)0(2) + 0(−1) + 1(1) 0(0) + 0(1) + 1(0) 0(2) + 0(−3) + 1(3)

=

2 0 2−1 1 −31 0 3


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2 0 2−1 1 −31 0 3

1 0 00 1 00 0 1

=

2(1) + 0(0) + 2(0) 2(0) + 0(1) + 2(0) 2(0) + 0(0) + 2(1)−1(1) + 1(0)− 3(0) −1(0) + 1(1)− 3(0) −1(0) + 1(0)− 3(1)1(1) + 0(0) + 3(0) 1(0) + 0(1) + 3(0) 1(0) + 0(0) + 3(1)

=

2 0 2−1 1 −31 0 3


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DefinitionLet M be a square matrix of order n and I be the identity matrixof order n. If there exists a matrix M−1 (read "M inverse") suchthat

M−1M = MM−1 = I

then M−1 is called the multiplicative inverse of M or, moresimply, the inverse of M. If no such matrix exists, then M is saidto be a singular matrix.


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Example: The matrices[

3 −4−2 3

]and

[3 42 3

]are inverses of

each other because[3 −4−2 3

] [3 42 3

]=

[1 00 1

]and [

3 42 3

] [3 −4−2 3

]=

[1 00 1

]


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Example: Since[

2 2−1 −1

] [1 1−1 −1

]=

[0 00 0

]We conclude

that[

2 2−1 −1

]and

[1 1−1 −1

]are not inverses of each other.

(In fact, these matrices have no inverses).


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To find the inverse of a square matrix M,

1 Form the augmented matrix

[M |I ]

2 Use row operations to transform [M |I ] into [I |B ]

3 The matrix B is the inverse of M; in other words, M−1 = B


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To find the inverse of a square matrix M,1 Form the augmented matrix

[M |I ]

2 Use row operations to transform [M |I ] into [I |B ]

3 The matrix B is the inverse of M; in other words, M−1 = B


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Example: Let

M =

[2 −61 −2

]Find the inverse of M, if it exists.

M−1 =

[−1 3−1

2 1

]


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Example: Let

M =

[3 16 2

]Find the inverse of M, if it exists.

[3 16 2

∣∣∣∣1 00 1

]−2R1+R2→R2−−−−−−−−−→

[3 10 0

∣∣∣∣ 1 0−2 1

]Here we have a problem: the row with zeros on the leftindicates that our matrix M has no inverse.

During the process of finding the inverse of M, if a row resultswith all zeros on the left of the vertical bar (the M side), then Mhas no inverse. In this case, M is called a singular matrix.


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Example: Let

M =

[3 16 2

]Find the inverse of M, if it exists.[

3 16 2

∣∣∣∣1 00 1

]

−2R1+R2→R2−−−−−−−−−→[3 10 0

∣∣∣∣ 1 0−2 1




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Example: Let

M =

[3 16 2


3 16 2

∣∣∣∣1 00 1

]−2R1+R2→R2−−−−−−−−−→

[3 10 0

∣∣∣∣ 1 0−2 1




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Example: Let

M =

[3 16 2


3 16 2

∣∣∣∣1 00 1

]−2R1+R2→R2−−−−−−−−−→

[3 10 0

∣∣∣∣ 1 0−2 1

]

Here we have a problem: the row with zeros on the leftindicates that our matrix M has no inverse.



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Example: Let

M =

[3 16 2


3 16 2

∣∣∣∣1 00 1

]−2R1+R2→R2−−−−−−−−−→

[3 10 0

∣∣∣∣ 1 0−2 1




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Cryptography

Matrix inverses can provide a simple and effective procedurefor encoding and decoding messages.

To begin, assign the numbers 1-26 to the letters in thealphabet. Also assign the number 0 to a blank to provide forspace between words.


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Blank A B C D E F G H I J K L M0 1 2 3 4 5 6 7 8 9 10 11 12 13

N O P Q R S T U V W X Y Z14 15 16 17 18 19 20 21 22 23 24 25 26

For example, the sequence

19 5 3 18 5 20 0 3 15 4 5

corresponds to the (plaintext) message “SECRET CODE”.


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Any matrix whose elements are positive integers and whoseinverse exists can be used as an encoding matrix.

Forexample, to use the 2× 2 matrix

A =

[4 31 1

]to encode the preceeding message, first we divide the numbersin the sequence into groups of 2 and use these groups as thecolumns of a matrix B with 2 rows.

B =

[19 3 5 0 15 55 18 20 3 4 0

]


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Any matrix whose elements are positive integers and whoseinverse exists can be used as an encoding matrix. Forexample, to use the 2× 2 matrix

A =

[4 31 1

]to encode the preceeding message,

first we divide the numbersin the sequence into groups of 2 and use these groups as thecolumns of a matrix B with 2 rows.

B =

[19 3 5 0 15 55 18 20 3 4 0

]


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Any matrix whose elements are positive integers and whoseinverse exists can be used as an encoding matrix. Forexample, to use the 2× 2 matrix

A =

[4 31 1

]to encode the preceeding message, first we divide the numbersin the sequence into groups of 2 and use these groups as thecolumns of a matrix B with 2 rows.

B =

[19 3 5 0 15 55 18 20 3 4 0

]


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Then we multiply on the left by A:

AB =

[4 31 1

] [19 3 5 0 15 55 18 20 3 4 0

]

=

[91 66 80 9 72 2024 21 25 3 19 5

]Thus the coded message (the ciphertext) is

91 24 66 21 80 25 9 3 72 19 20 5

This message can be decoded by putting it back into matrixform and multiplying on the left by the decoding matrix A−1


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AB =

[4 31 1

] [19 3 5 0 15 55 18 20 3 4 0

]=

[91 66 80 9 72 2024 21 25 3 19 5

]

Thus the coded message (the ciphertext) is

91 24 66 21 80 25 9 3 72 19 20 5



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AB =

[4 31 1

] [19 3 5 0 15 55 18 20 3 4 0

]=

[91 66 80 9 72 2024 21 25 3 19 5

]Thus the coded message (the ciphertext) is

91 24 66 21 80 25 9 3 72 19 20 5



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We have

C =

[91 66 80 9 72 2024 21 25 3 19 5

]and A =

[4 31 1

]

A−1 =

[1 −3−1 4

]To decipher the ciphertext, we multiply:

A−1C =

[1 −3−1 4

] [91 66 80 9 72 2024 21 25 3 19 5

]

=

[19 3 5 0 15 55 18 20 3 4 0

]


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We have

C =

[91 66 80 9 72 2024 21 25 3 19 5

]and A =

[4 31 1

]

A−1 =

[1 −3−1 4

]

To decipher the ciphertext, we multiply:

A−1C =

[1 −3−1 4

] [91 66 80 9 72 2024 21 25 3 19 5

]

=

[19 3 5 0 15 55 18 20 3 4 0

]


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We have

C =

[91 66 80 9 72 2024 21 25 3 19 5

]and A =

[4 31 1

]

A−1 =

[1 −3−1 4

]To decipher the ciphertext, we multiply:

A−1C =

[1 −3−1 4

] [91 66 80 9 72 2024 21 25 3 19 5

]

=

[19 3 5 0 15 55 18 20 3 4 0

]


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N O P Q R S T U V W X Y Z14 15 16 17 18 19 20 21 22 23 24 25 26

P =

[19 3 5 0 15 55 18 20 3 4 0

]

This gives the sequence

19 5 3 18 5 20 0 3 15 4 5

And this corresponds to the plaintext message “SECRETCODE”

.


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N O P Q R S T U V W X Y Z14 15 16 17 18 19 20 21 22 23 24 25 26

P =

[19 3 5 0 15 55 18 20 3 4 0

]This gives the sequence

19 5 3 18 5 20 0 3 15 4 5

And this corresponds to the plaintext message “SECRETCODE”

.


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N O P Q R S T U V W X Y Z14 15 16 17 18 19 20 21 22 23 24 25 26

P =

[19 3 5 0 15 55 18 20 3 4 0

]This gives the sequence

19 5 3 18 5 20 0 3 15 4 5

And this corresponds to the plaintext message “SECRETCODE”.


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