section 2.1 operations with matrices

Section 2.1Operations with Matrices

Notation:

If A is a matrix with m rows and n columns, then we say A is an mxn matrix and we can write: A = [ aij ]

Terminology:

Column matrix (column vector) – A matrix with just one column.

Row matrix (row vector) – A matrix with just one row.

21 3 4 6 9

2100

1164

77

9 4 8

Equality of MatricesTwo matrices A and B are said to be equal if aij = bij for all possible i and j.

Addition/Subtraction of Matrices, Scalar Multiplication

Ex. Let

(a) 2A =

(b) A + B =

(c) A – 2B =

1 213 5

A

10 17 4

B

Addition/Subtraction of Matrices, Scalar Multiplication

Ex.4 0 50

1 3 15 101 2 3

5 6 7 26 2 12

Let

You might think that AB =

Actually, AB =

1 3 5 3, .

2 2 1 0A B

5 9,

2 0

2 38 6

Matrix Multiplication

Dot Products

= a1b1 + a2b2 + a3b3 + . . . + anbn

1

2

1 2 3 3n

n

bb

a a a a b

b


Dot Products

= a1b1 + a2b2 + a3b3 + . . . + anbn

Note: In a dot product the row matrix must come first and the column matrix must come second.

1

2

1 2 3 3n

n

bb

a a a a b

b


Examples:

=

=

21

3 1 4 257

1

2 1 5 23

(3)(2) + (1)(1) + (–4)(5) + (2)(7) = 1

(2)(1) + (1)(2) + (5)(3) = 19



Some dot products don't make sense.

= 2

3 1 4 2 15

(3)(2) + (1)(1) + (–4)(5) + (2)( )

To compute AB, we must multiply each row in A with each column in B.

If A is an mxn matrix and B is an nxp matrix, then C = AB is an mxp matrix and

cij = ai1b1j + ai2b2j + ai3b3j + . . . + ainbnj

1

n

ik kjk

a b



5 41 6

1 32 7

2 223 50

A B


1 2 51 1 10 3 1

2 3 41 2 11 1 0

9 –62 4

A B63

4 –7 3

55 66 77 88 991 3 4 7 3 2 1 6 83 0 11 9 7 10 5 4 9

21 12 90 7 4

Ex. Determine if the following matrix multiplications can be done. If they can, give the dimensions of the resulting matrix. But do not actually perform the multiplication.

(a)



(b)

1 344 5

1 3 4 78 7

3 0 11 910 4

1 0



(c) 1 344 5

1 3 4 78 7

3 0 11 910 4

1 0



(d) 8 8 4

1 25 1 113

3 49 2 4



(e)

1 344 5

3 0 11 98 7

10 4


Ex. Use a calculator to perform the following matrix multiplication.

3 11 12 10 21 4 5 1 7

17 2 20 5 9


Ex. Let and . Compute AB and then compute BA.


1 23 4

A

1 23 4

B

A system of many linear equations can be handled by one matrix equation: AX = B.

3x – y – z = 5 4x + 2y + 6z = –2 → 5x + 3y – 8z = 7


3 1 1 54 2 6 25 3 8 7

xyz

Solve AX = B for X.


Ex. Solve I + X = A for X, where I = and A = 1 00 1

3 710 9

A system of linear equations can be written as a linear combination of column vectors.

Ex. Write the following system as a linear combination of column vectors:3x – 2y + z = 3–x + 3y + 2z = 4 x – y = 3

Section 2.2Properties of Matrix Operations

Def. The transpose of an mxn matrix A, denoted AT, is an nxm matrix defined by the following: If A = [ aij ] then AT = [ aji ].

Ex. Compute the transpose of the following matrices.

1 2 3 45 6 7 8

A

3 820 5

B

1 3 53 7 85 8 6

C

Def. A matrix A is symmetric if A = AT.

Recall that for matrices A and B, it is not always true that AB = BA.

Another property which is not always true for matrices (but we’d like to be true):If AC = BC then A = B.

Ex. Let , , and .

Compute AC and BC. Verify that AC = BC, but A ≠ B.

1 35 9

A

1 413 5

B

1 22 4

C

Properties

Let A, B, and C be matrices and let s and t be scalars.

1. A + B = B + A

2. A + (B + C) = (A + B) + C

3. s(tA) = (st)A

4. s(A + B) = sA + sB

5. (s + t)A = sA + tA

6. A(BC) = (AB)C

7. A(B + C) = AB + AC

8. (A + B)C = AC + BC

The zero matrix - O

The identity matrix - I

Properties

Let A, B, and C be matrices and let s and t be scalars.

9. A + O = A

10. A + (-A) = O

11. If sA = O then either s = 0 or A = O

12. AI = A

13. IA = A

Ex. Verify number 12 and 13 above for 1 2 34 5 67 8 9

A

Another property which is not always true for matrices:

If AB = O (the zero matrix) then A = O or B = O.

Ex. Let and

Verify that AB = O, but A ≠ O and B ≠ O

1 22 4

A

2 61 3

B

Powers of a (square) matrix.

Ak = A·A·A·····A (k factors of A multiplied together).

Ex. Let

A2 =

A3 =

A39 =

1 23 4

A

Ex. Let

A0 = (guess)

A-1 = (guess)

1 23 4

A

Section 2.3The Inverse of a Matrix

For scalars, a-1 = 1/a .

The actual definition of a-1 is:

If a ≠ 0 then there is a unique number denoted a-1 so that a · a-1 = 1 and a-1 · a = 1.

For scalars, a-1 = 1/a .

The actual definition of a-1 is:

If a ≠ 0 then there is a unique number denoted a-1 so that a · a-1 = 1 and a-1 · a = 1.

If A is a square matrix with ____ , then there is a unique matrix denoted A-1 so thatA · A-1 = I and A-1 · A = I.

Terminology

If A-1 exists, then we say that A is invertible (non-singular).

If no such A-1 exists then we say that A is singular (non-invertible).

Ex. Let and . Show that B = A-1.1 21 1

A

1 21 1

B

Solve each of the following systems of equations, then do so all at once.

2x – 4y = –6 2x – 4y = 10 3x + y = 5 3x + y = 8

2 4 63 1 5

2 4 103 1 8

1 2 33 1 5

1 2 53 1 8

1 2 30 7 14

1 2 50 7 7

1 2 30 1 2

1 2 50 1 1

1 0 10 1 2

1 0 30 1 1

R2 –3R1→R2

1/2 R1→R1 1/2 R1→R1

R2 –3R1→R2

1/7 R2→R2 1/7 R2→R2

R1 +2R2→R1 R1 +2R2→R1

(1, 2) (3, 1)

Solving both at once: Solving each separately:

2 4 6 103 1 5 8

1 2 3 53 1 5 8

1 2 3 50 7 14 7

1 2 3 50 1 2 1

1 0 1 30 1 2 1

R2 –3R1→R2

1/2 R1→R1

1/7 R2→R2

R1 +2R2→R1

(1, 2) and (3, 1)

2 4 63 1 5

1 2 33 1 5

1 2 30 7 14

1 2 30 1 2

1 0 10 1 2

R2 –3R1→R2

1/2 R1→R1

1/7 R2→R2

R1 +2R2→R1

(1, 2)

2 4 103 1 8

1 2 53 1 8

1 2 50 7 7

1 2 50 1 1

1 0 30 1 1

(3, 1)

Now back to computing A-1.

Let's compute A-1 if we have

Let Our job is to determine the values of a, b, c, d.

Now,

Multiply the two matrices on the left to get:

Since the two matrices are equal, the corresponding components of these two matrices must be equal. This gives us four equations in four unknowns:

2a + c = 13a + 2c = 02b + d = 03b + 2d = 1

We can separate this into two sets of two equations with two unknowns:

2 13 2

A

1 .a b

Ac d

2 1 1 03 2 0 1

a bc d

2 2 1 03 2 3 2 0 1

a c b da c b d

2a + c = 1 2b + d = 03a + 2c = 0 3b + 2d = 1

Let's compute A-1 if we have

2a + c = 1 2b + d = 03a + 2c = 0 3b + 2d = 1

We could solve these two systems separately and determine a, b, c, d that way. But since the coefficient matrix of each is the same

(each system has a coefficient matrix of ) we can solve them

both at the same time with one augmented matrix: This reduces to

Therefore a = 2, b = – 1, c = – 3, d = 2.

And finally,

2 13 2

A

2 13 2

2 1 1 03 2 0 1

1 0 2 1

0 1 3 2

1 2 13 2

a bA

c d

To generalize what we have just done:

To find A-1, form the augmented matrix [ A | I ] and place it into reduced row echelon form to get [ I | A-1 ].

Ex. Use an augmented matrix to find the inverse matrix for each of the following:

1 1 01 0 16 2 3

B

1 2 03 1 22 3 2

C

Properties:

1. (A-1)-1 = A

2. (AB)-1 = B-1A-1

Ex. Show that (AB)-1 = B-1A-1.

More properties.

(Recall that it is not always true that if AC = BC then A = B.)If C is invertible then whenever AC = BC it is always true that A = B.

Ex. Show that if C is invertible then AC = BC implies that A = B.

Recall that a system of linear equations can be written as AX = B. We can now solve this matrix equation.

Ex. Solve the following system by first writing it as a matrix equation, then use an inverse matrix to solve this equation.2x + 3y + z = -13x + 3y + z = 12x + 4y + z = -2

Ex. Solve the following system by first writing it as a matrix equation, then use an inverse matrix to solve this equation. x + y – 5z = 3 x – 2z = 12x – y – z = 0

Section 2.5Applications of Matrix Operations

Cryptography – encoding and decoding messages.

0=_ 5=E 10=J 15=O 20=T 25=Y1=A 6=F 11=K 16=P 21=U 26=Z2=B 7=G 12=L 17=Q 22=V3=C 8=H 13=M 18=R 23=W4=D 9=I 14=N 19=S 24=X

Ex. Use matrices of size 1x3 to write an encoded message for “Meet me Monday”.

Ex. Use matrix multiplication with the following matrix to encode “Meet me Monday”.

1 2 21 1 3

1 1 4A

We decipher a coded message by applying an inverse matrix to the row matrices containing the coded message.

Ex. The encrypted message 89 -35 -36 -15 7 8 106 -27 -59 -30 5 25was encoded using the matrix . Decipher the message. 1 1 0

1 0 16 2 3

Ex. The following encrypted message was encoded with a 2x2 matrix.67 30 63 7 55 77 100 53 14 -21 109 141 48 -14 50 -75 25 -23 47 2We know the last four characters of the original text are JAKE. What is the original text?

section 2.1 operations with matrices

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