linear algebra. session 2roquesol/math_304_fall_2018_session… · session 2. matrices. matrix...

Matrices. Matrix Algebra

Linear Algebra. Session 2

Dr. Marco A Roque Sol

09/04/2018

Dr. Marco A Roque Sol Linear Algebra. Session 2

Matrices, matrix algebra

Applications of systems of linear equationsProblem 2.1.

Find the point of intersection of the lines x − y = −2 and2x + 3y = 6 .SolutionThe intersection point is the solution of the linear system{

x − y = −22x + 3y = 6

Problem 2.2.

Find the point of intersection of the planes x − y = −2,2x − y − z = 3, and x + y + z = 6SolutionThe intersection point is the solution of the linear system

x − y = −22x − y − z = 3x + y + z = 6

Problem 2.3.

Find a quadratic polynomial p(x) such that p(1) = 4, p(2) = 3,and p(3) = 4

Solution

Suppose that p(x) = ax2 + bx + c , then

p(1) = a + b + c

p(2) = 4a + 2b + c

p(3) = 9a + 3b + c

The values for a, b, c are given by the solution of the linear systema + b + c = 4

4a + 2b + c = 39a + 3b + c = 4

Problem 2.4.

Electrical network . Determine the amount of current in eachbranch of the network.

Solution

To solve this problems, we will use three fundamental Laws comingfrom Physics, namely,

Kirchhofs law 1 ( Charge Conservation ):

At every node the sum of the incoming currents equals the sum ofthe outgoing currents.

Kirchhofs law 2 ( Energy Conservation ):

Around every loop the algebraic sum of all voltages is zero.

Ohm’s Law:

For every resistor the voltage drop E , the current i , and theresistance R satisfy E = iR

Thus, applying these three laws to the above circuit we have

Node A: i1 + i2 = i3

Node B: i3 = i1 + i2

Left loop: 10− 10i1 − 40i3 = 0

Right loop: 20− 20i2 − 40i3 = 0

In this way we have the system:

i3 − i1 − i2 = 0

10− 10i1 − 40i3 = 020− 20i2 − 40i3 = 0

Problem 2.5.

Trafic Flow . Determine the amount of traffic between each ofthe four intersectionsof of the following diagram

650 400

610 −→ A x1 −→ B 640 −→

←− 520 D ←− x3 C ←− 600

Solution

At each intersection, the incoming traffic has to match theoutgoing traffic.

Intersection A : x4 + 610 = x1 + 450

Intersection B : x1 + 400 = x2 + 640

Intersection C : x2 + 600 = x3

Intersection D : x3 = x4 + 520

Which is equivalent to the system:x4 − x1 + 160 = 0x1 − x2 − 240 = 0x2 − x3 + 600 = 0x3 − x4 − 520 = 0

Matrices

Let us start by solving an m × n system of linear equations

a11x1 + a12x2 + . . .+ a1nxn = b1a21x1 + a22x2 + . . .+ a2nxn = b2

......

am1x1 + am2x2 + . . .+ amnxn = bm

where aij are given coefficients, b′ms are given right-hand side, andx ′ns are the unknowns. In this way, we can introduce new arrays ofnumbers to study the linear system

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

x1x2...xm

b1b2...bm

In this way, we have the following

Definition

An m× n matrix A , is an array of complex numbers ( m-rows andn-columns ), denoted by

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

= (aij)m×n

In this context, an element in the i-row and j-column is of thematrix A denoted by aij .

An n-dimensional vector v, can be represented as a 1× n matrix(row vector) or as an n × 1matrix (column vector):

v =(x1, x2, x3, · · · , xn

)−→

(x1 x2 x3 · · · xn

v =(x1, x2, x3, · · · , xn

)−→

x1x2x3...xn

An m × n matrix A = (aij) can be regarded as a column ofn-dimensional row vectors or as a row of m-dimensional columnvectors:

, vi =(ai1 ai2 ai3 · · · ain

w1 w2 w3 · · · wn

), wj =

a1ja2ja3j...

Associated with any m × n matrix A, we have the following basicmatrices:

a) Transpose

Is the ( n ×m ) matrix, denoted by AT , and defined by

a11 a21 . . . am1

a12 a22 . . . am2...

a1n a2n . . . amn

=(aTij

= (aji )m×n

b) Complex Conjugate

Is the ( m × n ) matrix, denoted by A, and defined by

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

= (aij)m×n = (aij)m×n

c) Adjoint

Is the ( m × n ) matrix, denoted by A∗ = AT

, and defined by

A∗ =

a11 a21 . . . am1

a12 a22 . . . am2...

a1n a2n . . . amn

=(a∗ij)n×m = (aji )m×n

Basic Matrix Operations

Let A = (aij)m×n and B = (bij)m×n be two matrices, then wedefine

1) A = B ⇐⇒

aij = bij ; i = 1, 2, ...,m, j = 1, 2, ..., n

2) Addition

A± B = (aij ± bij)m×n

e) Scalar Multiplication

rA = (raij)m×n

In particular we have the case of

Vector algebraLet

a =(a1, a2, a3, · · · , an

b =(b1, b2, b3, · · · , bn

)be n-dimensional vectors, and r ∈ R

Vector sum

a + b =(a1 + b1, a2 + b2, a3 + b3, · · · , an + bn

)Scalar multiple

ra =(ra1, ra2, ra3, · · · , ran

)Dr. Marco A Roque Sol Linear Algebra. Session 2

Zero vector

0, 0, 0, · · · , 0)

Negative of a vector

−a =(−a1,−a2,−a3, · · · ,−an

)Vector difference

a− b =(a1 − b1, a2 − b2, a3 − b3, · · · , an − bn

Given n-dimensional vectors, {v1, v2, v3, · · · , vk} and scalars{r1, r2, r3, · · · , rk}, the expression

r1v1 + r2v2 + r3v3 + · · ·+ rkvk

is called a linear combination of vectors v1, v2, v3, · · · , vk.

Also, vector addition and scalar multiplication are called linearoperations

Definition. The dot product of n-dimensional vectors

x =(x1, x2, x3, · · · , xn

)and y =

(y1, y2, y3, · · · , yn

)is given by

x · y = x1y1 + x2y2 + · · ·+ xnyn

The dot product is also called the scalar product .

Matrix Multiplication

Let A and B, m × p and p × n matrices respectively

AB = (cij)m×n

p∑k=1

aikbkj

(AB)ij = cij =

. . . . . .. . . . . .ai1 ai2 . . . aip

. . ....

. . . b1j . . .

. . . b2j . . .

. . .... . . .

. . . bpj . . .

That is, matrices are multiplied row by column :

(∗ ∗ ∗ ∗∗ ∗ ∗ ∗

)∗ ∗ ∗∗ ∗ ∗∗ ∗ ∗∗ ∗ ∗

(∗ ∗ ∗∗ ∗ ∗

2× 4 4× 3 2× 3

From another point of view, we have that the matrices A and Bcan be seen as

a11 a12 · · · a1pa21 a22 · · · a2p

am1 am2 · · · amp

b11 b12 · · · b1nb21 b22 · · · b2n

...bp1 bp2 · · · bpn

w1, w2, . . . , wp

v1 ·w1 v1 ·w1 · · · v1 ·wp

v2 ·w1 v2 ·w1 · · · v2 ·wp...

vm ·w1 vm ·w1 · · · vm ·wp

In general, when AB is defined, not necessarily BA is also defined,but even in that case, we have in general

AB 6= BA

Example 2.1

Let A and B the matrices defined by

1 −2 10 2 −12 1 1

2 1 −11 −1 02 −1 1

Find A + B, A− B, 3A AB, BA

Solution

A + B =

3 −1 01 1 −14 0 2

A− B =

−1 −3 2−1 3 −10 2 0

6 3 −3−3 3 06 −3 3

1 −2 10 2 −12 1 1

2 1 −11 −1 02 −1 1

2 2 00 −1 −17 0 −1

2 1 −11 −1 02 −1 1

1 −2 10 2 −12 1 1

0 −3 01 0 24 −5 4

Example 2.2

Let C and D the matrices defined by

2 11 −12 −1

(1 −2 10 2 −1

)Find CD and DC.Solution

2 11 −12 −1

(1 −2 10 2 −1

2 −2 11 −4 22 −6 3

(1 −2 10 2 −1

) 2 11 −12 −1

(2 20 −1

)6= CD

Example 2.3

Using matrix operations rewrite the linear system

......

am1x1 + am2x2 + . . .+ amnxn = bm

in terms of matrices.

Solution

Starting with the system

......

am1x1 + am2x2 + . . .+ amnxn = bm

and choosing

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

x1x2...xm

b1b2...bm

we get

a11x1 + a12x2 + . . .+ a1nxna21x1 + a22x2 + . . .+ a2nxn

...am1x1 + am2x2 + . . .+ amnxn

b1b2...bm

= B =⇒ AX = B

Example 2.4

(x1, x2, x3, · · · , xn

y1y2y3· · ·yn

xkyk = x · y

y1y2y3· · ·yn

( x1, x2, x3, · · · , xn)

y1x1 y1x2 · · · y1xny2x1 y2x2 · · · y2xn

...ynx1 ynx2 · · · ynxn

Example 2.5

(1 1 −10 2 1

) 0 3 1 1−2 5 6 01 7 4 1

(−3 1 3 0−3 17 16 1

2× 3 3× 4 2× 4

0 3 1 1−2 5 6 01 7 4 1

( 1 1 −10 2 1

3× 4 2× 4 undefined

Properties of matrix multiplication:

(AB)C = A(BC ) (associative law)

(A + B)C = AC + BC (distributive law 1)

C (A + B) = CA + CB (distributive law 2)

(rA)B = A(rB) = r(AB) (associative law)

Any of the above identities holds provided that matrix sums andproducts are well defined.

Types of Matrices An m × n matrix A = (aij)m×n is a

1) Zero matrix if aij = 0; i = 1, 2, ...,m, j = 1, 2, ..., n

2) Square Matrix if m = n.

2 −2 11 −4 22 −6 3

(3 75 −4

3) Identity matrix (n× n) (In) if aij = δij where δij =

{1 i = j0 i 6= j

A = In =

1. . .

I1 = (1)

(1 00 1

1 0 00 1 00 0 1

1 0 0 00 1 0 00 0 1 00 0 0 1

4) Symetric Matrix (n × n) if AT = A or aij = aji ;i = 1, 2, ...,m, j = 1, 2, ..., n

5) Triangular Matrix (n × n)

5a) Upper Triangular Matrix (U) if uij = 0, i > j

a11 · · · · · · a1n

a22. . .

5b) Lower Triangular Matrix (L) if lij = 0, i < j

a22 0. . .

... · · · · · · ann

6) Diagonal Matrix (n × n) (D) if aij = Dij where Dij = diδij

d1 · · · · · ·

d2 00 . . .

... · · · · · · dn

Notation

A diagonal matrix D is going to be denoted byD = diag(d1, d2, · · · , dn)

Diagonal matrices

Theorem

Let A = diag(s1, s2, · · · , sn), B = diag(t1, t2, · · · , tn)then

A + B = diag(s1 + t1, s2 + t2, · · · , sn + tn)

rA = diag(rs1, rs2, · · · , rsn)

AB = diag(s1t1, s2t2, · · · , sntn)

(AB = BA, diagonal matrices always commute)

Theorem

Let D = diag(d1, d2, · · · , dm) and A be an m × n matrix. Thenthe matrix DA is obtained from A by multiplying the ith row by difor i = 1, 2, ...,m

⇒ DA =

d2v2...

Thus, for instance we have

7 0 00 1 00 0 2

a11 a12 a13a21 a22 a23a31 a32 a33

7a11 7a12 7a13a21 a22 a23

2a31 2a32 2a33

Theorem

Let D = diag(d1, d2, · · · , dn) and A be an m × n matrix. Then thematrix AD is obtained from A by multiplying the jth column by djfor j = 1, 2, ..., n

w1,w2, · · · ,wn

)⇒ AD =

(d1w1, d2w2, · · · , dnwn

)Thus, for instance we have

a11 a12 a13a21 a22 a23a31 a32 a33

7 0 00 1 00 0 2

7a11 a12 2a137a21 a22 2a237a31 a32 2a33

7) Invertible Matrix (n × n) If A is a square matrix (n × n) andthere exists an n × n matrix B such that

AB = BA = In

The matrix B is denoted by A−1 and is called the Inverse Matrixand A is called invertible or nonsingular matrix. Matrices that donot have an inverse are called singular or noninvertible.OBS

A−1 is the notation for the inverse of A, but keep in mind that

A−1 6= 1

A−1A = AA−1 = In

Example 2.6

(1 10 1

), B =

(1 −10 1

), C =

(−1 00 1

(1 10 1

) (1 −10 1

(1 00 1

(1 −10 1

) (1 10 1

(1 00 1

(−1 00 1

) (−1 00 1

(1 00 1

Thus A−1 = B, B−1 = A, and C−1 = C

Inverse matrix

Let Mn(R) denote the set of all n × n matrices with real entries.We can add, subtract, and multiply elements of Mn(R).However, eventhough, we know that the division does not exist ingeneral, for a subset of Mn(R) can be defined as follow. If A andB are n × n matrices and B is an invertible matrix, then

A/B := AB−1

Basic properties of inverse matrices

1) If B = A−1, then A = B−1. In other words, if A is invertible, sois A−1, and A = (A−1)−1 .

2) The inverse matrix (if it exists) is unique. Moreover, ifAB = CA = I for some n × n matrices B and C , thenB = C = A−1.

(B = IB = (CA)B = C (AB) = CI = C )

3) If the n × n matrices, A, B, are invertible, so is AB, and(AB)−1 = B−1A−1

4) Similarly (A1A2 · · ·Ak)−1 = A−1k Ak−1 · · ·A−12 A−11

linear algebra. session 2roquesol/math_304_fall_2018_session… · session 2. matrices. matrix...

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