Systems of First Order Linear Equations
Ordinary Differential Equations
Dr. Marco A Roque Sol
12/01/2015
Dr. Marco A Roque Sol
Ordinary Differential Equations
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Matrix Multiplication
Let A and B, m × p and p × n matrices respectively
AB = (cij )m×n
where
cij =
p
X
aik bkj
k=1


...
...
. . .

...


(AB)ij = cij = a

a
.
.
.
a
i1
i2
in


..
...
.
Dr. Marco A Roque Sol
. . . b1j
. . . b2j
.
. . . ..
. . . bnj
Ordinary Differential Equations

...
. . .


. . .
...
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OBS
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 7.1
Let A and B the matrices

1 −2

A= 0 2
2 1
Find A + B,
A − B,
defined by



1
2 1 −1
−1 B = 1 −1 0 
1
2 −1 1
3A
AB,
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Solution

 

1 −2 1
2 1 −1
A + B = 0 2 −1 + 1 −1 0  =
2 1
1
2 −1 1



 

1 −2 1
2 1 −1
A − B = 0 2 −1 − 1 −1 0  =
2 1
1
2 −1 1

3 −1 0
1 1 −1
4 0
2

−1 −3 2
−1 3 −1
0
2
0

 

1 −2 1
3 −6 3
3A = 3 0 2 −1 = 0 6 −3
2 1
1
6 3
3
Dr. Marco A Roque Sol
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AB =
BA =



1 −2 1
2 1 −1
0 2 −1 1 −1 0  =
2 1
1
2 −1 1



2 1 −1
1 −2 1
1 −1 0  0 2 −1 =
2 −1 1
2 1
1
Dr. Marco A Roque Sol


2 2
0
0 −1 −1
7 0 −1


0 −3 0
1 0 2 6= AB
4 −5 4
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Example 7.2
Let C and D the matrices defined by


2 1
1 −2 1


C = 1 −1
D=
0 2 −1
2 −1
Find CD and DC.
Solution
Since C and D are 3 × 2 and 2 × 3 matrices respectively, then CD
is a well defined 3 × 3 matrix and DC is a well defined 2 × 2 matrix




2 1
2 −2 1
1 −2 1
CD = 1 −1
= 1 −4 2
0 2 −1
2 −1
2 −6 3
Dr. Marco A Roque Sol
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DC =
1 −2 1
0 2 −1


2 1
1 −1 =
2 −1
2 2
6= CD
0 −1
Example 7.3
Using matrix operations rewrite the linear system
a11 x1 + a12 x2 + . . . + a1n xn =
a21 x1 + a22 x2 + . . . + a2n xn =
..
.
b1
b2
..
.
am1 x1 + am2 x2 + . . . + amn xn = bm
in terms of matrices.
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Solution
Starting with the system
a11 x1 + a12 x2 + . . . + a1n xn =
a21 x1 + a22 x2 + . . . + a2n xn =
..
.
b1
b2
..
.
am1 x1 + am2 x2 + . . . + amn xn = bm
and choosing


a11 a12 . . . a1n
 a21 a22 . . . a2n 


A=

..


.
am1 am2 . . . amn
Dr. Marco A Roque Sol


x1
 x2 
 
X= . 
 .. 

b1
 b2 
 
B= . 
 .. 
xm
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
bm
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we get

a11 x1 + a12 x2 + . . . + a1n xn
a21 x1 + a22 x2 + . . . + a2n xn
..
.


AX = 

am1 x1 + am2 x2 + . . . + amn xn

b1
  b2 
  
 =  ..  = B =⇒ AX = B
  . 


bm
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.
Dr. Marco A Roque Sol
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

2 −2 1
A = 1 −4 2 ;
2 −6 3
3 7
B
5 −4
3) Identity
matrix (I) (n × n) if aij = δij where
1 i =j
δij =
, (Kronecker Delta
0 i 6= j
https://en.wikipedia.org/wiki/Kronecker_delta )

1


A=I=


Dr. Marco A Roque Sol
0

..





1
0
.
1
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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 = aij = 0,
a11 · · · · · ·

a22

U=
..
.


0
Dr. Marco A Roque Sol
a1n
i >j


.. 

. 
ann
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5b) Lower Triangular Matrix (L) if lij = uij = 0,



L=



a11
a22
0
..
..
.
i <j
···





.
···
ann
6) Diagonal Matrix(n × n) (D) if aij = dij where dij = Di δij

.. 
D ··· ···
.
 1



D


2
D=

..


.


..
. · · · · · · Dn
0
0
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7) Invertible Matrix (n × n) If A is a square matrix (n × n) and
there exists an n × n matrix B such that
AB = BA = I
The matrix B is denoted by A−1 and is called the Inverse Matrix
and A is called invertible or nonsingular matrix. Matrices that do
not have an inverse are called it singular or noninvertible.
OBS
A−1 is the notation for the inverse of A, but keep in mynd that
A−1 6=
Dr. Marco A Roque Sol
1
A
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There are various ways to compute A−1 from A, assuming that it
exists.
One way is the cofactor expansion. Associated with each
element aij of a given matrix is the minor Mij from, which is the
determinant ( not defined yet !!!) of the matrix
Mij is obtained by deleting the ith row and jth column of the
original matrix that is, the row and column containing aij .
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Also associated with each element aij , is the cofactor Cij defined
by the equation
Cij = (−1)n Mij
If B = A−1 , then it can be shown that the general element bij is
given by
bij =
Cij
detA
However, one way to do it is using elementary row operations.
There are three such operations:
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1. Interchange of two rows.
2. Multiplication of a row by a nonzero scalar.
3. Addition of any multiple of one row to another row.
The transformation of a matrix by a sequence of elementary row
operations is referred to as row reduction or Gaussian
elimination. Starting with the matrix A we build the m × 2n
Augmented Matrix
[A|I]
and using elementary row operations we tranforme it into
I|A−1
Dr. Marco A Roque Sol
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Example 7.4
Find the inverse of


1 −1 −1
A = 3 −1 2 
2 2
3
Solution
First of all, let’s build the augmented matrix


1 −1 −1 1 0 0




A = 3 −1 2 0 1 0


2 2
3 0 0 1
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(a) Obtain zeros in the off-diagonal positions in the first column by
adding (3) times the first row to the second row and adding (2)
times the first row to the third row.


1 −1 −1 1 0 0




5 −3 1 0
A = 0 2


0 4
5 −2 0 1
(b) Obtain a 1 in the diagonal position in the second column by
multiplying the second row by 1/2 .


1 −1 −1 1
0 0




A = 0 1 5/2 −3/2 1/2 0


0 4
5 −2
0 1
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(c) Obtain zeros in the off-diagonal positions in the second column
by adding the second row to the first row and adding (4) times the
second row to the third row


1 0 3/2 −1/2 1/2 0




A = 0 1 5/2 −3/2 1/2 0


0 0 −5 4
−2 1
(d) Obtain a 1 in the diagonal position in the third column by
multiplying the third row by (15).


1 0 3/2 −1/2 1/2
0




A = 0 1 5/2 −3/2 1/2
0 


0 0 1 −4/5 2/5 −1/5
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(e) Obtain zeros in the off-diagonal positions in the third column
by adding (3/2) times the third row to the first row and adding
(5/2) times the third row to the second row.


1 0 0 7/10 −1/10 3/10




−1/2
1/2 
A = 0 1 0 1/2


0 0 1 −4/5
2/5
−1/5
so, the inverse matrix A−1 is given by

A−1



7/10 −1/10 3/10
7 −1 3
1 
−1/2
1/2  =
5 −5 5 
=  1/2
10
−4/5
2/5
−1/5
−8 4 −2
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Matrix Functions .
We sometimes need to consider vectors or matrices whose
elements are functions of a real variable t. In that case, we write




x1 (t)
a11 (t) · · · a1n (t)
x2 (t)



.. 
X(t) =  .  = A(t) =  ...
. 
 .. 
am1 (t)
amn (t)
xn (t)
respectively.
Continuity
The matrix A(t) is said to be continuous at t = t0 or on an
interval α < t < β if each element of A(t) is a continuous function
at the given point or on the given interval.
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Diffrentiability
Similarly, A(t) is said to be differentiable if each of its elements is
differentiable, and its derivative dA(t)/dt is defined by
daij (t)
dA(t)
=
dt
dt
m×n
that is, each element of dA(t)/dt is the derivative of the
corresponding element of A(t).
Integrability
In the same way, the integral of a matrix function is defined as
Z
b
Z
A(t)dt =
a
Dr. Marco A Roque Sol
b
aij (t)dt
a
m×n
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Example 7.5
Consider the matrix
A(t) =
Find A0 (t) and
Rπ
0
sin(t)
1
0
cos(t)
A(t)dt.
Solution
(sin(t))0
(1)0
cos(t)
0
=
A (t) =
(0)0
(cos(t))0
0
−sin(t)
0
Z
0
π
Rπ
R π
sin(t)dt R 0 1dt
2 π 2 /2
0
R
A(t)dt =
=
π
π
π
0
0 0dt
0 cos(t)dt
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Elementary calculus extend easily to matrix functions; in particular,
d (CA)
dA
=C
;
dt
dt
C = constant
d (A + B)
dA dB
=
+
dt
dt
dt
d (AB)
dB dA
=A
+
B
dt
dt
dt
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Systems of Linear Algebraic Equations .
A set of n simultaneous linear algebraic equations in n variables
a11 x1 + a12 x2 + . . . + a1n xn = b1
a21 x1 + a22 x2 + . . . + a2n xn = b2
..
..
.
.
an1 x1 + an2 x2 + . . . + ann xn = bnn
can be written as
AX = b
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If b = 0, the system is said to be homogeneous; otherwise, it is
nonhomogeneous.
If the matrix A is invertible, hence A−1 exists, and therefore we
have
X = A−1 b
In particular, the homogeneous problem AX = b, corresponding to
b = 0, has only the trivial solution 0.
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On the other hand, if A is singular, A−1 does not exist, so the
homogeneous system
AX = 0
has (infinitely many) nonzero solutions in addition to the trivial
solution.
Solving a Linear System
For solving particular systems, we can form the augmented matrix
Dr. Marco A Roque Sol
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
a11 a12 . . . a1n b1


a

 21 a22 . . . a2n b2 
.
[A|b] = 
..

.

.
.


an1 an2 . . . ann bn

We now perform row operations on the augmented matrix so as to
transform A into an upper triangular matrix.
[U|b̄]
Once this is done, it is easy to see whether the system has
solutions, and to find them if it does.
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Example 7.6
Solve the system of equations
x1 − 2x2 + 3x3 = 7
−x1 + x2 − 2x3 = −5
2x1 − x2 − x3 =
4
Solution
The augmented matrix for the system is


1 −2 3 7




−1 1 −2 −5


2 −1 −1 4
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We now perform row operations on the augmented matrix with a
view to introducing zeros in the lower left part of the matrix.
(a) Add the first row to the second row, and add (−2) times the
first row to the third row.

1 −2 3


0 −1 1

0 −3 −7
Dr. Marco A Roque Sol

7


2 

−10
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(b) Multiply the second row by −1.

1 −2 3


0 1 −1

0 3 −7
(c) Add (−3) times the second

1 −2


0 1

0 0
Dr. Marco A Roque Sol

7


−2 

−10
row to the third row.

3 7


−1 −2

−4 −4
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(d) Divide the third row by −4.

1 −2 3


0 1 −1

0 0
1

7


−2

1
The matrix obtained in this manner corresponds to the system of
equations
x1 − 2x2 + 3x3 = 7
x2 − x3 = −2
x3 =
1
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From the last of equations we have
x3 = 2,
x2 = −2 + x3 = −1,
x3 = 7 + 2x2 − 2x3 = 2
Thus, we obtain


2
X = − 1
1
Now, since the solution is unique, we conclude that the coefficient
matrix is nonsingular.
Dr. Marco A Roque Sol
Ordinary Differential Equations