Engineering Mathematics
Chapter 8: Matrices
by
Dr.Adnan Daraghmeh
Department of Mathematics
An-Najah National University
2021-2022
1
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Dr.Adnan Daraghmeh
Lecture Notes - Page 2 of 66
Section 8.1: Matrix Algebra
ˆ Definition: A matrix is any rectangular array of numbers or functions


a11 a12 · · · a1n
 a21 a22 · · · a2n 


 ..
..
.. 
.
.
 .
.
.
. 
am1 am2 · · · amn
ˆ We denote a matrix by a capital boldfaced letter such as A, B, C, or X.
ˆ The numbers or functions are called entries or elements of the matrix.
ˆ If a matrix has m rows and n columns we say that its size is m by n (written m × n).
ˆ An n × n matrix is called a square matrix or a matrix of order n.
ˆ The entry in the ith row and j th column of an m × n matrix A is written aij .
ˆ An m × n matrix A is then abbreviated as
A = aij
m×n
ˆ For an n × n square matrix, the entries a11 , a22 , · · · , ann are called the main diagonal
entries.
ˆ Definition: Column and Row Vectors
An n × 1 matrix,


a1
 a2 
 
 .. 
.
an
is called a column vector.
An 1 × n matrix,
a1 a2 · · · an
is called a row vector.
Matrices, 2021
Dr.Adnan Daraghmeh

3 −8 π
−3 7
6


9 10
Example 1: Let A =  4
 3 −8 −5
2
3
6
Lecture Notes - Page 3 of 66
−3 
5


12 
5 −6 2



11 
 and B = −8 9 0 . Find
4 
−3 5 4
7
13
a) The size of A.
solution:
b) The size of B.
solution:
c) a22 , b22 , a53 and b42
solution:
Example 2: Find a matrix A = aij
solution:
3×4

 2i
i + 2j
such that aij =

3−j
,i > j
,i = j
,i < j
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Dr.Adnan Daraghmeh
Lecture Notes - Page 4 of 66
ˆ Definition: Equality of Matrices
Two m × n matrices A and B are equal if aij = bij for each i and j.
Example 1: Find the value(s) of x and y
1 x
1 y−2
a)
=
.
y −3
y 2 −3
solution:
b)
2
x
3
16 3
.
=
9 2x
y 2 −8
solution:
Definition:
ˆ Matrix Addition: If A and B are m × n matrices, then their sum is
A + B = aij + bij m×n
ˆ Scalar Multiple of a Matrix:
If k is a real number, then the scalar multiple of a matrix A is


ka11 ka12 · · · ka1n
 ka21 ka22 · · · ka2n 


kA =  .
= kaij m×n

.
.
.
..
..
.. 
 ..
kam1 kam2
· · · kamn
ˆ The difference of two m × n matrices is defined in the usual manner:
A − B = A + (−B)
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Lecture Notes - Page 5 of 66
1 3
4 −2
−1 6 7
Example 1: Given A =
,B =
and C =
. Find if possible
5 −3
7 8
2 −3 9
a) A + B and B + A
solution:
b) A − B and B − A.
solution:
c) A − C and C + B.
solution:
d) 3A − 2B
solution:
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Dr.Adnan Daraghmeh
Lecture Notes - Page 6 of 66
Properties of Matrix Addition and Scalar Multiplication :
Suppose A, B, and C are m × n matrices and k1 and k2 are scalars. Then
ˆ A + B = B + A and A − B 6= B − A
ˆ A+ B+C = A+B +C
ˆ k1 k2 A = k1 k2 A
ˆ 1A = A
ˆ k1 A + B = k1 A + k1 B
ˆ k1 + k2 A = k1 A + k2 A
1 3
4 −2
Example 2: Find the matrix X such that 5X +
=2
+ 3X
5 −3
7 8
solution:
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Dr.Adnan Daraghmeh
Lecture Notes - Page 7 of 66
Matrix Multiplication:
ˆ Let A be a matrix having m rows and p columns, and let B be a matrix
having p rows and n columns. The product AB is the m × n matrix

a11
 a21

AB =  .
 ..
a12
a22
..
.
am1 am2


b11 b12
a1p
b21 b22
a2p 

..
..   ..
.
.  .
bp1 bp2
· · · amp
···
···
..
.

+a1p bpn
+a2p bpn 



am1 b11 + am2 b21 + · · · +amp bp1 + · · · +am1 b1n + am2 b2n + · · · +amp bpn
!
p
X
aik bkj
a11 b11 + a12 b21 +
 a21 b11 + a22 b21 +

=
..

.
=

· · · b1n
· · · b2n 

.. 
..
. . 
· · · bpn
k=1
···
···
+a1p bp1 +
+a2p bp1 +
···
···
+a11 b1n + a12 b2n +
+a21 b1n + a22 b2n +
..
.
···
···
m×n
ˆ The product C = AB is defined only when the number of columns in the
matrix A is the same as the number of rows in B. The dimension of the
product can be determined from
Am×n × Bn×p = Cm×p
−1 6 7
4 −2
1 3
. Find if possible
and C =
,B =
Example 1: Given A =
2 −3 9
7 8
5 −3
a) AB and BA
solution:
Matrices, 2021
Dr.Adnan Daraghmeh
Lecture Notes - Page 8 of 66
b) AC and CA.
solution:




3 −8 4 −3
5 −6 2
−3 7

6 12 

−8 9
0
. If C = AB, find the

4
9
10
11 
Example 2: Let A = 

 and B = −3 5

4
 3 −8 −5 4 
10 6 −2
2
3
6 13
value of c42
solution:
Matrices, 2021
Dr.Adnan Daraghmeh
Lecture Notes - Page 9 of 66
Properties Matrix Multiplication:
ˆ If A is an m × p matrix, B a p × r matrix, and C an r × n matrix, then the
product
A(BC) = (AB)C
is an m × n matrix.
ˆ If B and C are both r × n matrices and A is an m × r matrix, then
A(B + C) = AB + AC
Furthermore, if the product (B + C)A is defined, then
(B + C)A = BA + CA
ˆ If n is an integer number and A is a square matrix, then
An = AAA
| {z· · · A}
n−times
−1 6
4 −2
1 3
. Find if possible
and C =
,B =
Example 1: Given A =
2 −3
7 8
5 −3
a) A3
solution:
b) (AB)C and A(BC).
solution:
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Dr.Adnan Daraghmeh
Lecture Notes - Page 10 of 66
Transpose of a Matrix:
ˆ The transpose of the m × n matrix

a11 a12
 a21 a22

A= .
..
 ..
.
am1 am2
is the n × m matrix AT given by

a11
 a12

AT =  .
 ..
a21
a22
..
.
a1n a2n
···
···
..
.

a1n
a2n 

.. 
. 
· · · amn
···
···
..
.

am1
am2 

.. 
. 
· · · amn
In other words, the rows of a matrix A become the columns of its transpose
AT .
Properties of Transpose: Suppose A and B are matrices and k a scalar.
Then
T
ˆ AT
=A
T
ˆ
A+B
ˆ
A+B+C
ˆ
AB
ˆ
ABC
ˆ
kA
T
T
= AT + B T + C T
= B T AT
T
T
= AT + B T
= C T B T AT
= kAT
Example 1: Given A =
a)
T
AT
solution:
1 3
4 −2
−1 6
,B =
and C =
. Find
5 −3
7 8
2 −3
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b)
c)
d)
e)
A+B+C
solution:
T
2CB
solution:
T
3B
solution:
T
A2
solution:
Dr.Adnan Daraghmeh
T
Lecture Notes - Page 11 of 66
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Lecture Notes - Page 12 of 66
Special Matrices
ˆ Zero Matrix :
A matrix that consists of all zero entries and is denoted by 0. For example,


0 0 0
0
0 = 0 0 0 , 0 =
0
0 0 0
are zero matrices. If A and 0 are n × m matrices, then
A+0=A
A + (−A) = 0
ˆ Triangular Matrix :
A a square matrix A = (aij)n×n is said to be a triangular matrix if all its
entries above the main diagonal are zeros or if all its entries below the main
diagonal are zeros.
If aij = 0 whenever i < j, then the matrix is called lower triangular matrix

7
5

−6

3
−5
|
lower

0 0
0 0
2 0
0 0

4 9
0 0

1 8 −1 0
7 −3 2 6
{z
}
triangular
matrix
If aij = 0 whenever i > j, then the matrix is called upper triangular matrix

3
0

0

0
0
|
upper

4 −6 1
9
2 −5 7
6

0 9
2 −1

0 0 −1 8 
0 0
0 −6
{z
}
triangular
matrix
ˆ Diagonal Matrix :
A a square matrix A = (aij)n×n is said to be a diagonal matrix if all its
entries not on the main diagonal are zeros.
If aij = 0 for i 6= j, then the matrix is called a diagonal matrix.


3 0 0
0 2 0 
0 0 9
|
{z
}
diagonal
matrix
Matrices, 2021
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Lecture Notes - Page 13 of 66
ˆ Scalar Matrix:
A diagonal matrix A is called scalar matrix, when the entries aii are all
equal.


3 0 0
0 3 0 
0 0 3
|
{z
}
scalar
matrix
ˆ Identity Matrix:
A scalar matrix is called identity matrix, denoted by In , when the entries
of the main diagonal are all equal to one.


1 0 ··· 0
0 1 · · · 0


In =  . . .
.. 
.
.
.
. .
. .
|
0 0 ··· 1
{z
}
identity
matrix

1 0 0
I3 = 0 1 0
0 0 1

is an identity matrix of order 3.
For any m × n matrix A it is readily verified that
Im A = AIn = A
ˆ Symmetric and Skew-Symmetric Matrix:
An n × n matrix A is said to be symmetric if AT = A
T
An n× n matrix
 A is said to be skew-symmetric if A = −A
1 2 7
A = 2 5 8 is symmetric, why?
7 8 6


0
1 −2
3  is skew-symmetric, why?
B = −1 0
2 −3 0
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Lecture Notes - Page 14 of 66
Example 1: Determine whether each of the following statements is TRUE or FALSE.
a) A − B A + B = A2 − B 2
(T) (F)
solution:
b)
2
A + B = A2 + 2AB + B 2
solution:
(T)
(F)
c) If A 6= 0 and AB = 0, then B = 0
solution:
d) If A is a square matrix, then C =
solution:
(T)
1
2
A−AT
(F)
is a skew-symmetric matrix
e) If A is an m × n matrix, then AAT is a symmetric matrix
solution:
(T)
(F)
(T)
(F)
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Lecture Notes - Page 15 of 66
f ) If AB defined and A has a row of zeros, then AB has a row of zeros
solution:
(T)
(F)
g) Symmetric and upper triangular matrix must be a diagonal matrix
solution:
(T)
(F)
(T)
(F)
h) If AB = 0, then A = 0 or B = 0
solution:
i) If AC = BC, then A = B
solution:
(T)
(T)
(F)
(F)


2 1
j) Suppose A = 6 3 Verify that the matrix B = AAT is symmetric
2 5
solution:
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Lecture Notes - Page 16 of 66
Section 8.2: Systems of Linear Algebraic Equations
ˆ A system of linear equations (or linear system) is a collection of one or
more linear equations involving the same set of variables.
a11 x1 + a12 x2 + · · · + a1n xn = b1
a21 x1 + a22 x2 + · · · + a2n xn = b2
..
.
am1 x1 + am2 x2 + · · · + amn xn = bm
ˆ The matrix (general) form of this linear system is: AX = b
   

a11 a12 · · · a1n
x1
b1
 a21 a22 · · · a2n   x2   b2 

   
 ..
..
..   ..  =  .. 
.
.
 .
.
.
.  .   . 
am1 am2 · · · amn
xn
bm
|
{z
} | {z } | {z }
A
X
b
A: The coefficients matrix
X: The variables matrix
b: The constants matrix.
ˆ Homogeneous Systems :
The system AX = b is called homogeneous system, if the matrix (vector)
b is zero.
The system AX = b is called non-homogeneous system, if the matrix (vector)
b is non-zero.
ˆ Augmented Matrix
The augmented matrix ( or the matrix of the system) of the system AX = b
is defined as


a11 a12 · · · a1n b1


 a21 a22 · · · a2n b2 
A |b =  .
..
..
.. 
 ..
.
···
.
. 
am1 am2 · · · amn bm
Example 1: Write the linear system represented by the augmented matrix
1 −3 5 2
4 7 −1 8
solution:
Matrices, 2021
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Lecture Notes - Page 17 of 66
Example 2: Find the augmented matrix ( or the matrix of the system)
x1 − 5x3 = −1
2x1 + 8x2 = 7
x2 + 9x3 = 1
solution:
Elementary Row Operations on A Matrix :
ˆ Interchange any two rows i and j : Ri ←→ Rj
ˆ Multiply the ith row by a nonzero constant c : cRi
ˆ Multiply the ith row by c and add to the j th row : cRi + Rj −→ Rj
Example 1: Apply elementary row operations on the matrix: A =
a)
1
2 R1
b) −5R1 + R2
solution:
c)
1
11 R2
d) 2R2 + R1
2 −4
5 1
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Lecture Notes - Page 18 of 66
Elimination Methods:
ˆ We say that two matrices are row equivalent if one can be obtained from
the other through a sequence of elementary row operations.
ˆ The procedure of carrying out elementary row operations on a matrix to
obtain a row-equivalent matrix is called row reduction.
ˆ Row-Echelon Form (REF): We row-reduce the matrix until we arrive at a
row-equivalent matrix in row-echelon form:
1. The first nonzero entry in a nonzero row is a 1.
2. In consecutive nonzero rows, the first entry 1 in the lower row appears
to the right of the 1 in the higher row.
3. Rows consisting of all zeros are at the bottom of the matrix
ˆ Reduced Row-Echelon Form (RREF): We row-reduce the matrix until we
arrive at a row-equivalent matrix in reduced row-echelon form:
1. The first nonzero entry in a nonzero row is a 1.
2. In consecutive nonzero rows, the first entry 1 in the lower row appears
to the right of the 1 in the higher row.
3. Rows consisting of all zeros are at the bottom of the matrix
4. A column containing a first entry 1 has zeros everywhere else.
Example 1: Determine whether the matrix is in row echelon form, reduced row echelon
form, both, or neither


1 5 0 2
a) 0 1 0 −1 
0 0 0 0


1 0 0 7
b) 0 1 0 −1 
0 0 1 0
0 0 1 −6 2 2
c)
0 0 0 0 1 4
0 0 1 −6 0 −6
d)
0 0 0 0 1 4


0 0 1 2 3 6
e) 0 1 0 0 0 4 
0 0 1 0 0 7
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Lecture Notes - Page 19 of 66
Elimination Methods for Solving Linear Systems:
To solve a linear system using an augmented matrix, we shall use either Gaussian
elimination or the Gauss–Jordan elimination method.
ˆ In the Gaussian elimination, we stop when we have obtained an augmented
matrix in row-echelon form.
ˆ In the Gauss–Jordan method, we stop when we have obtained an augmented
matrix in reduced row-echelon form.
ˆ A linear system of equations is said to be consistent if it has at least one
solution (a unique solution) or infinitely many solutions.
ˆ A linear system of equations is said to be inconsistent if it has no solutions.
Example 1: Solve the linear system
a)
3x1 + 6x2 = 3
x1 − 4x2 = 7
solution:
b)
3x1 + 6x2 = 3
−x1 − 2x2 = −1
solution:
c)
3x1 + 6x2 = 3
−x1 − 2x2 = 2
solution:
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Lecture Notes - Page 20 of 66
Example 2: Verify that x1 = 14 + 7t, x2 = 9 + 6t, x3 = t, where t is any real number, is a
solution of the system
2x1 − 3x2 + 4x3 = 1
x1 − x2 − x3 = 5
solution:
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Dr.Adnan Daraghmeh
Example 3: Solve the linear system using
1. Gaussian elimination.
2. Gauss–Jordan elimination.
a)
2x1 + 6x2 + x3 = 7
x1 + 2x2 − x3 = −1
5x1 + 7x2 − 4x3 = 9
solution:
Lecture Notes - Page 21 of 66
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b)
x + 3y − 2z = −7
4x + y + 3z = 5
2x − 5y + 7z = 19
solution:
Lecture Notes - Page 22 of 66
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

 
2
2
2
4
3
1 ,b =  0 
Example 4: Let A =  1
−2 −4 −2
−2
a) Write A in REF.
solution:
b) Find all solutions to AX = b ( if exist)
solution:
Lecture Notes - Page 23 of 66
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Lecture Notes - Page 24 of 66
Example 5: Use Gauss–Jordan elimination to solve the system
i1 − i2 − i3 = 0
i1 R1 + i2 R2 = E
i2 R2 − i3 R3 = 0
when R1 = 10 ohms, R2 = 20 ohms, R3 = 10 ohms, and E = 12 volts. solution:
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Example 6: Solve
x1 + x2 = 1
4x1 − x2 = −6
2x1 − 3x2 = 8
solution:
Lecture Notes - Page 25 of 66
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Lecture Notes - Page 26 of 66
Homogeneous Systems:
ˆ A homogeneous system AX = 0 is always consistent and either possesses
only the trivial solution X = 0 or possesses the trivial solution along with
infinitely many nontrivial solutions.
ˆ A homogeneous system AX = 0 possesses nontrivial solutions if the number
m of equations is less than the number n of variables (m < n).
ˆ If X1 is a solution of AX = 0, then so is cX1 for any constant c.
ˆ If X1 and X2 are solutions of AX = 0, then so is X1 + X2 .
Example 1: Solve
2x1 − 4x2 + 3x3 = 0
x1 + x2 − 2x3 = 0
solution:
Matrices, 2021
Dr.Adnan Daraghmeh
Lecture Notes - Page 27 of 66
Section 8.3: Rank of a Matrix
ˆ Rank of a Matrix by Row Reduction :
a) rank(A) = the number of nonzero rows in B.
b) A linear system of equations AX = b is consistent if and only if
rank(A) = rank(A|b)


1 1 −1 3
Example 1: Given A = 2 −2 6 8. Find rank(A)
3 5 −7 8
solution:
Matrices, 2021
Dr.Adnan Daraghmeh
Lecture Notes - Page 28 of 66
Section 8.4: Determinants
A determinant of an n×n matrix A is said to be a determinant of order n denoted
by |A| or det(A) (a number).
ˆ Determinant of a 1 × 1 Matrix:
If A = (a)1×1 , then det(A) = a
ˆ Determinant
ofa 2 × 2 Matrix:
a
a
a11 a12
If A =
, then det(A) = 11 12 = a11 × a21 − a12 × a22
a22 a21
a22 a21 2×2
1 2
4 −2
Example 1: Given A =
,B =
and C = −3 . Find
4 −3
−5 −1
a) det(A)
solution:
b) det(B)
solution:
c) det(C)
solution:
d) det(AT )
solution:
e) det A + B
solution:
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Lecture Notes - Page 29 of 66
f ) det 2A − 3B
solution:
g) det AB
solution:
Example 2: Find the value(s) of x such that
solution:
x−2 x
3 −8
=
3
x
0 2
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Dr.Adnan Daraghmeh
Determinant of a n × n Matrix:
Let

a11 a12
 a21 a22

A= .
..
 ..
.
am1 am2
···
···
..
.
Lecture Notes - Page 30 of 66

a1n
a2n 

..  = aij n×n
. 
· · · amn
be an n × n matrix.
ˆ The cofactor of aij is the determinant
Cij = (−1)i+j Mij
where Mij is the determinant of the submatrix obtained by deleting the
ith row and the jth column of A. The determinant Mij is called a minor
determinant.
ˆ The matrix of cofactors corresponding to the entries of A denoted by cof (A)
and defined as:


C11 C21 · · · Cn1
 C12 C22 · · · Cn2 


cof (A) =  .
..
.. 
..
 ..
.
.
. 
C1n C2n
· · · Cnn
ˆ For each 1 ≤ i ≤ n, the cofactor expansion of det(A) along the ith row is
det(A) = ai1 Ci1 + ai2 Ci2 + · · · + ain Cin
.
ˆ For each 1 ≤ j ≤ n, the cofactor expansion of det(A) along the j th column is
det(A) = a1j C1j + a2j C2j + · · · + anj Cnj
.
Matrices, 2021
Dr.Adnan Daraghmeh
Lecture Notes - Page 31 of 66


2 −4 7
Example 1: Given A = 6 2 3.
1 5 3
a) Find the matrix of cofactors corresponding to the entries of A ( find cof (A))
solution:
b) Evaluate the determinant of A using cofactor expansion along the first row.
solution:
c) Evaluate the determinant of A using cofactor expansion along the third column.
solution:
Matrices, 2021
Dr.Adnan Daraghmeh
Example 2: Evaluate the determinant of the matrix


6 5 0
a) A = −1 8 −7
−2 4 0
solution:

5
−1
b) B = 
1
1
solution:
1
0
1
0
2
2
6
0

4
3

1
4
Example 2: Evaluate
2 2 0 0 −2
1 1 6 0
5
a) 1 4 2 −1 −1
2 0 1 −3 3
0 1 0 0
1
solution:
Lecture Notes - Page 32 of 66
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0
0
−2 0
b) 8 −1
−1 2
2
2
solution:
Dr.Adnan Daraghmeh
Lecture Notes - Page 33 of 66
0 3 0
0 2 0
0 −7 2
2 3 2
3 6 4
Example 3: Find the values of λ that satisfy the given equation.
a)
b)
−3 − λ
10
=0
2
5−λ
solution:
1−λ
0
−1
1
2−λ 1 =0
3
3
−λ
solution:
Matrices, 2021
Dr.Adnan Daraghmeh
Lecture Notes - Page 34 of 66
Section 8.5: Properties of Determinants
ˆ Determinant of a Transpose
If AT is the transpose of the n × n matrix A, then
det(AT ) = det(A)
2 −4
Example: Given A =
. Find det(A) and det(AT )
6 2
solution:
ˆ Two Identical Rows
If any two rows (columns) of an n × n matrix A are the same, then
det(A) = 0


6 2 2
Example: Given A = 4 2 2. Find det(A)
9 2 2
solution:
ˆ Zero Row or Column
If all the entries in a row (column) of an n × n matrix A are zero, then
det(A) = 0


6 2 0
Example: Given A = 4 3 0. Find det(A)
9 5 0
solution:
Matrices, 2021
Dr.Adnan Daraghmeh
Lecture Notes - Page 35 of 66
ˆ Interchanging Rows
If B is the matrix obtained by interchanging any two rows (columns) of an
n × n matrix A, , then
det(B) = −det(A)
a1 a2 a3
Example: Given that b1 b2 b3 = 5
c1 c2 c3
a3 a2 a1
a) Find b3 b2 b1
c3 c2 c1
solution:
a1 b1 c1
b) Find a2 b2 c2
a3 b3 c3
solution:
ˆ Constant Multiple of a Row
If B is the matrix obtained from an n × n matrix A by multiplying a row
(column) by a nonzero real number k, then
det(B) = kdet(A)
a1 a2 a3
Example: Given that b1 b2 b3 = 5
c1 c2 c3
2a1 a2 a3
a) Find 6b1 3b2 3b3
2c1 c2 c3
solution:
2a1 4b1 2c1
b) Find 5a2 10b2 5c2
a3 2b3 c3
solution:
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Dr.Adnan Daraghmeh
Lecture Notes - Page 36 of 66
ˆ Constant Multiple of a Matrix
If A is an n × n matrix and k real number , then
det(kA) = k n det(A)
Example: Let A and B are both 3 × 3 matrices such that |A| = −4 and |B| = 21 . Find
a) |2A|
solution:
b) |4B|
solution:
c) |(4B T )|
solution:
ˆ Determinant of a Matrix Product
If A and B are both n × n matrices, then
det(AB) = det(A) · det(B)
Example: Let A and B two 3 × 3 matrices such that |A| = −4 and |B| = 12 . Find
a) |AB|
solution:
b) |(4BA)T |
solution:
c) |A + B|
solution:
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Dr.Adnan Daraghmeh
Lecture Notes - Page 37 of 66
ˆ Determinant Is Unchanged
Suppose B is the matrix obtained from an n × n matrix A by multiplying
the entries in a row (column) by a nonzero real number k and adding the
result to the corresponding entries in another row (column). Then
det(B) = det(A)
Example: Given that
a b
=5
c d
a
b
2a + c 2b + d
solution:
a) Find
a b − 3a
c d − 3c
solution:
b) Find
a1 a2 a3
Example: Given that b1 b2 b3 = 5
c1 c2 c3
−a1
−a2
−a3
b1
b2
b3
a) Find
c1 − a1 c2 − a2 c3 − a3
solution:
a1 − 2b1 + 3c1 a2 − 2b2 + 3c2 a3 − 2b3 + 3c3
b1
b2
b3
b) Find
c1
c2
c3
solution:
Matrices, 2021
Dr.Adnan Daraghmeh
Lecture Notes - Page 38 of 66
ˆ Determinant of a Triangular Matrix
Suppose A is an n × n triangular matrix (upper or lower). Then
det(A) = a11 × a22 × · · · × ann
, where a11 , a22 , · · · , ann are the entries on the main diagonal of A.
Example: Find
2 0 0 0 0
9 1 0 0 0
a) 3 4 2 0 0
2 −5 1 −3 0
4 1 7 0 1
solution:
2 0 0
0
0 6 0
0
b)
0 0 −2 0
0 0 0 −5
solution:

2
1
Example 1: Evaluate the determinant of the matrix A = 
0
3
solution:
9
3
1
1
1
7
6
4

8
4

5
2
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Dr.Adnan Daraghmeh
Lecture Notes - Page 39 of 66


a a+1 a+2
Example 2: Evaluate the determinant of the matrix A =  b b + 1 b + 2 
c c+1 c+2
solution:
1 1 1 1
a b c d
Example 3: Evaluate 2 2 2 2
a b c d
a3 b3 c3 d3
solution:
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Dr.Adnan Daraghmeh
1 1 1
Example 4: Show that a b c = (b − a)(c − a)(c − b).
a2 b2 c2
solution:
Lecture Notes - Page 40 of 66
Matrices, 2021
Dr.Adnan Daraghmeh
Lecture Notes - Page 41 of 66
Example 5: Determine whether each of the following statements is TRUE or FALSE.
a) Suppose A is an n × n matrix such that A2 = I. Then |A| = ±1
solution:
(T)
b) Suppose A is an n×n matrix such that A2 = A. Then |A| = 0 or |A| = 1
solution:
c) If A and B are n × n matrices, then det(AB) = det(BA)
solution:
(T)
f ) If A is a 5 × 5 skew-symmetric matrix, then |A| = 0
solution:
(T)
(F)
(T)
(T)
(F)
d) If A is an n × n skew-symmetric matrix, then |A| = 0 or |A| = 1
solution:
e) If A is an n × n symmetric matrix, then |A| = 0
solution:
(F)
(F)
(T)
(F)
(F)
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Lecture Notes - Page 42 of 66
Section 8.6: Inverse of a Matrix
Non-singular Matrices and det(A)
ˆ An n × n matrix A is non-singular (invertible) if and only if det(A) 6= 0
ˆ An n × n matrix A is singular (non-invertible) if and only if det(A) = 0
Example 1: determine whether the given matrix is singular or non-singular.
1 12
a)
2 32
solution:


1 −2 3
b) 5 4 7
2 −4 6
solution:

1 2 1
0 0 3
c) 
3 2 1
1 1 1
solution:

1
0

0
0


2 −3 4
Example 2: Find all values of x ∈ R such that the matrix is non-singular 6 1 9
4 x 8
solution:
Matrices, 2021
Dr.Adnan Daraghmeh
Lecture Notes - Page 43 of 66
The Inverse of a Matrix
ˆ Inverse of a Matrix
Let A be an n × n matrix. If there exists an n × n matrix B such that
AB = BA = I
where I is the n × n identity, then the matrix A is said to be nonsingular or
invertible. The matrix B is said to be the inverse of A.
ˆ An n × n nonsingular matrix A has an inverse denoted by A−1 such that
AA−1 = A−1 A = I
ˆ An n × n singular matrix A has no inverse.
ˆ Note that the symbol −1 in the notation A−1 is not an exponent; in other
words, A−1 is not a reciprocal.
ˆ If A is nonsingular, then its inverse is unique.
ˆ Properties of the Inverse
−1
a) A−1
=A
−1
b) AB
= B −1 A−1
T
−1 = A−1
c) AT
1 −1
A
k
1
e) If A is non-singular matrix, then det(A−1 ) =
det(A)
d) If k is a non-zero constant, then (kA)−1 =
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Lecture Notes - Page 44 of 66
Finding the Inverse
ˆ Adjoint Method:
Let A be an n × n matrix. The matrix that is the transpose of the matrix
of cofactors is called the adjoint of A and is denoted by adj(A).
T
adj(A) = cof (A)
ˆ Let A be an n × n matrix. If det(A) 6= 0, then
A−1 =
1 adj(A)
det(A)
ˆ For a 2 × 2 non-singular matrix A =
A
−1
1
=
det(A)
a11 a12
, then
a21 a22
a22 −a12
−a21 a11


a11 a12 a13
ˆ For a 3 × 3 non-singular matrix A = a21 a22 a23 , then
a31 a32 a33
A−1


C
C21 C31
1  11
C12 C22 C32 
=
det(A)
C13 C23 C33
Example 1: Find the inverse of the matrix (if it exist)
1 4
a) A =
2 10
solution:
Matrices, 2021
Dr.Adnan Daraghmeh


2 2 0
b) A = −2 1 1
3 0 1
solution:
Example 2:
a) If
A−1
=
4 3
, what is A ?
3 2
solution:
b) If
(2A)−1
solution:
=
5
3
, what is (3A) ?
−3 −2
Lecture Notes - Page 45 of 66
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Dr.Adnan Daraghmeh
Lecture Notes - Page 46 of 66
Finding the Inverse
ˆ Row Operations Method:
If an n × n matrix A can be transformed into the n × n identity matrix I
by a sequence of elementary row operations, then A is non-singular. The
same sequence of operations that transforms A into the identity I will also
transform I into A−1 .
Row
A|I −−−−−−−→ I|A−1
Operations
ˆ If the matrix to the left of the vertical bar has a row of zeros, we can stop
at this point and conclude that A is singular and has no inverse.
Example 1: Find

2 0
a) A = −2 3
−5 5
solution:
the inverse of the matrix (if it exist)

1
4
6
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Dr.Adnan Daraghmeh


1 −1 −2
5
b) A = 2 4
6 0
3
solution:

2 1
4 3
c) A = 
2 1
0 1
solution:
3
6
4
2

1
2

1
1
Lecture Notes - Page 47 of 66
Matrices, 2021

1 0
0 0
d) A = 
0 0
0 1
solution:
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0
1
0
0

0
0

1
0
Lecture Notes - Page 48 of 66
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Lecture Notes - Page 49 of 66
Using the Inverse to Solve Linear Systems
ˆ The linear system
a11 x1 + a12 x2 + · · · + a1n xn = b1
a21 x1 + a22 x2 + · · · + a2n xn = b2
..
.
an1 x1 + an2 x2 + · · · + ann xn = bn
can be written compactly as

a11 a12
 a21 a22

A= .
..
 ..
.
an1 an2
a matrix equation AX = b, where

 
 
· · · a1n
x1
b1
 x2 
 b2 
· · · a2n 

 
 
..  , X =  ..  , b =  .. 
..
.
.
.
. 
· · · ann
xn
bn
ˆ If A is non-singular, then the system AX = b can be solved by multiplying
both of the equations by A−1
X = A−1 b
ˆ A non-homogeneous system AX = b has:
a) A unique solution if and only if det(A) 6= 0 (A non-singular)
b) No solution or infinitely many solution if and only if det(A) = 0 (A
singular)
ˆ A homogeneous systems AX = 0 of n linear equations in n variables has:
a) Only the trivial solution if and only if A is non-singular (det(A) 6= 0)
b) A nontrivial solution if and only if A is singular (det(A) = 0).
Example 1: Find the pair (a, b) such that the system

    
1 b a
x
3
2 1 −1 , y  = 6
1 1 −2
z
3
has unique solution.
solution:
Matrices, 2021
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Lecture Notes - Page 50 of 66
Example 2: Use the inverse of the coefficient matrix to solve the system
2x1 − 9x2 = 15
3x1 + 6x2 = 16
solution:
Example 3: Use the inverse of the coefficient matrix to solve the system
2x1 +
x3 = 2
−2x1 + 3x2 + 4x3 = 4
−5x1 + 5x2 + 6x3 = −1
solution:
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Lecture Notes - Page 51 of 66
Example 4: The sequence of elementary row operations
R3 ←→ R2 ,
R1 + 2R3 −→ R1 ,
R3 + 3R2 −→ R3 ,
were applied on a 3 × 3 identity matrix to obtain matrix A.
a) Find A−1 .
solution:
b) Find A.
solution:
R2 − 2R1 −→ R2
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Lecture Notes - Page 52 of 66
Example 4: The 3 × 3 identity matrix was obtained by applying the following sequence
of elementary row operations on a 3 × 3 matrix A
R1 −→ R2 ,
R1 −2R2 −→ R1 ,
R1 +R3 −→ R1 ,
−6R3 −→ R3 ,
.
a) Compute det(A).
solution:


4
b) Solve the system AX = b, where b = 11 .
3
solution:
1
R3 +2R2 −→ R3 ,
2
−R2 +R1 −→ R2
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Lecture Notes - Page 53 of 66
Section 8.7: Cramer’s Rule
Cramer’s Rule
Let A be the coefficient matrix of the system:
a11 x1 + a12 x2 + · · · + a1n xn = b1
a21 x1 + a22 x2 + · · · + a2n xn = b2
..
.
an1 x1 + an2 x2 + · · · + ann xn = bn

a11
 a21

A= .
 ..
a12
a22
..
.
an1 an2

 
 
a1n
x1
b1
 x2 
 b2 
a2n 

 
 
..  , X =  ..  , b =  .. 



.
.
.
· · · ann
xn
bn
···
···
..
.
If det(A) 6= 0, then the solution of this system is given by
x1 =
det(A1 )
det(A2 )
det(An )
, x2 =
, · · · , xn =
det(A)
det(A)
det(A)
where Ak is the same as the matrix A except that the k th column of A has been
replaced by the entries of the column matrix
 
b1
 b2 
 
b=.
 .. 
bn
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Example 1: Use Cramer’s rule to solve the system
2x1 − 9x2 = 15
3x1 + 6x2 = 16
solution:
Example 2: Use Cramer’s rule to solve the system
3x1 + 2x2 + x3 = 7
x1 − x2 + 3x3 = 3
5x1 + 4x2 − 2x3 = 1
solution:
Lecture Notes - Page 54 of 66
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Lecture Notes - Page 55 of 66
Example 3: Use Cramer’s rule to determine the solution of the system
(2 − k)x1 +
kx2 = 4
kx1 + (3 − k)x2 = 3
For what value(s) of k is the system inconsistent? solution:
Matrices, 2021
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Lecture Notes - Page 56 of 66
Section 8.8: The Eigenvalue Problem
Eignvalues and Eigenvectors
ˆ Let A be an n × n matrix. A number λ is said to be an eigenvalue of A if
there exists a nonzero solution vector X of the linear system
AX = λX
The solution vector X is said to be an eigenvector corresponding to the
eigenvalue λ.
ˆ Eigenvalues and eigenvectors are also called characteristic values and characteristic vectors, respectively.
ˆ To find the eigenvalues of a n × n matrix A
a) Find the nth -degree characteristic polynomial in λ:
P (λ) = det A − λI
b) The eigenvalues of A are the roots of the characteristic equation
det A − λI = 0
c) To find an eigenvector corresponding to an eigenvalue λ, we simply
solve the system of equations
A − λI X = 0
by applying Gauss-Jordan elimination to the augmented matrix
A − λI|0
ˆ When an n×n matrix A possesses n distinct eigenvalues λ1 , λ2 , · · · λn , it can be
proved that a set of n linearly independent eigenvectors K1 , K2 , · · · , Kn can
be found. However, when the characteristic equation has repeated roots, it
may not be possible to find n linearly independent eigenvectors for A.
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Lecture Notes - Page 57 of 66




1
0 −1 −3
3
3
Example 1: Verify that X = −1 is an eigenvector of the 3×3 matrix A =  2
1
−2 1
1
solution:




2
0 −1 −3
3
3 , find
Example 2: If X = −2 is an eigenvector of the 3 × 3 matrix A =  2
2
−2 1
1
the corresponding eigenvalue.
solution:
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Lecture Notes - Page 58 of 66
4 2
Example 3: Find the eigenvalues and eigenvectors of A =
5 1
solution:
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Lecture Notes - Page 59 of 66


1
2
1
Example 4: Find the eigenvalues and eigenvectors of A =  6 −1 0 
−1 −2 −1
solution:
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Lecture Notes - Page 60 of 66


9 1 1
Example 5: Find the eigenvalues and eigenvectors of A = 1 9 1
1 1 9
solution:
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Lecture Notes - Page 61 of 66


4
6
6
3
2
Example 6: Find the eigenvalues and eigenvectors of A =  1
−1 −5 −2
solution:
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Lecture Notes - Page 62 of 66
Example 7:
a) Find the eigenvalues and eigenvectors of A =
solution:
6 −1
b) Find the eigenvalues of A =
5 4
solution:
3 4
−1 7
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Lecture Notes - Page 63 of 66


1
2
1
Example 8: Given the matrix A =  6 −1 0 
−1 −2 −1
a) Find the characteristic equation of A.
solution:
b) Find the eigenvalues of A.
solution:
c) Find an eigenvector corresponding to the largest eigenvalue
solution:
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Lecture Notes - Page 64 of 66


−1 4 0
Example 9: Given the matrix A =  1 1 3
0 1 0
a) Find the characteristic equation of A.
solution:
 
3

b) Given that X = 3 is an eigenvectors of A, find the eigenvalue associated with
1
X.
solution:
c) Find all eigenvalues of A.
solution:
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Lecture Notes - Page 65 of 66
Let A be an n × n matrix with eigenvalues λ1 , λ2 , · · · , λn
ˆ The set of all eigenvalues of A is called spectrum of A and denoted by
σ(A) = {λ1 , λ2 , · · · , λn }
ˆ The determinant of A is the product of its eigenvalues.
det(A) = λ1 λ2 · · · λn
ˆ The eigenvalues of the transpose AT are the same as the eigenvalues of A.
σ(AT ) = {λ1 , λ2 , · · · , λn }
ˆ Triangular and Diagonal Matrices:
The eigenvalues of an upper triangular, lower triangular, and diagonal matrix are the main diagonal entries.
ˆ Eigenvalues and Eigenvectors of A−1 :
Let A be a non-singular matrix. If λ is an eigenvalue of A with corresponding
1
eigenvector X, then
is an eigenvalue of A−1 with the same corresponding
λ
eigenvector X.
ˆ Eigenvalues and Singular Matrices:
a) λ = 0 is an eigenvalue of A if and only if A singular.
b) A matrix A is non-singular if and only if the number 0 is not an eigenuvalue of A.
Example 1: Let λ1 = 3,
the eigenvalues of a 2×2 matrix A with corresponding
λ2 = 4are
0
1
. Find the eigenvalues of A−1 and the corresponding
,X2 =
eigenvectors X1 =
1
2
eigenvectors.
solution:
Matrices, 2021
Dr.Adnan Daraghmeh
Lecture Notes - Page 66 of 66
Example 2: Let λ1 = 11, λ2 = λ3 = 8 be the eigenvalues of a 3 × 3 matrix A
a) Find the characteristic polynomial of A.
solution:
b) Find σ(AT ).
solution:
c) Find σ(A−1 ).
solution:
d) Determine whether A is singular or non-singular.
solution:


2 0 0
Example 3: Given the matrix A = 1 −3 0
1 1 4
a) Find the characteristic polynomial of A.
solution:
b) Find σ(AT ).
solution:
c) Find σ(A−1 ).
solution:
d) Determine whether A is singular or non-singular.
solution: