1
Lecture 3
Gaussian elimination and Gauss-Jordan
elimination for
(i) Underdetermined systems
(ii) Overdetermined systems
Instructor: Dr. Rehana Naz
Linear Algebra
2
Number of equations <Number of unknown
Underdetermined systems may or may not be consistent.


If an underdetermined system is consistent, it must
have infinitely many solutions.
We can find solutions of underdetermined
homogeneous and nonhomogeneous systems only
by Gaussian elimination and Gauss-Jordan
elimination methods.
Instructor: Dr. Rehana Naz
Linear Algebra
Example 1:
3
Solve the following system via Gaussian elimination
(nonhomogeneous system)
2𝑥 − 3𝑦 − 𝑧 + 2𝑤 + 3𝑣 = 4
4𝑥 − 4𝑦 − 𝑧 + 4𝑤 + 11𝑣 = 4
2𝑥 − 5𝑦 − 2𝑧 + 2𝑤 − 𝑣 = 9
2𝑦 + 𝑧 + 4𝑣 = −5
Solution: For this system Number of equations <Number of
unknown
The augmented matrix is
2 −3 −1 2 3
4
4 −4 −1 4 11 4
2 −5 −2 2 −1 9
0 2
1 0 4 −5

Instructor: Dr. Rehana Naz
Linear Algebra
Example 1:
4
We use elementary row operations to transform this matrix into row
echelon form.
(1/2)R1 → R1
3
1
3
1 −
−
1
2
2
2
2
4 −4 −1 4 11 4
2 −5 −2 2 −1 9
0 2
1 0 4 −5
R2 +(-4) R1 → R2, R3 +(-2)R1 → R3
3
1
3
1 −
−
1
2
2
2
2
0 2
1 0 5 −4
0 −2 −1 0 −4 5
0 2
1 0 4 −5

Instructor: Dr. Rehana Naz
Linear Algebra
Example 1:
5
(1/2)R2 → R2
3
1
1 −
−
2
2
1
0 1
2
0 −2 −1
0 2
1
R3 +2R2 → R3, R4 +(-2)R2 → R4
3
1
1 −
−
2
2
1
0 1
2
0 0
0
0 0
0
Instructor: Dr. Rehana Naz
1
0
0
0
1
0
0
0
3
2
2
5
−2
2
−4 5
4 −5
3
2
2
5
−2
2
1
1
−1 −1
Linear Algebra
Example 1:
6
R4 +R3→ R4
3
1 −
2
1
3
−
1
2
2
2
1
5
0 1
0
−2
2
2
0 0
0 0 1 1
0 0
0 0 0 0
This yields row echelon form of augmented matrix.
 Its associated system is
3
1
3
𝑥− 𝑦− 𝑧+𝑤+ 𝑣 =2
2
2
2
1
5
𝑦 + 𝑧 + 𝑣 = −2
2
2
𝑣=1
 Now above system
can
be Naz
solved by
back-substitution
method:
Instructor: Dr.
Rehana
Linear
Algebra
Example 1:
7
Solving for leading variables, we obtain
3
1
3
𝑥 = 𝑦+ 𝑧−𝑤− 𝑣+2
2
2
2
1
5
𝑦 =− 𝑧− 𝑣−2
2
2
𝑣=1
 Substituting the bottom equation into those above yields
3
1
1
𝑥 = 𝑦+ 𝑧−𝑤+
2
2
2
1
9
𝑦=− 𝑧−
2
2
𝑣=1

Instructor: Dr. Rehana Naz
Linear Algebra
Example 1:
8
And substituting the second equation into the top yields
25 1
𝑥 =− − 𝑧−𝑤
4 4
9 1
𝑦=− − 𝑧
2 2
𝑣=1
 Assign arbitrary values to the free variables. If we assign z=s and
w=t, the general solution is given by
25 1
𝑥 =− − 𝑠−𝑡
4 4
9 1
𝑦=− − 𝑠
2 2
𝑣 = 1, 𝑧 = 𝑠, 𝑤 = 𝑡
 Infinite many solutions exist in this example.

Instructor: Dr. Rehana Naz
Linear Algebra
Example 2:
9
Solve by Gauss-Jordan elimination (homogeneous
system)
2𝑥1 + 2𝑥2 − 𝑥3 + 𝑥5 = 0
−𝑥1 − 𝑥2 + 2𝑥3 − 3𝑥4 + 𝑥5 = 0
𝑥1 + 𝑥2 − 2𝑥3 − 𝑥5 = 0
𝑥3 + 𝑥4 + 𝑥5 = 0
 Solution: The augmented matrix
2
2 −1 0
1 0
−1 −1 2 −3 1 0
1
1 −2 0 −1 0
0
0
1
1
1 0

Instructor: Dr. Rehana Naz
Linear Algebra
Example 2:
10
can be expressed in following reduced row-echelon form
1 1 0 0 1 0
0 0 1 0 1 0
0 0 0 1 0 0
0 0 0 0 0 0
 The associated linear system is
𝑥1 + 𝑥2 + 𝑥5 = 0
𝑥3 + 𝑥5 = 0
𝑥4 = 0
 Solving for leading variables yield
𝑥1 = −𝑥2 − 𝑥5
𝑥3 = −𝑥5
𝑥4 = 0

Instructor: Dr. Rehana Naz
Linear Algebra
Example 2:
11

Thus general solution is
𝑥1 = −𝑠 − 𝑡, 𝑥2 = 𝑠, 𝑥3 = −𝑡 , 𝑥4 = 0, 𝑥5 = 𝑡
 Thus we have infinite many solutions for this case.
 Note that the trivial solution is obtained when s=0,
t=0.
Instructor: Dr. Rehana Naz
Linear Algebra
Example 3:
12
Suppose that the augmented matrix for a system of
linear equations has been reduced by row operations to
the given reduced row-echelon form. Solve the system.
1 −6 0 0 3 −2
0 0 1 0 4 7
0 0 0 1 5 8
0 0 0 0 0 0
 Solution: The corresponding system of equations is
𝑥1 − 6𝑥2 + 3𝑥5 = −2
𝑥3 + 4𝑥5 = 7
𝑥4 + 5𝑥5 = 8

Instructor: Dr. Rehana Naz
Linear Algebra
Example 3:
13
Here leading variables are 𝑥1 , 𝑥3 , 𝑥4 and free
variables are 𝑥2 and 𝑥5 . Solving for leading variables
in terms of free variables gives
𝑥1 = 6𝑥2 − 3𝑥5 − 2
𝑥3 = −4𝑥5 + 7
𝑥4 = −5𝑥5 + 8
 If we assign free variables 𝑥2 and 𝑥5 arbitrary values s
and t. The general solution is given by formulas
𝑥1 = 6𝑠 − 3𝑡 − 2, 𝑥2 = 𝑠, 𝑥3 = −4𝑡 + 7, 𝑥4
= −5𝑡 + 8, 𝑥5 = 𝑡

Instructor: Dr. Rehana Naz
Linear Algebra
Example 4:
14


Solve the following system via Gaussian elimination
3𝑥1 − 𝑥2 + 𝑥3 − 4𝑥4 = 2
6𝑥1 + 3𝑥2 − 𝑥3 − 4𝑥4 = 3
9𝑥1 + 2𝑥2 − 8𝑥4 = 6
Solution: The augmented matrix is
3 −1 1 −4 2
6 3 −1 −4 3
9 2
0 −8 6
Instructor: Dr. Rehana Naz
Linear Algebra
Example 4:
15
R2 +(-2) R1 → R2, R3 +(-3)R1 → R3
3 −1 1 −4 2
0 5 −3 4 −1
0 5 −3 4
0
R3 –R2→ R3
3 −1 1 −4 2
0 5 −3 4 −1
0 0
0
0
1
is in row-echelon form.
Instructor: Dr. Rehana Naz
Linear Algebra
Example 4:
16
The associated linear system is
3𝑥1 − 𝑥2 + 𝑥3 − 4𝑥4 = 2
5𝑥2 − 3𝑥3 + 4𝑥4 = −1
0𝑥1 + 0𝑥2 + 0𝑥3 + 0𝑥4 = 1
Last equation can never be satisfied; the system has
no solution and hence is inconsistent.

Instructor: Dr. Rehana Naz
Linear Algebra
Number of equations >Number of unknown
17
Overdetermined systems are usually (but not
necessarily) inconsistent.
 We can find solutions of overdetermined
homogeneous and nonhomogeneous systems
only by Gaussian elimination and Gauss-Jordan
elimination methods.

Instructor: Dr. Rehana Naz
Linear Algebra
Example 5:
18
Solve following overdetermined system by Gauss-Jordan
elimination
𝑥1 + 2𝑥2 = 5
2𝑥1 + 3𝑥2 = 1
5𝑥1 + 2𝑥2 = 1
 Solution: The augmented matrix
1 2 5
2 3 1
5 2 1
can be expressed in following reduced row echelon form
1 0 0
0 1 0
0 0 Linear
1 Algebra
Instructor: Dr. Rehana Naz

Example 5:
19

The associated linear system is
𝑥1 = 0
𝑥2 = 0
0. 𝑥1 + 0. 𝑥2 = 1

This last equation can never be satisfied; hence this
linear system is inconsistent.
Instructor: Dr. Rehana Naz
Linear Algebra
Section 2.2: Evaluating Determinants by Row
Reduction
20

This method is important since it is the most computationally efficient way to find
the higher order determinants.
Instructor: Dr. Rehana Naz
Linear Algebra
Section 2.2: Evaluating Determinants by Row
Reduction
21
Instructor: Dr. Rehana Naz
Linear Algebra
Example 1: Consider 2×2 matrix
22
1.
1.
2
6
3
1

𝐴=

Where det(A)=2-18= -16

Solution:
Multiply first row of A by 2 and call resulting matrix as B,
4 6
= 4 − 36 = −32
6 1

det 𝐵 =

=2(-16)=2det(A)
if we interchange any two rows then sign of determinant changes i.e. det(B)=-det(A)

det 𝐵 =
6 1
= 18 − 2 = 16 = −det(𝐴)
2 3
Instructor: Dr. Rehana Naz
Linear Algebra
Example 1: Consider 2×2 matrix
23
1.
A multiple of second is added to first row then det(B)=det(A)

2R1+R2 → R2

det 𝐵 =
2 3
= 14 − 30 = −16 = det(𝐴)
10 7

Determinants that are Zero: The determinant of a matrix will be zero if
1.
An entire row is zero.
2.
Two rows or columns are equal.
3.
A row or column is a constant multiple of another row or column
Instructor: Dr. Rehana Naz
Linear Algebra
24
Example 2: Evaluate the following
determinants by inspection
a.
3
𝐴= 4
3
−2
7
−2
5
−1
5
b.
2
𝐵= 5
0
−1
8
0
3
−2
0
c.
1
𝐶= 4
3
−2
7
−6
5
−1
15
Instructor: Dr. Rehana Naz
Linear Algebra
25
Example 2: Evaluate the following
determinants by inspection

Solution:
(a)
det(A)=0 since first and third rows are equal.
(b)
Third row of matrix B consists of zeros only thus det(B)=0.
(c)
The third row of matrix C is multiple of first row therefore det(C)=0.
Instructor: Dr. Rehana Naz
Linear Algebra
26
Instructor: Dr. Rehana Naz
Linear Algebra
Elementary Matrices
27

An elementary matrix results from performing a single elementary row operation
on an identity matrix 𝐼𝑛 .

𝑑𝑒𝑡(𝐼𝑛 ) = 1
Instructor: Dr. Rehana Naz
Linear Algebra
Example 3
28

The following determinants of elementary matrices, are evaluated by inspection
1
0
0
0

0
3
0
0
0 0
0 0
=3
1 0
0 1
The second row of 𝐼4 was multiplied by 3.
Instructor: Dr. Rehana Naz
Linear Algebra
Example 3
29
0
0
0
1

0
1
0
0
0
0
1
0
1
0
= −1
0
0
The first and last rows of 𝐼4 were interchanged.
Instructor: Dr. Rehana Naz
Linear Algebra
Example 3
30
1
0
0
0

0
1
0
0
0 7
0 0
=1
1 0
0 1
7 times the last row of 𝐼4 was added to the first row.
Instructor: Dr. Rehana Naz
Linear Algebra
Example 4
31

Evaluate det(A) by row reduction where
0
𝐴= 3
2
Instructor: Dr. Rehana Naz
1
−6
6
5
9
1
Linear Algebra
Example 4
32

Solution: We will reduce A to row-echelon form which is upper triangular.
0 1
𝐴 = 3 −6
2 6
Instructor: Dr. Rehana Naz
5
9
1
Linear Algebra
Example 4
33
1
= −3 0
2
−2 3
1 5 (Taking 3 as common from first row)
6 1
𝑅3 +(-2) 𝑅1 → 𝑅3
1 −2
= −3 0 1
0 10
Instructor: Dr. Rehana Naz
3
5
−5
Linear Algebra
Example 4
34
𝑅3 +(-10) 𝑅1 → 𝑅3
1 −2
3
= −3 0 1
5
0 0 −55
1 −2 3
= −3 (−55) 0 1 5 (Taking -55 as common from third row)
0 0 1
= −3 −55 1 = 165
Instructor: Dr. Rehana Naz
Linear Algebra
Example 5
35

Use row reduction to compute the determinant of
2 3
0 4
𝐴=
2 −1
0 −4
Instructor: Dr. Rehana Naz
3
3
−1
−3
1
−3
−3
2
Linear Algebra
Example 5
36
2 3
3
1
0 4
3 −3
𝐴 =
2 −1 −1 −3
0 −4 −3 2
Taking 2 common from first row
1 3/2 3/2 1/2
0
4
3
−3
=2
2 −1 −1 −3
0 −4 −3
2
Instructor: Dr. Rehana Naz
Linear Algebra
Example 5
37
𝑅3 − 2𝑅1 → 𝑅3
1
0
=2
0
0
3/2 3/2 1/2
4
3
−3
−4 −4 −4
−4 −3
2
Taking 4 common from second row
Instructor: Dr. Rehana Naz
Linear Algebra
Example 5
38
1 3/2 3/2 1/2
0
1
3/4 −3/4
= 2 (4)
0 −4 −4
−4
0 −4 −3
2
𝑅3 + 4𝑅2 → 𝑅3 , 𝑅4 + 4𝑅2 → 𝑅4
1 3/2 3/2 1/2
0
1
3/4 −3/4
= 2 (4)
0
0
−1
−7
0
0
0
−1
Instructor: Dr. Rehana Naz
Linear Algebra
Example 5
39
Taking -1 common from third and fourth rows we have
1 3/2
0
1
= 2 4 −1 (−1)
0
0
0
0
3/2 1/2
3/4 −3/4
1
−7
0
1
Compute the determinant of the upper triangular matrix
𝐴 = 2 4 −1 −1 1 = 8
Instructor: Dr. Rehana Naz
Linear Algebra
Example 6
40
Find the determinant of a 4x4 matrix
1
2
𝐴=
0
7
Instructor: Dr. Rehana Naz
0
7
6
3
0
0
3
1
3
6
0
−5
Linear Algebra
Example 6
41
Solution:
1
2
𝐴 =
0
7
0
7
6
3
0 3
0 6
3 0
1 −5
Observe that this can be converted to lower triangular form if we can make first two
entries (3 and 6) of last column 𝐶4 as zero. Is it possible?
Instructor: Dr. Rehana Naz
Linear Algebra
Example 6
42
The answer is yes, we can do just in one step i.e. 𝐶4 − 3𝐶1 → 𝐶4 yields
1
2
=
0
7
0
7
6
3
0
0
3
1
0
0
0
−26
now it is in lower triangular form and thus determinant is simply sum of diagonal elements
=(1)(7)(3)(-26)=-546
Instructor: Dr. Rehana Naz
Linear Algebra
43
Instructor: Dr. Rehana Naz
Linear Algebra