CHAPTER 2
System of Linear Equations
1. Introduction
Study of a linear system of equations is classical. First let’s consider a system having
only one equation:
2x + 3y + 4z = 5
(2.1)
Indeterminates x, y, z are referred to as unknowns of the equation. Both (1, 1, 0) and
(−1, 1, 1) satisfy Equation (2.1). In fact, assign whatever real numbers you like to x
and y and find z from (2.1) by substituting the values of x and y (you have assigned).
Geometrically, the collection of all the solutions of Equation (2.1) is a plane in the
(real) space R3 .
Suppose a, b, c, d ∈ R are fixed. Then, the collection of all (α, γ, β) ∈ R3 satisfying
equation
ax + by + cz = d
(2.2)
is geometrically a plane in the space R3 if at least one of a, b, c is nonzero. More
generally, suppose, a1 , a2 , . . . , an , b ∈ R. Then, the collection of all (α1 , α2 , . . . , αn ) ∈
Rn satisfying
n
X
ai x i = b
(2.3)
i=1
n
is called a hyper plane of R . Every such point in Rn is called a solution. The
collection of all the solution is called the solution set of the equation.
By a system of linear equations we mean a finite set of linear equations in finitely
many indeterminates. For instance, the following is a system of two linear equations:
2x + 3y + 4z = 5
x + y + z = 2.
(2.4)
By a solution of this system we mean a solution of the first equation which is also
a solution of the second equation. If Ai (i = 1, 2) is the set of solutions of the i-th
equation, then the set of solutions of the system is A1 ∩ A2 . In this example, we know
that for each i, Ai is geometrically a plane. Thus the solution of system (2.4) is the
intersection of two planes. We had learnt in school that two planes in R3 intersect
11
12
2. SYSTEM OF LINEAR EQUATIONS
in a line unless they are parallel or conincident. Further, we learn certain adhoc
methods to solve a linear system of equations (applicable to a restricted situations);
for instance, Cramer’s rule.
Suppose F is a field. While reading for the first time you may assume that F = R.
The set of m × n matrices with coefficients from F is denoted by Mm×n (F).
A system of m linear equations in n unknowns x1 , x2 , . . . , xn written as
a11 x1 + a12 x2 + · · · + a1n xn = b1
a21 x1 + a22 x2 + · · · + a2n xn = b2
·····················
···
am1 x1 + am2 x2 + · · · + amn xn = bm
(2.5)
where ai,j ∈ F for (1 ≤ i ≤ m, 1 ≤ j ≤ n), called coefficients and b1 , b2 , . . . , bm ∈ F,
called constant terms of the system.
System (2.5) can be described by the following matrix equation:
AX = B
(2.6)
where

 
 
b1
x1
a11 a12 a1n
 b2 
 x2 
 a21 a22 a2n 

 
 
A=
 · · · · · · · · ·  ∈ Mm×n (F), B = · · · ∈ Mm×1 (F), and X = . . .
am1 am2 amn
bm
xn

is the matrix (or column) of unknowns. A is called the coefficient matrix and B
the matrix (or column) of constants. The matrix (A|B) ∈ Mm×(n+1) (F) obtained by
attaching the column B with A, is called the augmented matrix of the system. By a
solution of AX = B where A and B are as above, we mean an element Y ∈ Mn×1 (F)
such that AY = B. The collection of all the solutions is called the solution set of
AX = B. Thus, the solution set of AX = B is a subset of Mn×1 (F). Observe that
every element of Mn×1 (F) defines an element of Fn uniquely and vice versa. Thus,
often we express a solution of AX = B as an n-tuple rather than an n-column.
2. Three Types of Solutions and Examples
Before we talk more about a general system of linear equations, we look at the following three systems with real coefficients.
a) x1 + x2 + x3 = 3, x1 + 2x2 + 3x3 = 6, x2 + 2x3 = 1;
b) x1 + x2 + x3 = 3, x1 + 2x2 + 3x3 = 6, x1 + x2 + 2x3 = 4;
c) x1 + x2 + x3 = 3, x1 + 2x2 + 3x3 = 6, x2 + 2x3 = 3.
2. THREE TYPES OF SOLUTIONS AND EXAMPLES
13
We seek solution of these systems in R3 . It is not difficult to verify that system a) has
no solution (subtracting the second equation from the sum of the first and the third
equation, we get 0 = 2); system b) has only one solution, namely, (1, 1, 1) (Cramme’s
rule applicable) and system c) has infinitely many solutions. Indeed, for any real
number λ, the tuple (λ, 3 − 2λ, λ) is a solution of system c). Since Crammer’s rule is
applicable when the number of unknowns and the number of equations are same, and
the determinant of the coefficient matrix is nonzero, we require a method to deal with
a system of linear equations. In fact, in general, we have three types of solution set
to a system of linear equations; namely, an empty set, a singleton set or an infinite
set (if the field is infinite).
Here wewrite matrices
A,
X
and B for

 each
 system of linear equations above.
x1
3
1 1 1
In a), A = 1 2 3 , B = 6, X = x2  so that
1
0 1 2
x3


1 1 1 | 3
(A|B) = 1 2 3 | 6 .
0 1 2 | 1
 


 
1 1 1
3
x1
In b), A = 1 2 3 , B = 6, X = x2  so that
x3
1 1 2
4


1 1 1 | 3
(A|B) = 1 2 3 | 6 .
1 1 2 | 4
 


 
x1
3
1 1 1





In c), A = 1 2 3 , B = 6 , X = x2  so that
3
0 1 2
x3


1 1 1 | 3
(A|B) = 1 2 3 | 6 .
0 1 2 | 3
If the coefficient matrix A belongs to Mm×n (F), then the column of unknowns has
n components so that a solution of the system is an n-tuple which is geometrically
a point in Fn . In this case, column B of constants must have m components. If B
is the zero column, then the system (namely, AX = 0) of linear equations is called a
homogenous system. A system which is not homogenous is called a non-homogenous
system. A homogeneous system always has a solution (namely, the zero column, i.e.,
xi = 0 for each i). When scalars are complex numbers geometric description of the
solution set is not what we have when scalars are reals.
14
2. SYSTEM OF LINEAR EQUATIONS
Warning: Before we write the matrix equation we should arrange the unknowns in
the same order in every equation of the system. If an unkown is missing from an
equation the corresponding coefficient is assumed to be zero.
3. Preliminaries related to Matrices
Recall that the collection R of all the real numbers is not just a set. There are
two binary operations (namely, addition and multiplications) along with certain nice
properties (namely, commutativity, associativity, distributivity of multiplication over
addition, difference of numbers and division by nonzero real numbers.) If you are used
to the concept of a field then you may assume that F is an arbitrary field. Otherwise,
F can be assumed to be R or C (complex numbers) or Q (rational numbers). However,
mostly we will use R. We will not deal with the question of finding integer solutions
of a system of linear equations. During the process of finding solution of a system
of linear equations, we require to perform difference operation, division by nonzero
number other than addition; so we always assume that the coefficients are from a
field (you may assume F = R or C).
To have a geometric idea we consider R as a straight line and R2 as plane and
R3 as the (3-dimensional) space. If x̄ = (x1 , x2 , . . . , xn ), ȳ = (y1 , y2 , . . . , yn ) ∈ Rn and
a ∈ R, define x̄ + ȳ = (x1 + y1 , x2 + y2 , . . . , xn + yn ) and ax̄ = (ax1 , ax2 , . . . , axn ).
Recall that Mm×n (F) is the collection of all matrices having m rows and n columns
with coefficients from F. Matrices A, B of same size can be added component-wise
and the sum is denoted by A + B. Matrix A can be premultiplied to B if number of
columns of A is equal to the number of rows of B; in fact, the coefficient in the i-th
row and the j-th column of AB is given by
n
X
aik bkj
k=1
if (i, k)-th coefficient of A ∈ Mm×n (F) is aik and (k, j)-th coefficient of B ∈ Mn×p
is bkj , and AB ∈ Mm×p (F). Recall that addition of matrices is commutative, but
multiplication is not.
A matrix having equal number of rows and column is called a square matrix. Suppose A ∈ Mm×m is a square matrix. Then tr(A) is denotes the sum of the coefficient
in the pricipal diagonal (top-left corner to the bottom-right corner); symbolically,
tr(A) =
m
X
ai,i .
(2.7)
i=1
Definition of det(A) is by induction. If m = 1 (i.e., A is an 1 × 1 matrix), then
det(A) = A. Induction hypothesis is that we know the definition for square matrices
4. ROW REDUCED ECHELON (RRE) MATRIX
15
of sizer n × n for 1 ≤ n ≤ m − 1. Define for any j (1 ≤ j ≤ m),
det(A) =
n
X
aij bij
(2.8)
i=1
where bij = (−1)i+j det(Ai,j ) and Ai,j is the matrix obtained from A by simply removing the i-th row and j-th column. Quantity det(A) is independent of choice of i in the
definition. Since we assume readers to have studied the concept of determinant, we
will not justify this statement. Another fact is that in the definition of determinant,
role of i and j can be interchanged. For instance,
yaha shaayad j hona chahiye
a b
det
= ad − bc
c d


a1,1 a1, 2 a1,3
a2,2 a2,3
a2,1 a2,3


= a1,1 · det
det a2, 1 a2,2 a2,3
− a1,2 · det
+
a3,2 a3,3
a3,1 a3,3
a3,1 a3,2 a3,3
a2,1 a2,2
a1,3 · det
a3,1 a3,2
= a1,1 (a2,2 a3,3 − a2,3 a3,2 ) − a1,2 (a2,1 a3,3 − a1,3 a3,2 ) +
a1,3 (a2,1 a3,2 − a2,2 a3,1 )
a3,1 hona chahiye
Reader can try to verify the following identities:
(1) tr(A + B) = tr(A) + trB;
(2) tr(AB) = tr(BA);
(3) det(AB) = detA · detB.
It is hard to verify the last identity.
4. Row Reduced Echelon (RRE) Matrix
A matrix is called row reduced echelon (in short RRE) if the following properties
hold:
(1) Every zero row is below every nonzero row.
(2) The leading coefficient of every nonzero row is 1.
(3) A column which contains leading nonzero entry (which is 1) of a row has all
other coefficients equal to zero.
(4) Suppose the matrix has r nonzero rows (and remaining m − r rows are zero).
If leading nonzero entry of i-th row (1 ≤ i ≤ r) occurs in the ki -th column,
then k1 < k2 < · · · < kr .
For any matrix, the row rank is always equal to the column
rank.
16
2. SYSTEM OF LINEAR EQUATIONS

1 0 2
Consider the following matrices: a) 0 0 0is not RRE since it has a zero row
0 1 0


1 0 1
(namely, 2-nd row) preceding a nonzero row (namely, 3-rd row), b) 0 2 0 is
0 0 0
notRRE since
 the second row is nonzero but its leading non zero entry is 2 (6= 1),
1 1 2
c) 0 1 1 is not RRE since the leading nonzero coefficient of the second row is
0 0 0
in the secondcolumn but
 the second column has another nonzero coefficien (in the
1 0 2
first row), d) 0 1 3 is not RRE since the leading nonzero coefficient of the third
0 0 1
rowis in thethird columnand thethird column has another nonzero coefficients ,
1 0 2
0 1 2
e) 0 1 3 is RRE, f) 1 0 3 is not RRE since in this there are two nonzero
0 0 0
0 0 0
rows (so that r = 2) and k1 = 2 and k2 = 1 violating k1 < k2 .

5. Preliminaries related to Equivalence Relation
Goes to apprendix
If a reader is aware of the concept of an equivalence relation on a set, then they can
either skip this section or continue to read to revisit the same.
Let A be a nonempty set. A subset R of A × A (the two fold cartesian product
of A) is called a relation on A. If (x, y) ∈ R, then we write xRy and read it as “x is
related to y.” For instance, < is a relation on R. We write x < y for x, y ∈ R if x is
less than y. Define relation ∼
=n y if n divides x − y.
=n in Z: x ∼
(1) A relation R on A is called reflexive if xRx for every x ∈ A. E.g., Equality
on C, ≥ on R, ≤ on R, ∼
=n on Z.
(2) A relation R on A is called symmetric if xRy ⇐⇒ yRx for any x, y ∈ A.
E.g. Equality on C, 6= on C, ∼
=n on Z.
(3) A relation R is called transitive if xRy, yRz =⇒ xRz for any x, y, z ∈ A.
E.g., < on R, > on R, Equality on C, ∼
=n on Z.
A relation on A which is reflexive, symmetric and transitive is called an equivalence
relation on A. We are going to describe an equivalence relation (namely, equivalence)
on Mn (F), n × n matrices with coefficients from F.
6. ELEMENTARY ROW OPERATIONS
17
6. Elementary row operations
An elementary row operation is a map from Mm×n (F) to itself which is any one
of the following three types:
(1) Multiplying the i-th row by a nonzero scalar λ denoted by Ri → λRi .
(2) Interchanging the i-th row and the j-th row denoted by Ri ↔ Rj .
(3) For i 6= j, replacing the i-th row by the sum of the i-th row and µ multiple
of the j-th row denoted by Ri → Ri + µRj .
A row operation is a map from Mm×n (F) to itself which is a composition of finitely
many elementary row operations.
a b
. Then, for λ 6= 0 in F, R1 → λR1 produces
Example 1. Let A =
c d
c d
λa λb
and, for µ ∈ F, R1 → R1 + µR2 produces
, R1 ↔ R2 produces
a b
c d
a + λc b + λd
.
c
d
Every elementary row operation is invertible. Indeed,
(1) inverse of Ri → λRi is Ri → 1/λRi ;
(2) inverse of Ri ↔ Rj is Ri ↔ Rj (self inverse, i.e., inverse of itself).
(3) inverse of Ri → Ri + µRj is Ri → Ri − µRj .
Let ρ be an elementary row operation. Then we have det(A) 6= 0 ⇒ det(ρ(A)) 6= 0.
Indeed, if ρ is Ri → λRi , then det(ρ(A)) = λdet(A); if ρ is Ri ↔ Rj (i 6= j), then
det(ρ(A)) = −detA; if ρ is Ri → Ri + µRj (i 6= j), then det(ρ(A)) = det(A).
Proposition 6.1. Denote an elementary row operation by ρ. If A ∈ Mm×n (R)
then
ρ(A) = ρ(I) · A
where I is the m × m identity matrix.
Thus, if you know the effect of ρ on the identity matrix, then you know its effect on
A (namely, by simply premultiplying A by ρ(I)). Readers are suggested to verify this
fact in each case.
A square matrix E of size n is called an elementary matrix if there is an elementary
row operation ρ such that E = ρ(In ), where In is the identity matrix of size n.
Use the above rule successively on a matrix A and get the following two statements
for finitely many elementary row operations ρ1 , ρ2 , . . . , ρs (by induction):
(1) (ρs ◦ · · · ◦ ρ2 ◦ ρ1 )(A) = ρs (I) · · · ρ2 (I)ρ1 (I)A.
(2) (ρs ◦ · · · ◦ ρ2 ◦ ρ1 )(A) = (ρs ◦ · · · ◦ ρ2 ◦ ρ1 )(I)A.
18
2. SYSTEM OF LINEAR EQUATIONS
7. Row Equivalence and Row Equivalent Matrices
A matrix A is said to be row equivalent to the matrix B if there are finitely
many elementary row operations ρ1 , ρ2 , . . . , ρs such that B = (ρs ◦ · · · ◦ ρ2 ◦ ρ1 )(A).
Proposition 7.1. Row equivalence is an equivalence relation.
(1) A is row equivalent to itself. Recall that the identity map is composition of
two elementary row operations (for instance, composition of Ri ↔ Rj with
itself) or R1 → 1 · R1 .
(2) A is row equivalent to B if and only if B is row equivalent to A. If B =
−1
−1
(ρs ◦ · · · ◦ ρ2 ◦ ρ1 )(A), then A = (ρ−1
1 ◦ · · · ◦ ρs−1 ◦ ρs )(B) and inverse of an
elementary row operation is an elementary row operation.
(3) If A is row equivalent to B and B is row equivalent to C then A is row
equivalent to C. If B = (ρs ◦· · ·◦ρ2 ◦ρ1 )(A) and C = (ρs+t ◦· · ·◦ρs+2 ◦ρs+1 )(B)
then C = (ρs+t ◦ · · · ◦ ρ2 ◦ ρ1 )(A).
If A is row equivalent to B we will often write A ∼ B.
Theorem 7.2. Every matrix is row equivalent to a unique row reduced echelon
matrix.
Corollary 7.2.1. Every equivalence class of row equivalent matrices contains a
unique row reduced echelon matrix.
Proof. One write an algorithm to find the RRE row equivalent to the given
matrix:
Step 1: Appy interchange of rows to push down the zero rows to the end of the matrix
so that no zero row is before a nonzero row.
Step 2: Find the first nonzero column (from left) (suppose it is k1 ).
Step 3: Again apply interchange of rows to push up a row whose leading nonzero
coefficient occurs in the first nonzero column (i.e. in the k1 -th column), to the first
row. Divide the first row by the leading nonzero coefficient so that the leading nonzero
coefficient becomes 1.
Step 4. Next apply apply Ri → Ri − µR1 for suitable values of i and µ so that the
first nonzero column has nonzero coefficient only in the first row.
Step5: Find the first nonzero column (from left) when we ignore the first row (suppose
it is k2 ). Apply interchange of rows to push up a row whose leading nonzero coefficient
occurs in column k2 , to the second row. Divide the second the second row by the
leading nonzero coefficient.
Step 6: Apply Ri − µR2 for suitable values of i and µ so that k2 column has nonzero
coefficients only in the second row.
Step 7: Find the first nonzero row when we ignore the first two rows. continue. When
there is no nonzero rows left, stop.
Reader can try to show that two distince RRE matrices cannot be row equivalent.
This will prove uniqueness part of the statment.
8. APPLICATION-I OF ROW REDUCTION: INVERSE OF A MATRIX
19
The RRE matrix row equivalent to A is called row reduced echelon form of A.


0 0 4 1
0 3 0 1 

Example 2. We find the RRE form of 
0 0 0 0  .
0 4 2 0




0 0 4 1
0 3 0 1
0 3 0 1
0 0 4 1 



Apply R3 ↔ R4 . Get 
0 4 2 0. Apply R1 ↔ R2 . Get 0 4 2 0.
0 0 0 0
0 0 0 0


0 1 0 1/3
0 0 4 1 

Next apply R1 → 1/3R1 . Get 
0 4 2 0 . Apply R3 → R3 − 4R1 , get
0 0 0 0




0 1 0 1/3
0 1 0 1/3
0 0 4


1 

. Appy R2 → 1/4R2 , get 0 0 1 1/4 . Apply R3 → R3 −
0 0 2 −4/3
0 0 2 −4/3
0 0 0
0
0 0 0
0




0 1 0
1
0 1 0 1
0 0 1 1/4 
0 0 1 1/4



2R2 , get 
0 0 0 −11/6. Apply, R3 → −6/11R3 , get 0 0 0 1 . Apply
0 0 0 0
0 0 0
0


0 1 0 0
0 0 1 0 

R1 → R1 − R3 and R2 → R2 − 1/4R3 get 
0 0 0 1, which is RRE.
0 0 0 0
Remark 1. There are more than one ways of getting the row reduced echelon
form of a matrix.
8. Application-I of Row Reduction: Inverse of a Matrix
Algorithm to find the row reduced echelon form of a matrix, also determines whether
a square matrix is invertible and if it is invertible, finds its inverse.
Theorem 8.1.
(1) Suppose R is row reduced echelon and row equivalent to
A. Then A is invertible if and only if R is invertible.
(2) An n × n RRE matrix is invertible if and only if it is the identity matrix In .
(3) Suppose (A| I) is row equivalent to (R| B) and R is row reduced echelon.
Then A is invertible if and only if R = I and in this case B = A−1 .
Proof. Proofs of the first two statements are left to the reader. Suppose ρ is a
composition of finitely many elementary row operations such that ρ(A) = R. Then
20
2. SYSTEM OF LINEAR EQUATIONS
ρ(I) = B. Since R is an RRE, R and hence A is invertible if and only if R = I. Thus
ρ(I) · A = I so that A−1 = ρ(I) = B.


1 1 1
Example 3. Suppose A = 1 2 1. Apply elementary row operations on the
1 2 3
matrix (A|I) to find (R|B) where R is the RRE
 to A.
 matrix row equivalent
1 1 1 | 1 0 0
Apply R2 → R2 − R1 and then R3 → R1 on 1 2 1 | 0 1 0 and get
1 2 3 | 0 0 1


1 1 1 | 1 0 0
0 1 0 | −1 1 0. Then apply R1 → R1 − R2 and R3 → R3 − R1 and get
0 1 2 | −1 0 1


1 0 1 | 2 −1 0
0 1 0 | −1 1 0. Then apply R3 → 1/3R3 and then R1 → R1 − R3 and
0 0 2 | 0 −1 1


1 0 0 | 2 −1/2 −1/2
1
0 . Conclude that that A is invertible and A−1 =
get 0 1 0 | −1
0 0 1 | 0 −1/2 1/2


2 −1/2 −1/2
−1
1
0 .
0 −1/2 1/2
9. Application-II of Row Reduction: Rank of a Matrix
Suppose A is an n × n matrix with coefficients from F. A matrix obtained from A by
deleting certain numbers of rows and columns is called a submatrix of A. An integer
r ≥ 0 but r ≤ n is called the rank of A if there is an invertible submatrix of A of
size r × r but there is no invertible submatrix of A of size (r + 1) × (r + 1). If A is
invertible A has full rank n. The rank of A is zero if and only if A = 0.
The rank of a matrix can be found by using the process of row reduction. We
record a theorem charcterizing the rank of a matrix whose proof involves only the
concept of RRE matrix (or it follows in the section on row rank of Chaper 3).
Theorem 9.1.
(1) If A and B are row equivalent, then they have the same
rank.
(2) If A is an RRE matrix, then the rank of A is the number of nonzero rows.
(3) The rank of a matrix is the number of nonzero rows of the RRE form of the
matrix.
10. APPLICATION-III OF ROW REDUCTION: SOLVING SYSTEMS OF EQUATIONS
21



1 1 1 1 1
1


1
 is 3. The Rank of 1 2 1 3 1
0 1 0 2 0 
0
1 0 1 −1 1
0


1 0 1 −1 1
0 1 0 2 0 

is two. Student must show that this matrix is row equivalent to 
0 0 0 0 0 
0 0 0 0 0
which is row reduced echelon (so that the rank of the given matrix is 2).

0
0
Example 4. The rank of 
0
0
0
3
0
4
4
0
0
2
Remark 2. The rank of a matrix is a very important parameter. We will see its
use in several places in the remainder of the book.
10. Application-III of Row Reduction: Solving Systems of Equations
A homogeneous system of linear equations, AX = 0, always has a solution (for
instance, xi = 0 ∀ i). Suppose we have the following system of linear equations:
AX = B. Observe the following statements:
Theorem 10.1. If (A| B) is row equivalent to (A0 | B 0 ), then the two systems AX =
B and A0 X = B 0 have the same solutions.
Proof. Observe, ρ(P Q) = ρ(I) (P Q) = (ρ(I) P )Q = ρ(P ) Q. Thus, we have
AX = B if and only ρ(A)X = ρ(B) for any composition ρ of finitely many elementary
row operations.
Remark 3. If A is a row reduced echelon matrix then one can determine whether
AX = B has a solution and when there is a solution, can find all solutions just by
inspection.
Theorem 10.2. The system of linear equations AX = B has a solution if and only
if rank(A) = rank(A|B). This system has a unique solution if and only if rank(A) is
equal to the number of unknowns.
Proof. By Theorem 10.1, we may assume, A is a row reduced echelon matrix.
If the i-th row of A is zero but i-th row of B is nonzero (i.e., bi 6= 0), then AX = B
has no solution because the i-th equation of AX = B is 0 = bi which is absurd. Thus
we have, rank(A) < rank(A|B) =⇒ AX = B has no solution.
Next suppose rank(A) = rank(A|B). Assume that A has r nonzero rows and the
leading nonzero coefficient of the i-th row is in the ki -th column. Then by assumption,
bi = 0 for i > r. Let S = {1, 2, . . . , n} \ {k1 , k2 , . . . , kr }. Assign arbitrary values to
each xi for i ∈ S and find xkj from j-th equation (1 ≤ j ≤ r).
Call xk1 , . . . xkr dependent unknowns and xi with i ∈ S as free unknowns or independent unknowns.
22
2. SYSTEM OF LINEAR EQUATIONS
If rank(A) = number of unknowns, then there are no free unknowns so that
AX = B has a unique solution. Conversely, assume that AX = B has a unique
solution. In particular, there are no free unknowns. Thus, n − r = 0 =⇒ r = n. Now we discuss the three system of equations in Section 2 of the Chapter.
 
 

x1
3
1 1 1
Example 5. In a), A = 1 2 3 , B = 6, X = x2  and (A|B) =
1
0 1 2
x3


1 1 1 | 3
1 2 3 | 6 .
0 1 2 | 1
Apply elementary row operations
 on (A|B) to
 find the row reducedechelon form of the
1 0 −1 | 0
1 1 1 | 3
R1 →R1 −R2 ,R3 →R3 −R2
R2 →R2 −R1



0 1 2 | 3 .
0 1 2 | 3
first block. (A|B)
∼
∼
0 0 0 | −2
0 1 2 | 1
Since the third row of the first block is zero but the third entry of the second column
is nonzero the system has no solution.


 

 
1 1 1
3
x1





Example 6. In b), A = 1 2 3 , B = 6 , X = x2  and(A|B) =
x3
1 1 2
4


1 1 1 | 3
1 2 3 | 6  .
1 1 2 | 4
In this case,
(A|B)


1 1 1 | 3
R2 →R2 −R1 ,R3 →R3 −R1
0 1 2 | 3
∼
0 0 1 | 1


1 0 −1 | 0
R1 →R1 −R2
0 1 2 | 3
∼
0 0 1 | 1


1 0 0 | 1
R1 →R1 +R3 ,R2 →R2 −2R3
0 1 0 | 1  .
∼
0 0 1 | 1
Thus, r = 3 = n = m, ki = i for each i = 1, 2, 3 so that there is no independent
unknown. In fact the solution is x1 = x2 = x3 = 1 (unique).
10. APPLICATION-III OF ROW REDUCTION: SOLVING SYSTEMS OF EQUATIONS
23
 

 
x1
3
1 1




2 3 , B = 6 , X = x2  and
3
1 2
x3



1 1 1 | 3
1 | 3
R →R −R
3 | 6 2 ∼2 1 0 1 2 | 3
0 1 2 | 3
2 | 3


1 0 −1 | 0
R1 →R1 −R2 ,R3 →R3 −R2
0 1 2 | 3  .
∼
0 0 0 | 0

1

Example 7. In c), A = 1
0

1 1

1 2
(A|B) =
0 1
In this case, n = m = 3, r = 2, k1 = 1, k2 = 2. Thus, x3 is the independent unknown
so assign any arbitrary value to it, say x3 = λ. Then x1 − λ = 0 and x2 + 2λ = 3.
Thus the general solution of the system is (λ, 3 − 2λ, λ). Thus, there are infinitely
many solutions.
Example 8. Find λ ∈ R and µ ∈ R such that the following system of two
equations in two unknowns has (i) a solution, and (ii) has a unique solution.
x + λy = 1,
2x + y = µ.
1 λ 1
The augmented matrix of the system is
. Applying R2 → R2 −2R1 we have
2 1 µ
1
λ
1
. Thus, the system has no solution if 1 − 2λ = 0 and µ − 2 6= 0.
0 1 − 2λ µ − 2
In other words, the system is consistent (i.e., there is a solution) when either λ 6= 1/2
or µ = 2. If λ 6= 1/2, then there is no free unknown and hence there is a unique
solution (for every µ ∈ R). Finally, when λ = 1/2 and µ = 2, the system has an
infinitely many solution.
Remark 4. If a linear system of equations has more than one solutions then it
has infinitely many solutions. (However, this is not valid over a finite field.)