3
(12)
(13)
(14)
(15)
(16)
(17)
(18)
Lecture 3 (Row equivalence, row reduction)
Students should describe how determinant of a square matrix would change if an elementary
row operation is applied (in each of the three cases).
If B is obtained from A by an elementary row operation then find how to get A from B.
Students have to describe the inverse operations in each cases. Students should note that an
elementary row operation is an invertible map from Mm×n (R) to itself.
Denote an elementary row operation by ρ. If A ∈ Mm×n (R) then ρ(A) = ρ(I) A (premultiplication of A by ρ(I)) where I is the m × m identity matrix. Students should verify
this fact in each case. If ρ is an elementary row operation then ρ(I) is referred to as an
elementary matrix.
Use the above rule successively on a matrix and get the following two statements for any finitely
many elementary row operations ρ1 , ρ2 , . . . , ρs :
(a) (ρs ◦ · · · ◦ ρ2 ◦ ρ1 )(A) = (ρs ◦ · · · ◦ ρ2 ◦ ρ1 )(I)A.
(b) (ρs ◦ · · · ◦ ρ2 ◦ ρ1 )(A) = ρs (I) · · · ρ2 (I)ρ1 (I)A.
A matrix A is said to be row equivalent to the matrix B if there is a finitely many elementary
row operations ρ1 , ρ2 , . . . , ρs such that B = (ρs ◦ · · · ◦ ρ2 ◦ ρ1 )(A).
Observe that row equivalence is an equivalence relation.
(a) 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).
(b) A is row equivalent to B if and only if B is row equivalent to A. If B = (ρs ◦· · ·◦ρ2 ◦ρ1 )(A),
−1
−1
then A = (ρ−1
1 ◦ · · · ◦ ρs−1 ◦ ρs )(B) and inverse of an elementary row operation is an
elementary row operation.
(c) 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.
One can write an algorithm to show, “Every matrix is row equivalent to a unique row reduced
echelon matrix.” This statement is same as “Every equivalence class of row equivalent matrices
contains a unique row reduced echelon 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 ).
Step3: Again apply interchange of rows to push up a row whose leading nonzero coefficient
occurs in 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.
0 0 4 1
0 3 0 1
0 0 4 1




0 3 0 1 

 Apply R3 ↔ R4 . Get 0 3 0 1. Apply R1 ↔ R2 . Get 0 0 4 1.Next
0 4 2 0 
0 4 2 0 
0 0 0 0 
0 0 0 0
0 0 0 0
0 4 2 0




0 1 0 1/3
0 1 0 1/3
0 0 4 1 
0 0 4
1 



apply R1 → 1/3R1 . Get 
0 4 2 0 . Apply R3 → R3 − 4R1 , get 0 0 2 −4/3.
0 0 0 0
0 0 0
0




0 1 0 1/3
0 1 0
1
0 0 1 1/4 


. Apply R3 → R3 − 2R2 , get 0 0 1 1/4 .
Appy R2 → 1/4R2 , get 
0 0 2 −4/3
0 0 0 −11/6
0 0 0
0
0 0 0
0
4


0 1 0 1
0 0 1 1/4

Apply, R3 → −6/11R3 , get 
0 0 0 1 . Apply R1 → R1 − R3 and R2 → R2 − 1/4R3 get
0 0 0 0


0 1 0 0
0 0 1 0 


0 0 0 1, which is RRE.
0 0 0 0
Remark: There are more than one ways of getting the row reduced echelon form of a matrix.
Lecture 4 (Rank and invertibility of Matrices)
(19) Following applications of row equivalence of matrices are to be discussed:
(a) Finding rank of a matrix. If A is row equivalent to the row reduced echelon matrix R then
the number of nonzero rows of R is called the rank of A. In particular, the rank of a row
reduced echelon matrix is the number of nonzero rows. We will see later that the row rank
of a matrix (i.e.
the rank of the matrix.
 dimension
 of the row space) is

0 0 4 1
1 1 1 1 1
0 3 0 1 
1 2 1 3 1 



The rank of 
0 0 0 0 is 3. The Rank of 0 1 0 2 0 is two. Student must
0 4 2 0
1 0 1 −1 1


1 0 1 −1 1
0 1 0 2 0 

show that this matrix is row equivalent to 
0 0 0 0 0 which is row reduced echelon
0 0 0 0 0
(so that the rank of the given matrix is 2).
Remark: We will see use of rank of a matrix.
(b) Determining whether a square matrix is invertible. In case it is invertible finding the
inverse. Theorem: Suppose R is row reduced echelon and row equivalent to A. Then A
is invertible if and only if R is the identity matrix. In fact,
Algorithm: 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 .
If (ρk ◦ · · · ◦ ρ2 ◦ ρ1 )(A) = I, then (ρk ◦ · · · ◦ ρ2 ◦ ρ1 )(I) · A = I. Therefore A−1 =
(ρk ◦ · · · ◦ ρ2 ◦ ρ1 )(I).


1 1 1
Example: Suppose A = 1 2 1. Apply elementary row operations on the matrix
1 2 3
(A|I) to find (R|B) where R is the RRE matrix
A.
 row equivalent to
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 get
0 0 2 | 0 −1 1




2 −1/2 −1/2
1 0 0 | 2 −1/2 −1/2
0 1 0 | −1
1
0 .
1
0 . Conclude that A−1 = −1
0 −1/2 1/2
0 0 1 | 0 −1/2 1/2
Remark: Algorithm to find row reduced echelon form of a matrix, also determines whether
a square matrix is invertible and if it is invertible, it can be used to find the inverse.
(c) Solving a system of linear equations. Obviously, 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:
5
i) If (A| B) is row equivalent to (A� | B � ), then the two systems AX = B and A� X = B �
have the same solutions.
Reasoning: Since ρ(P Q) = ρ(P )Q we have AX = B if and only ρ(A)X = ρ(B) for any
composition ρ of finitely many elementary row operations.
ii) If A is row reduced echelon matrix then one can determine whether AX = B has a
solution and what are all solutions just by inspection.
Reasoning: Suppose A is a row reduced echelon matrix. If there is a row i of (A|B) such
that i-th row of A a zero row but i-th row of B nonzero then AX = B has no solution.
Suppose AX = B has a solution. Assume that A has r nonzero rows and the leading
nonzero coefficient of i-th row is in the ki -th coefficients. Call xk1 , . . . xkr dependent unknowns. Assign arbitrary values to all xi for i �∈ {k1 , k2 , . . . , kr } then write xk1 , . . . , xkr as
a linear combination as the independent unknowns.
Discuss enough examples.
(We will discuss later, if the system is homogenous then the dimension of the solution
space is n − r.)