To define the rank of a matrix, we have to know about sub-matrices and minors of a matrix.

**Rank of a Matrix**

To define the rank of a matrix, we have to know about sub-matrices
and minors of a matrix.

Let *A *be a given matrix. A matrix obtained by deleting some
rows and some columns of *A *is called a **sub-matrix **of *A*. A matrix is a sub-matrix of itself
because it is obtained by leaving zero number of rows and zero number of
columns.

Recall that the determinant of a square sub-matrix of a matrix
is called a minor of the matrix.

The **rank of a matrix ***A *is defined as the order of a highest order
non-vanishing minor of the matrix *A*. It is denoted by the symbol *ρ** *(*A*). The rank of
a zero matrix is defined to be 0.

i.
If a matrix contains at-least one non-zero element, then *ρ** *( *A*) ≥ 1.

ii.
The rank of the identity matrix *I _{n} *is

iii.
If the rank of a matrix *A *is *r*, then there exists
at-least one minor of *A *of order *r *which does not vanish and
every minor of *A *of order *r *+1 and higher order (if any) vanishes.

iv.
If *A *is an *m *× *n *matrix, then *ρ** *(*A*) ≤ min{*m*, *n*} = minimum of *m*, *n*.

v.
A square matrix *A *of order *n *has inverse if and
only if *ρ** *( *A*) = *n*.

Find the rank of each of the following matrices:

**Solution**

(i) Let A =. Then A is a matrix of order 3× 3. So ρ(A) ≤ min {3, 3} = 3. The highest order of minors of A is 3 . There is only one third order minor of A .

It is = 3 (6− 6) − 2 (6−6) + 5 (3 − 3) = 0. So, ρ(A) < 3.

Next consider the
second-order minors of A .

We find that the second order minor = 3 − 2 = 1 ≠ 0 . So ρ(A) = 2 .

(ii) Let A = . Then A is a matrix of order 3×4 . So ρ(A) ≤ min {3, 4} = 3.

The highest order of minors of A is 3 . We search for a non-zero
third-order minor of A . But

we find that all of them vanish. In fact, we have

So, ρ(*A*) < 3. Next, we search for a non-zero second-order minor of A .

We find that = -4+9 =5 ≠ 0 . So, ρ(A) = 2 .

Finding the rank of a matrix by searching a highest order
non-vanishing minor is quite tedious when the order of the matrix is quite
large. There is another easy method for finding the rank of a matrix even if
the order of the matrix is quite high. This method is by computing the rank of
an equivalent row-echelon form of the matrix. If a matrix is in row-echelon
form, then all entries below the **leading diagonal **(it is the line
joining the positions of the diagonal elements *a*_{11} , *a*_{22}
, *a*_{33} ,L. of the matrix) are
zeros. So, checking whether a minor is zero or not, is quite simple.

Find the rank of the following matrices which are in row-echelon
form :

**Solution**

(i) Let A = . Then A is a matrix of order 3 3 × and ρ(A) ≤ 3

The third order minor |A| = = (2) (3)( 1) = 6 ≠ 0 . So, ρ(A) = 3 .

**Note that there are
three non-zero rows. **

(ii) Let A = . Then A is a matrix of order 3× 3 and ρ(A)
≤ 3.

The only third order minor is |A| = = (-2) (5) (0) = 0
. So ρ(A) ≤ 2 .

There are several second order minors. We find that there is a second order minor, for example, = (-2)(5) = -10 ≠ 0 . So, ρ(A) = 2.

**Note that there are
two non-zero rows. The third row is a zero row.**

(iii) Let A = . Then A is a matrix of order 4 × 3 and ρ(A) ≤ 3.

The last two rows are zero rows. There are several second order minors. We find that there is a second order minor, for example, = (6) (2) = 12 ≠ 0 . So, ρ(A) = 2.

**Note that there are
two non-zero rows. The third and fourth rows are zero rows.**

We observe from the above example that the rank of a matrix in row
echelon form is equal to the number of
non-zero rows in it. We state this observation as a theorem without proof.

**Theorem
1.11**

The rank of a matrix in row echelon form is the number of
non-zero rows in it.

The rank of a matrix which is not in a row-echelon form, can be
found by applying the following result which is stated without proof.

**Theorem
1.12**

The rank of a non-zero matrix is equal to the number of non-zero
rows in a row-echelon form of the matrix.

Find the rank of the matrix by reducing it to a row-echelon form.

Let A = . Applying elementary row operations, we get

The
last equivalent matrix is in row-echelon form. It has two non-zero rows. So, ρ
(A)= 2.

Find the rank of the matrix by reducing it to a row-echelon form.

Let
A be the matrix. Performing elementary row operations, we get

The
last equivalent matrix is in row-echelon form. It has three non-zero rows. So,
ρ(*A*) = 3 .

Elementary
row operations on a matrix can be performed by pre-multiplying the given matrix
by a special class of matrices called elementary matrices.

An **elementary matrix **is defined as a matrix which is obtained from an
identity matrix by applying only one elementary transformation.

If we are dealing with matrices with three rows, then all
elementary matrices are square matrices of order 3 which are obtained by
carrying out only one elementary row operations on the unit matrix *I _{3}.
*Every elementary row operation that is carried out on a given matrix

For instance, let us consider the matrix A =

Suppose that we do the transformation R_{2} → R_{2 }+ λR_{3} on A, where
λ ≠ 0 is a constant. Then, we get

The matrix is an elementary matrix, since we have

Pre-multiplying A by , we get

From (1) and (2), we get

So, the effect of applying the elementary transformation R_{2}
→ R_{2 }+ λR_{3} on A is the same as that of pre-multiplying
the matrix A with the elementary matrix

Similarly, we can show that

(i) the effect of applying the elementary transformation R_{2}↔
R_{3 }on A is the same as that of
pre-multiplying the matrix A with the elementary matrix

(ii) the effect of applying the elementary transformation R_{2 }→
R_{2}λ on A is the same as that of
pre-multiplying the matrix A with the elementary matrix

We state the following result without proof.

Every non-singular matrix can be transformed to an identity
matrix, by a sequence of elementary row operations.

As an illustration of the above theorem, let us consider the
matrix *A *=

Then, |A| = 12+ 3 = 15
≠ 0. So, A is non-singular. Let us
transform A into I_{2} by a sequence of
elementary row operations. First, we search for a row operation to make
a_{11} of A as 1. The elementary row operation needed for this is R_{1}
→ (1/2) R_{1}. The corresponding elementary matrix is

Next, let us make all elements below a_{11} of E_{1}A
as 0. There is only one element a_{21}.

The elementary row operation needed for this is R_{2} → R_{2} + (−3) R_{1} .

The corresponding elementary matrix is E_{2} =

Next, let us make a_{22} of E_{2}(E_{2}A)
as 1. The elementary row operation needed for this is

The corresponding elementary matrix is E_{3} =

Then, we get E_{3}(E_{2}(E_{1}A)) =

Finally, let us find an elementary row operation to make a_{12}
of E_{3}(E_{2}(E_{1}A)) as 0. The elementary row operation needed for this is R_{1}
→ R_{1} + (1/2) R_{2}. The corresponding elementary matrix is

We write the above sequence of elementary transformations in the
following manner:

Show that the matrix is non-singular and reduce it to the identity matrix by elementary row transformations.

Let A = .Then, |A| = 3 (0+2 ) – 1(2+5) + 4(4-0) = 6-7+16
≠ 0. So, A is non-singular. Keeping the identity matrix as our goal, we perform
the row operations sequentially on A as follows:

Tags : Definition, Theorem, Formulas, Solved Example Problems | Elementary Transformations of a Matrix , 12th Mathematics : UNIT 1 : Applications of Matrices and Determinants

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12th Mathematics : UNIT 1 : Applications of Matrices and Determinants : Rank of a Matrix | Definition, Theorem, Formulas, Solved Example Problems | Elementary Transformations of a Matrix

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