Properties of Determinant

- if
is obtained from
by interchanging two rows, then

- if
is obtained from
by
multiplying a row by
then

- if all the elements of one row is 0 then
- if is a square matrix having two rows equal then
- Let and be two matrices which differ from the matrix only in the row for some . If for then .
- if is obtained from by replacing the th row by itself plus times the th row, for then
- if is a triangular matrix then the product of the diagonal elements.
- If is an elementary matrix of order then .
- is invertible if and only if .
- If is an matrix then .
- , where recall that is the transpose of the matrix .

Let
be a transposition. Then by Proposition 14.2.4,
. Hence by the definition of determinant
and Example 14.2.14.2,
we have

**Proof of Part 2.** Suppose that
is obtained by multiplying the
row of
by
. Then
and
for
.
Then

**Proof of Part 3.** Note that
. So, each term in the expression for determinant, contains one
entry from each row. Hence, from the condition that
has a row consisting of all zeros, the value of
each term is 0
. Thus,
.

**Proof of Part 4.** Suppose that the
and
row of
are equal. Let
be the matrix obtained from
by interchanging the
and
rows. Then
by the first part,
But the assumption implies that
. Hence,
.
So, we have
Hence,
.

**Proof of Part 5.** By definition and the given assumption, we have

**Proof of Part 6.** Suppose that
is obtained from
by
replacing the
th row by itself plus
times the
th row, for
.
Then
and
for
.
Then

**Proof of Part 7.** First let us assume that
is an upper triangular matrix.
Observe that if
is different from the identity permutation
then
. So, for every
,
there exists a positive integer
(depending on
)
such that
. As
is an upper triangular matrix,
for each
. Hence the result follows.

A similar reasoning holds true, in case is a lower triangular matrix.

**Proof of Part 8.** Let
be the identity matrix of order
.
Then using Part 7,
.
Also, recalling the notations for the elementary matrices given in
Remark 2.3.14, we have
(using Part 1)
(using Part 2) and
(using Part 6).
Again using Parts 1, 2 and 6, we get
.

**Proof of Part 9.** Suppose
is invertible. Then by Theorem 2.5.8,
is a product of elementary matrices. That is, there exist elementary matrices
such that
. Now a repeated application of Part 8 implies that
. But
for
. Hence,
.

Now assume that . We show that is invertible. On the contrary, assume that is not invertible. Then by Theorem 2.5.8, the matrix is not of full rank. That is there exists a positive integer such that . So, there exist elementary matrices such that . Therefore, by Part 3 and a repeated application of Part 8,

But for . Hence, . This contradicts our assumption that . Hence our assumption is false and therefore is invertible.

**Proof of Part 10.** Suppose
is not invertible. Then by Part 9,
. Also, the product matrix
is also not invertible. So,
again by Part 9,
. Thus,
.

Now suppose that
is invertible. Then by Theorem 2.5.8,
is a product of elementary matrices. That is, there exist elementary matrices
such that
. Now a repeated application of Part 8 implies that

**Proof of Part 11.** Let
. Then
for
.
By Proposition 14.2.4, we know that
. Also
.
Hence,

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- The result that implies that in the statements made in Theorem 14.3.1, where ever the word ``row" appears it can be replaced by ``column".
- Let
be a matrix satisfying
and
for
. Let
be the
submatrix of
obtained by removing the first row and the first column. Then it can be easily
shown that
. The reason being is as follows:

for every with is equivalent to saying that is a permutation of the elements . That is, . Hence,

We are now ready to relate this definition of determinant with the one given in Definition 2.6.2.

Then by Theorem 14.3.1.5,

We now compute for . Note that the matrix can be transformed into by interchanges of columns done in the following manner:

first interchange the and column, then interchange the and column and so on (the last process consists of interchanging the column with the column. Then by Remark 14.3.2 and Parts 1 and 2 of Theorem 14.3.1, we have . Therefore by (14.3.6),

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A K Lal 2007-09-12