# Properties of Determinant

THEOREM 15.3.1 (Properties of Determinant)   Let be an matrix. Then
1. if is obtained from by interchanging two rows, then
2. if is obtained from by multiplying a row by then
3. if all the elements of one row is 0 then
4. if is a square matrix having two rows equal then
5. Let and be two matrices which differ from the matrix only in the row for some . If for then .
6. if is obtained from by replacing the th row by itself plus times the th row, for then
7. if is a triangular matrix then the product of the diagonal elements.
8. If is an elementary matrix of order then .
9. is invertible if and only if .
10. If is an matrix then .
11. , where recall that is the transpose of the matrix .

Proof. Proof of Part 1. Suppose is obtained from by the interchange of the and row. Then for and for .

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 12 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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Remark 15.3.2
1. 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".
2. 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.

THEOREM 15.3.3   Let be an matrix. Then where recall that is the submatrix of obtained by removing the row and the column.

Proof. For , define two matrices

Then by Theorem 14.3.1.5,

 (15.3.6)

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