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Linear Algebra, Theory And Applications, 2012a

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3.1. BASIC TECHNIQUES AND PROPERTIES 79<br />

where the i th row of A 1 is (a 1 , ··· ,a n ) and the i th row of A 2 is (b 1 , ··· ,b n ) , all other rows<br />

of A 1 and A 2 coinciding with those of A. In other words, det is a linear function of each<br />

row A. The same is true with the word “row” replaced with the word “column”. In addition<br />

to this, if A and B are n × n matrices, then<br />

and if A is an n × n matrix, then<br />

det (AB) =det(A)det(B) ,<br />

det (A) =det ( A T ) .<br />

This theorem implies the following corollary which gives a way to find determinants. As<br />

I pointed out above, the method of Laplace expansion will not be practical for any matrix<br />

of large size.<br />

Corollary 3.1.9 Let A be an n×n matrix and let B be the matrix obtained by replacing the<br />

i th row (column) of A with the sum of the i th row (column) added to a multiple of another<br />

row (column). Then det (A) =det(B) . If B is the matrix obtained from A be replacing the<br />

i th row (column) of A by a times the i th row (column) then a det (A) =det(B) .<br />

Here is an example which shows how to use this corollary to find a determinant.<br />

Example 3.1.10 Find the determinant of the matrix<br />

⎛<br />

⎞<br />

1 2 3 4<br />

A = ⎜ 5 1 2 3<br />

⎟<br />

⎝ 4 5 4 3 ⎠<br />

2 2 −4 5<br />

Replace the second row by (−5) times the first row added to it. Then replace the third<br />

row by (−4) times the first row added to it. Finally, replace the fourth row by (−2) times<br />

the first row added to it. This yields the matrix<br />

B =<br />

⎛<br />

⎜<br />

⎝<br />

1 2 3 4<br />

0 −9 −13 −17<br />

0 −3 −8 −13<br />

0 −2 −10 −3<br />

and from the above corollary, it has the same determinant as A. Now using the corollary<br />

some more, det (B) = ( )<br />

−1<br />

3 det (C) where<br />

⎛<br />

⎞<br />

1 2 3 4<br />

C = ⎜ 0 0 11 22<br />

⎟<br />

⎝ 0 −3 −8 −13 ⎠ .<br />

0 6 30 9<br />

The second row was replaced by (−3) times the third row added to the second row and then<br />

thelastrowwasmultipliedby(−3) . Now replace the last row with 2 times the third added<br />

to it and then switch the third and second rows. Then det (C) =− det (D) where<br />

D =<br />

⎛<br />

⎜<br />

⎝<br />

1 2 3 4<br />

0 −3 −8 −13<br />

0 0 11 22<br />

0 0 14 −17<br />

⎞<br />

⎟<br />

⎠<br />

⎞<br />

⎟<br />

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