# lecture_13 - MA 265 LECTURE NOTES WEDNESDAY FEBRUARY 6...

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Unformatted text preview: MA 265 LECTURE NOTES: WEDNESDAY, FEBRUARY 6 Properties of Determinants (cont’d) Type II: Multiplying Rows and Columns. Let A be an n × n matrix. We explain how det( A ) changes if we apply an elementary operation of Type II. (1) Let B be that matrix obtained from A by multiplying the i th row by k . Then det( B ) = k det( A ). (2) Let B be that matrix obtained from A by multiplying the i th column by k . Then det( B ) = k det( A ). (3) Say that A has a row of zeros. Then det( A ) = 0. (4) Say that A has a column of zeros. Then det( A ) = 0. We explain why these statements are true. We begin by showing (1). We write the determinant of A as det( A ) = X σ ∈ S n ε ( σ ) a 1 σ (1) a 2 σ (2) ··· a nσ ( n ) . Let B be that matrix obtained from A by multiplying the i th row by k . If we write the i th row of A as r i = a i 1 a i 2 ··· a in = ⇒ k r i = k a i 1 k a i 2 ··· k a in ; We can express the determinant of B as det( B ) = X σ ∈ S n ε ( σ ) a 1 σ (1) ··· a i- 1 σ ( i- 1) ( k a iσ ( i ) ) a i +1 σ ( i +1) ··· a nσ ( n ) = k X σ ∈ S n ε ( σ ) a 1 σ (1) ··· a i- 1 σ ( i- 1) a iσ ( i ) a i +1 σ ( i +1) ··· a nσ ( n ) ! = k det( A ) . Now we prove the other statements. First we show (2). Let B be that matrix obtained from A by multiplying the i th column by k . Then B T is that matrix obtained from A T by multiplying the i th row by k . From (1), we see that det( B ) = det( B T ) = k det( A T ) = k det( A ). Now we show (3). Say that A has a row of zeros. Let B be that matrix obtained from A by multiplying this row by k = 0; then B = A . But from (1), we see that det( A ) = det( B ) = 0 · det( A ), so that det( A ) = 0. Finally we show (4). Say that A has a column of zeros. Let B be that matrix obtained from A by multiplying this column by...
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lecture_13 - MA 265 LECTURE NOTES WEDNESDAY FEBRUARY 6...

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