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Unit19

# Unit19 - Unit 19 Properties of Determinants Theorem 19.1...

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Unformatted text preview: Unit 19 Properties of Determinants Theorem 19.1. Suppose A and B are identical n × n matrices with the exception that one row (or column) of B is obtained by multiplying the cor- responding row (or column) of A by some nonzero constant c . Then det B = c (det A ) This theorem may be used in a variety of situations to make a seemingly difficult determinant (due to large numbers or fractions involved) into a much easier one, as the following examples demonstrate. Example 1 . Find det A where A = 42 5 1 84 0 0 63 5 2 Solution: The numbers in the first column have a common factor of 21. The arithmetic will be easier if we factor it out. That is, we have the matrix 42 5 1 84 0 0 63 5 2 = 21 × 2 5 1 21 × 4 0 0 21 × 3 5 2 so the matrix 42 5 1 84 0 0 63 5 2 can be obtained from the matrix 2 5 1 4 0 0 3 5 2 by multiplying the first column by 21. Thus, Theorem 19.1 tells us that det A = det 42 5 1 84 0 0 63 5 2 = 21 × det 2 5 1 4 0 0 3 5 2 1 and so, expanding along row 2, we have det A = 21 × (4 × (- 1) 2+1 × det bracketleftbigg 5 1 5 2 bracketrightbigg + 0 + 0) =- 84 × 5 =- 420 Notice: What we saw here is that we can use Theorem 19.1 by factoring a constant, c , out of every entry in some particular row or column of the matrix and then multiplying the determinant of the resulting matrix by the constant. For instance, for a 3 × 3 matrix, we can factor out a common factor in row 1 and find the determinant as det c × a 11 c × a 12 c × a 13 a 21 a 22 a 23 a 31 a 32 a 33 = c × det a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 Example 2 . Find det A where A = 39 26 104 3 6 12 32 Solution: We can see that det A is most easily found by expanding along column 2. However, we can simplify the arithmetic by factoring out common factors in some rows and columns. To begin with, we see that column 3 contains only even numbers (and 2 is the largest integer factor shared by these numbers) so we can factor out a 2 from this column: det A = det 39 26 104 3 6 12 32 = 2 × det 39 26 52 3 3 12 0 16 Now, we can see fairly easily that row 1 contains only multiples of 13, so we can factor 13 out of row 1 to get: det A = (13 × 2) × det 3 2 4 3 0 3 12 0 16 Also, we can factor a 3 out of column 1 and after that, a 4 out of row 3. 2 This gives: det A = 3 × 26 × det 1 2 4 1 0 3 4 0 16 = 4 × 3 × 26 × det 1 2 4 1 0 3 1 0 4 Finally, we expand along column 2. det A = 12 × 26 × parenleftbigg 2(- 1) 1+2 det bracketleftbigg 1 3 1 4 bracketrightbigg + 0 + 0 parenrightbigg = 312 × (- 2) × (4- 3) =- 624 × 1 =- 624 Example 3 . Find det B , where B = 1 2- 1 6 1- 2 3 1 3 1 4- 1 4 Solution: We know we will want to expand along column 3, but let’s see if we can get rid of the fractions inside the matrix. Notice that the (1we can get rid of the fractions inside the matrix....
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Unit19 - Unit 19 Properties of Determinants Theorem 19.1...

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