qr code vb.net source Rework example 1, this time using the second column, instead of the rst row. in .NET

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Rework example 1, this time using the second column, instead of the rst row.
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Now, by eq. 36, answer:
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Solution We ll follow the three steps as in example 1, as follows. 1. The elements of the second column are (for the determinant of example 1) a12 2; a22 1; a32 2
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* The value, D, of a given determinant of any order is the same regardless of the row or column we decide to use. Take, for example, the general 3rd-order determinant shown at the beginning of this section. If you carefully expand the determinant in terms of the elements of ANY row or ANY column, your answer should reduce to the same value D a11 a22 a33 a11 a23 a32 a12 a23 a31 a12 a21 a33 a13 a21 a32 a13 a22 a31 in every case. The same principle applies to determinants of any order.
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CHAPTER 3 Determinants and Equations
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Same procedure as in example 1, except now we use the elements of the second column, as follows.   6 2   3 a12 A12 a12 1 M12 a12 M12 2  4 3   3 5   4 a22 A22 a22 1 M22 a22 M22   4 3   3 5   a32 A32 a32 1 5 M32 a32 M32 2   6 2 By step 3, letting D be the value of the determinant, we have         6 2 3 5  2 3 5     D 2 6 2  4 3 4 3 and now, by eq. (36), D 2 18 8 9 20 2 6 30 79; same answer as in example 1: Example 3
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Find the value of the fourth-order determinant  4 3 1  2 1 4   6 0 0  1 3 7  3  2   5  2
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Solution Let us carry out the usual three-step solution, as follows. 1. In this particular case the easiest thing to do is to make use of the third row; this is because there are two zeros in the third row. We ll nd that the presence of the zeros will considerably reduce the work required to get the solution. Note that the elements of the third row are a31 6 2.  3 1   6 1 4   3 7  3   2   2 a32 0 a33 0 a34 5
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a31 A31 a31 1 4 M31 a31 M31 a32 A32 0 a33 A33 0
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a34 A34 a34 1 7 M34 a34 M34
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 4   5 2   1
3 1 3
 1   4   7
CHAPTER 3 Determinants and Equations
3. By step 3, letting D be the value  3 1   D 6 1 4  3 7 of the determinant, we have    4 3 1 3       2  5 2 1 4     1 3 7 2
We thus now have to nd the values of the two third-order determinants above. The results are shown below, in which both determinants have been expanded in terms of the elements of the rst column.   !    1 3 4 2 1 3 value of first      D 1 6 3  7 2   7 2  3 4 2  6 29 174 determinant:  1 D2 5 4 3   3 4  2  3 7   1 3   7 1 ! 1  5 45 225 4 value of second determinant:
The value of the given fourth-order determinant is, therefore, D D1 D2 174 225 51; final answer:
Problem 30 Find the value of the following determinant, by expanding in terms of the elements of the rst column.    3 6 1     7 4   5    1 2 3 Problem 31 Repeat problem 30, this time expanding in terms of the elements of the third row (answer must be same as that in problem 30). Problem 32 Find the value of the determinant   6    2   3
1 4 5
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