(New page: '''<u> Chapter 6: Determinants</u>'''<u></u> <u></u>I will show several problems where I find the determinant, illustrating the several methods of doing this. <u></u><u>6.1</u> <u>...) |
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| − | '''<u> Chapter 6: Determinants</u>'''<u></u> | + | '''<u> Chapter 6: Determinants</u>'''<u></u> |
| − | <u></u>I will show several problems where I find the determinant, illustrating the several methods of doing this. | + | <u></u>I will show several problems where I find the determinant, illustrating the several methods of doing this. |
| − | <u></u><u>6.1</u> | + | <u></u><u>6.1</u> |
| − | <u></u>2. For any 2x2 matrix A, det(A) = ad - bc, so det(A) = (2)(5) - (3)(4) = -2 . Since this is not 0, A is invertible. | + | <u></u>2. For any 2x2 matrix A, det(A) = ad - bc, so det(A) = (2)(5) - (3)(4) = -2 . Since this is not 0, A is invertible. |
| − | 5. Let's use Laplace Expansion and expand across the first column. Remember to alternate signs. | + | 5. Let's use Laplace Expansion and expand across the first column. Remember to alternate signs. |
| − | det(A) = (1)(2)(det(A<sub>11</sub>) + (-1)(5)(det(A<sub>12</sub>) + (1)(7)(det(A<sub>13</sub>)<sub></sub> | + | det(A) = (1)(2)(det(A<sub>11</sub>) + (-1)(5)(det(A<sub>12</sub>) + (1)(7)(det(A<sub>13</sub>)<sub></sub> |
| − | = (2 * 55) + (-5 * 0) + (7 * 0) = 110 The matrix is invertible | + | = (2 * 55) + (-5 * 0) + (7 * 0) = 110 The matrix is invertible |
| − | 6. Determinant of a upper- or lower-triangular matrix is simply the product of the diagonal entries. | + | 6. Determinant of a upper- or lower-triangular matrix is simply the product of the diagonal entries. |
| − | det(A) = (6)(4)(1) = 24 The matrix is invertible | + | det(A) = (6)(4)(1) = 24 The matrix is invertible |
| − | 8. For any 3x3 matrix A with column vectors '''''u, v, w''''', determinant of A is '''''u · v'''''x'''''w''' | + | 8. For any 3x3 matrix A with column vectors '''''u, v, w''''', determinant of A is '''''u · v'''''x'''''w''''' |
| − | + | det(A) = [1 1 3] · ([2 1 2] x [3 1 1]) | |
| − | = [1 1 3] · [-1 4 -1] | + | = [1 1 3] · [-1 4 -1] |
| − | = 0 The matrix is not invertible. | + | = 0 The matrix is not invertible |
| + | |||
| + | 41. Remember, det(A) = Σ (sgn P)(prod P). In this matrix, two nonzero patterns exist: (2 -> 3 -> 1 -> 2 -> 4), with 5 inversions, and (2 -> 3 -> 3 -> 2 -> 2), with 8 inversions.<br> | ||
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| + | det(A) = (-1)<sup>5</sup>(2 * 3 * 1 * 2 * 4) + (-1)<sup>8</sup>(2 * 3 * 3 * 2 * 2) | ||
| + | |||
| + | = (-48) + (72) = 24 | ||
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| + | |||
| + | |||
| + | <u>6.3</u> | ||
| + | |||
| + | <u></u>22. Cramer's Rule states that in the system A'''x''' = '''b''' , where A is an invertible n x n matrix, the components x<sub>i</sub> of the solution vector are x<sub>i</sub> = det(A'''<sub>b</sub>'''<sub>,i</sub>)/det(A), where A'''<sub>b</sub>'''<sub>,i</sub> is the matrix obtained by replacing the ith column of A with '''b''' | ||
| + | |||
| + | x<sub>1</sub> = det([[1 7][3 11]])/det(A) = (-10) / (5) = -2 | ||
| + | |||
| + | x<sub>2</sub> = det([[3 1][4 3]])/det(A) = (5) / (5) = 1 | ||
| + | |||
| + | '''x''' = [-2 1] | ||
Revision as of 06:31, 6 May 2011
Chapter 6: Determinants
I will show several problems where I find the determinant, illustrating the several methods of doing this.
6.1
2. For any 2x2 matrix A, det(A) = ad - bc, so det(A) = (2)(5) - (3)(4) = -2 . Since this is not 0, A is invertible.
5. Let's use Laplace Expansion and expand across the first column. Remember to alternate signs.
det(A) = (1)(2)(det(A11) + (-1)(5)(det(A12) + (1)(7)(det(A13)
= (2 * 55) + (-5 * 0) + (7 * 0) = 110 The matrix is invertible
6. Determinant of a upper- or lower-triangular matrix is simply the product of the diagonal entries.
det(A) = (6)(4)(1) = 24 The matrix is invertible
8. For any 3x3 matrix A with column vectors u, v, w, determinant of A is u · vxw
det(A) = [1 1 3] · ([2 1 2] x [3 1 1])
= [1 1 3] · [-1 4 -1]
= 0 The matrix is not invertible
41. Remember, det(A) = Σ (sgn P)(prod P). In this matrix, two nonzero patterns exist: (2 -> 3 -> 1 -> 2 -> 4), with 5 inversions, and (2 -> 3 -> 3 -> 2 -> 2), with 8 inversions.
det(A) = (-1)5(2 * 3 * 1 * 2 * 4) + (-1)8(2 * 3 * 3 * 2 * 2)
= (-48) + (72) = 24
6.3
22. Cramer's Rule states that in the system Ax = b , where A is an invertible n x n matrix, the components xi of the solution vector are xi = det(Ab,i)/det(A), where Ab,i is the matrix obtained by replacing the ith column of A with b
x1 = det([[1 7][3 11]])/det(A) = (-10) / (5) = -2
x2 = det([[3 1][4 3]])/det(A) = (5) / (5) = 1
x = [-2 1]
