University:
Whitman College
Course:
MATH-240 | Linear Algebra
Academic year:

2024
Views:

230

Pages:

2
Author:

catfever701r

EXAMPLE: Section 1.1, Problems 12, 13 and 28 12. Solve the system: x1 −3x2 +4x3 = −4 3x1 −7x2 +7x3 = −8 −4x1 +6x2 −x3 = 7 SOLUTION:     1 −3 4 −4 −3r1 +r2 →r2 1 −3 4 −4   4r1 +r3 →r3   →  0 2 −5 4   3 −7 7 −8  −4 6 −1 7 0 −6 15 −9 At this point, we could divide row 2 by 2, but that would introduce fractions. Alternatively, keep that for now.     1 −3 4 −4 1 −3 4 −4   3r2 +r3 →r3   →  0 2 −5 4   0 2 −5 4  3 0 0 0 0 −6 15 −9 Because the last row is translated to read: “0 = 3”, this system is inconsistent (has no solution). 13. Solve the system: x1 −3x3 = 8 2x1 +2x2 +9x3 = 7 x2 +5x3 = −2 SOLUTION:       1 0 −3 8 1 0 −3 8 1 0 −3 8   −2r1 +r2 →r2   r3 ↔r2   −2r2 +r3 →r2 2 2 9 7 0 2 15 −9 → → →      0 1 5 −2  0 1 5 −2 0 1 5 −2 0 2 15 −9    1 0 −3 8  0 1 5 −2    0 0 5 −5 1 r →r2 5 2 →    −5r3 +r2 →r2 1 0 −3 8 1 0 0 5   3r3 +r1 →r1  0 1 5 −2 0 1 0 3  →     0 0 1 −1 0 0 1 −1 Therefore, the solution is unique, and x1 = 5, x2 = 3, x3 = −1 (EXTRA PRACTICE: Verify that this is the solution by substitution of these numbers into the original equations!) 28. Find a, b, c, d so that the system below is consistent for all values of f and g. Assume that a 6= 0. ax + by = f cx + dy = g SOLUTION: Form the augmented matrix: " a b f c d g # c Multiply the first row by − (which is always possible since a 6= 0), and add to the a second row: # " a b f 0 d − bca g − fac For this system to be consistent for any f, g, we must have that d− bc ad − bc 6= 0 ⇒ 6= 0 ⇒ ad − bc 6= 0 a a Side Remark 1: If d − bca = 0, it is possible to have a consistent system- if g − fac is also 0. But in this case, the system is not consistent for all f, g. Side Remark 2: You might recall that ad − bc is a special number for the matrix " . It is called the determinant. a b c d #

Linear Algebra Example Section 1.1, Problems 12, 13 and 28

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