Homework 4 Solution

Evalute 8 6 4 6 3 0 4 3 det a2 b 23 det a 5 0 2 0 2 a

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Unformatted text preview: −7 = det A 5 y = x2 = −9 4 0 4 −2 0 2. Evalute −8 6 −4 6 −3 0 4 3 det A2 [B ] 23 = det A 5 0 2 0 2 a. [1 point] . . . using Expansion by Minors (at least once). Solution: The easiest way is to expand by minors along the third column, then use Sarrus’s Method: −9 4 0 4 −2 0 −8 6 −4 6 −3 0 0 −9 4 0 2 = 0 · | ∗ | − 0 · | ∗ | + (−4) · 4 −2 2 − 0 · | ∗ | 0 6 −3 2 2 = (−4) · [(−9)(−2)(2) + (4)(2)(6) + (0)(4)(−3) − (0)(−2)(6) − (4)(4)(2) − (−9)(2)(−3)] = (−4) · [36 + 48 − 32 − 54] = (−4)(−2) = 8. =⇒ 1 b. [1 point] . . . using Gaussian Elimination. Solution: A typical solution (avoiding integers when possible) is: −9 4 0 4 −2 0 −8 6 −4 6 −3 0 0 2 0 2 ==== === 1−3 −1 −2 4 4 −2 0 −8 6 −4 6 −3 0 0 2 0 2 1 2 −4 4 −2 0 === == (−1) −8 6 −4 −1 6 −3 0 0 2 0 2 1 2 −4 0 1 2 −4 0 ===== ===== 0 −10 16 2 0 −10 16 2 2 −4 1 == ===== (−1) = (−1) 0 2 −4 4 0 22 −36 0 3 +8 1 3 +2 2 0 −15 24 2 0 −15 24 2 4 −6 1 1 2 −4 0 1 2 −4 0 0 1 −2 2 0 2 −4 4 == ===== (−1)(−1)(2) = = = = (−1)(−1) ==== 0 −10 16 2 0 −10 16 2 2 ÷2 2↔3 0 −15 24 2 0 −15 24 2 1 2 −4 0 1 2 −4 0 0 1 −2 2 0 1 −2 2 = = = = = (−1)(−1)(2) ===== === ==== (−1)(−1)(2) 3 + 10 2 0 0 2 −10 0 0 −4 22 3−4 4 + 15 2 0 0 −6 32 0 0 −6 32 1 2 −4 0 0 1 −2 2 = (2) · [1 · 1 · 2 · 2] = 8. = = = = (−1)(−1)(2) ==== 0 0 2 −10 4 +3 3 000 2 2...
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This note was uploaded on 01/16/2014 for the course MAT 242 taught by Professor Kaliszewski during the Spring '02 term at ASU.

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