Homework 12.4-solutions

# 0 points find the value of the determinant 2 d x 3 1

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Unformatted text preview: (−2) = −60 . keywords: matrix, determinant, quadratic function, expansion by minors 003 10.0 points Find the value of the determinant 2 D= x −3 −1 y z −2 −1 2. D = 7x + 7y + 7z 10.0 points By evaluating the determinant, express f ( x) = 1 3 −4 . 1. D = −7x + 7y + 7z correct keywords: determinant 002 3 x −2 2 x2 0 −3 3. D = 7x − 7y − 7z 4. D = 7x + 7y − 7z 5. D = −7x − 7y − 7z mehmood (ajm4462) – Homework 12.4 – karakurt – (56295) Explanation: One way of computing the cross product 6. D = −7x − 7y + 7z Explanation: For any 3 × 3 determinant A B C a1 b1 c1 a2 b2 c2 =A 2 (−i − 2j − 3k) × (−3i + j − k) is to use the fact that b1 c1 b2 c2 i × j = k, j×k = i, k ×i = j, j×j = 0, k× k = 0. while −B a1 c1 a2 c2 −1 y −2 z −2 −1 a2 b2 −1 y b1 z −3 a1 3 +C . i× i = 0, For then Thus 2 D= =2 x + a × b = 5i + 8j − 7k . Alternatively, we can use the deﬁnition x z −3 −1 +3 x y −3 −2 = 2 (−y + 2z ) + (−x + 3z ) + 3 (−2x + 3y ) . i −1 −3 a×b = −2 1 = Consequently, j −2 1 −1 −3 i− −3 −1...
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## This homework help was uploaded on 02/19/2014 for the course M 56295 taught by Professor Odell during the Spring '10 term at University of Texas.

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