fina_ex_sol - SOLUTIONS FOR FINAL 1(15 points Consider 7 ×...

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Unformatted text preview: SOLUTIONS FOR FINAL 1 (15 points) : Consider 7 × 7 matrices A and B , such that det A = 2, and det B = − 5. (a) Compute det( A 2 B t ). (b) The matrix C is obtained from A by interchanging Columns 2 and 5. Compute det C . (c) The matrix D is obtained from B by subtracting Row 4 from Row 3. Compute det D − 1 . (a) det( A 2 B t ) = (det A ) 2 det B t = (det A ) 2 det B = 2 2 ( − 5) =- 20 . (b) Interchanging columns changes the sign of the determinant. Thus, det C = − det A =- 2 . (c) det D = det B = − 5, hence det D − 1 = (det D ) − 1 =- 1 / 5 . 2 (10 points) : Find the equation of a line, passing through the points (2 , , 1) and (0 , − 2 , 2). The line is the set of all points (2 , , 1) + ( (2 , , 1) − (0 , − 2 , 2) ) t = (2 , , 1) + (2 , 2 , − 1) t = (2 + 2 t, 2 t, 1 − t ) where t ∈ R . 3 (15 points) : Do the vectors v 1 = (1 , , 1 , − 2), v 2 = (0 , 2 , 1 , 1), v 3 = (1 , 4 , 3 , 0), v 4 = (0 , 2 , − 2 , 1), and v 5 = (2 , 2 , , 1) generate R 4 ? If yes, select a subset of { v 1 ,v 2 ,v 3 ,v 4 ,v 5 } that is a basis for R 4 . Let’s form a matrix A , whose columns are the vectors from our system: A = 1 0 1 2 2 4 2 2 1 1 3 − 2 0 − 2 1 0 1 1 The columns of A span R 4 iff rank A = 4. We compute the rank of A using Gaussian Elimination: A −→ R 3 → R 3 − R 1 ,R 4 → R 4 +2 R 1 1 0 1 2 0 2 4 2 2 0 1 2 − 2 − 2 0 1 2 1 5 −→ R 2 → R 2 / 2 1 0 1 2 0 1 2 1 1 0 1 2 − 2 − 2 0 1 2 1 5 −→ R 3 → R 3 − R 2 ,R 4 → R 4 − R 2 1 0 1 2 0 1 2 1 1 0 0 0 − 3 − 3 0 0 0 4 −→ R 3 →− R 3 / 3 ,R 4 → R 4 / 4 1 0 1 0 2 0 1 2 1 1 0 0 0 1 1 0 0 0 0 1 1 2 SOLUTIONS FOR FINAL −→ R 3 → R 3 − R 4 ,R 2 → R 2 −...
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This note was uploaded on 02/02/2012 for the course MATH 115A 262398211 taught by Professor Fuckhead during the Spring '10 term at UCLA.

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fina_ex_sol - SOLUTIONS FOR FINAL 1(15 points Consider 7 ×...

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