# sol7 - SOLUTIONS TO HOMEWORK 7 Section 4.1 1(a No Indeed...

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Unformatted text preview: SOLUTIONS TO HOMEWORK 7 Section 4.1 1. (a) No . Indeed, for any A ∈ M 2 × 2( F ) and c ∈ F , det( cA ) = c 2 det( A ), which may be different from c det( A ). (b) Yes . (c) No . A is invertible iff det( A ) negationslash = 0. (d) No – you have to take the absolute value of the determinant. (e) Yes . 3. (a) det parenleftbigg − 1 + i 1 − 4 i 3 + 2 i 2 − 3 i parenrightbigg = ( − 1 + i )(2 − 3 i ) − (1 − 4 i )(3 + 2 i ) = ( − 2 + 3 + 5 i ) − (3 + 8 − 10 i ) = − 10 + 15 i. 4. (a) area of parallelogram = vextendsingle vextendsingle vextendsingle det parenleftbigg u v parenrightbigg vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle det parenleftbigg 3 − 2 2 5 parenrightbigg vextendsingle vextendsingle vextendsingle = | 3 · 5 − ( − 2) · 2 | = 19 . 5. Consider A = parenleftbigg A 11 A 12 A 21 A 22 parenrightbigg . Then B = parenleftbigg A 21 A 22 A 11 A 12 parenrightbigg , and det B = A 21 A 21 − A 11 A 22 = − ( A 11 A 22 − A 21 A 21 ) = − det A. 9. Consider A = parenleftbigg A 11 A 12 A 21 A 22 parenrightbigg and B = parenleftbigg B 11 B 12 B 21 B 22 parenrightbigg . Then AB = parenleftbigg A 11 B 11 + A 12 B 21 A 11 B 12 + A 12 B 22 A 21 B 11 + A 22 B 21 A 21 B 12 + A 22 B 22 parenrightbigg , hence det( AB ) = ( A 11 B 11 + A 12 B 21 )( A 21 B 12 + A 22 B 22 ) − ( A 11 B 12 + A 12 B 22 )( A 21 B 11 + A 22 B 21 ) = A 11 A 22 B 11 B 22 + A 12 A 21 B 12 B 21 − A 11 A 22 B 12 B 21 − A 12 A 21 B 11 B 22 = ( A 11 A 22 − A 12 A 21 )( B 11 B 22 − B 12 B 21 ) = det A det B. 1 2 SOLUTIONS TO HOMEWORK 7 Section 4.2 3. k = 42 . Indeed, write X = a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 = R 1 R 2 R 3 , where R i is the i-th row of our matrix ( R 1 = ( a 1 , a 2 , a 3 ) etc.). In this notation, the matrix in the left hand side can be written as Y = 2 R 1 3 R 2 + 5 R 3 7 R 3 . By Theorem 4.3, det Y = det 2 R 1 3 R 2 7 R 3 + det 2 R 1 5 R 3 7 R 3 = 2 · 3 · 7 det R 1 R 2 R 3 + 2 · 5 · 7 det R 1 R 3 R 3 . By Corollary on p. 215, det R 1 R 3 R 3 = 0, hence det Y = 42 det X ....
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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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sol7 - SOLUTIONS TO HOMEWORK 7 Section 4.1 1(a No Indeed...

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