# 5.3 - -16 I didn’t check If it’s-16 how do you ﬁx it...

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Math 2940 Solutions, Fall 2011 Section 5 . 3 1 )(a) x 1 = det 2 4 1 5 2 4 3 5 det 2 4 2 5 1 4 3 5 x 2 = det 2 4 2 1 1 2 3 5 det 2 4 2 5 1 4 3 5 so x 1 = - 2, x 2 = 1. (b) x 1 = det 2 6 6 6 4 1 1 0 0 2 1 0 1 2 3 7 7 7 5 det 2 6 6 6 4 2 1 0 1 2 1 0 1 2 3 7 7 7 5 x 2 = det 2 6 6 6 4 2 1 0 1 0 1 0 0 2 3 7 7 7 5 det 2 6 6 6 4 2 1 0 1 2 1 0 1 2 3 7 7 7 5 x 3 = det 2 6 6 6 4 2 1 1 1 2 0 0 1 0 3 7 7 7 5 det 2 6 6 6 4 2 1 0 1 2 1 0 1 2 3 7 7 7 5 so x 1 = 3 / 4, x 2 = - 1 / 2, x 3 = 1 / 4. 7 ) If all the cofactors are 0 then the product of A and the cofactor matrix is 0, so the determinant of A is 0. On the other hand, A = ± 1 - 1 - 1 1 ² has cofactor matrix ± 1 1 1 1 ² but A is not invertible. 16 ) (a) We need to compute the determinant of ~ i ~ j ~ k 3 2 0 1 4 0 which is 10 ~ k and has magnitude 10. (c) The triangle’s area is 1 / 2 that of the parallelogram, 5. 21 ) The largest value of the determinant occurs when the columns are mutually perpendic- ular. Then its absolute value is the product of the lengths of the columns, L 1 L 2 L 3 L 4 . If all entries are ± 1, then each column has length 4 = 2 so the maximum determinant is 16. This is achieved by 1 1 1 1 1 1 - 1 - 1 1 - 1 - 1 1 1 - 1 1 - 1 . Actually, the determinant is either 16 or
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Unformatted text preview: -16, I didn’t check. If it’s-16 how do you ﬁx it? 28 ) sin φ cos θ ρ cos φ cos θ-ρ sin φ sin θ sin φ sin θ ρ cos φ sin θ ρ sin φ cos θ cos φ-ρ sin φ . Doing a cofactor expansion along the last column, our determinant is-ρ sin φ sin θ ³ ³ ³ ³ sin φ sin θ ρ cos φ sin θ cos φ-ρ sin φ ³ ³ ³ ³-ρ sin φ cos θ ³ ³ ³ ³ sin φ cos θ ρ cos φ cos θ cos φ-ρ sin φ ³ ³ ³ ³ which equals-ρ sin φ sin θ ((-ρ sin 2 φ ) sin θ-ρ cos 2 φ sin θ )-ρ sin φ cos θ ((-ρ sin 2 φ cos θ-ρ cos 2 φ cos θ ) = = ρ 2 sin φ sin 2 θ (sin 2 φ + cos 2 φ ) + ρ 2 sin φ cos 2 θ (sin 2 φ + cos 2 φ ) = ρ 2 sin φ sin 2 θ + ρ 2 sin φ cos 2 θ = ρ 2 sin φ....
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