psLinAlgebra3Ans

psLinAlgebra3Ans - Fall 2007 ARE211 Problem Set#08 Answer...

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Unformatted text preview: Fall 2007 ARE211 Problem Set #08- Answer key Third Lin Algebra Problem Set Due date: Nov 05 Problem 1 Let A = 3 3- 1 0- 2- 3 1 0- 6- 2 1 0 0 1 and B = - 1 0 0 0 2 0 0 0 3 0 0 0- 6 a) Show that A and B are nonsingular. b) Calculate A- 1 , B- 1 , ( AB )- 1 , and ( A T )- 1 . c) Solve the linear equation system Ax = 1- 1 and ( AB ) x = 1 1 Ans: a) emphShow that A and B are nonsingular. A matrix is nonsingular if it’s determinant is nonzero. det ( A ) = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 3 3- 1 0- 2- 3 1 0- 6- 2 1 0 0 1 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle (Develop after the fouth column) = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 3 3- 1- 2- 3 1- 6- 2 1 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle (Develop after the third column) 1 2 = (- 1) vextendsingle vextendsingle vextendsingle vextendsingle- 2- 3- 6- 2 vextendsingle vextendsingle vextendsingle vextendsingle + (- 1) vextendsingle vextendsingle vextendsingle vextendsingle 3 3- 6- 2 vextendsingle vextendsingle vextendsingle vextendsingle + vextendsingle vextendsingle vextendsingle vextendsingle 3 3- 2- 3 vextendsingle vextendsingle vextendsingle vextendsingle =- (4- 18)- (- 6 + 18) + (- 9 + 6) = 14- 12- 3 =- 1 negationslash = 0 Note that for a diagonal matrix the determinant is simply the product of the diagonal elements: det ( B ) = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle- 1 0 0 0 2 0 0 0 3 0 0 0- 6 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = 36 negationslash = 0 b) Let’s first calculate the inverse of A 3 3- 1 0 | 1 0 0 0- 2- 3 1 0 | 0 1 0 0- 6- 2 1 0 | 0 0 1 0 0 1 | 0 0 0 1 Add the 3 rd row to the 1 st row and*(-1) 3 rd row to the 2 nd row: ⇔ - 3 1 0 0 | 1 0 1 0 4- 1 0 0 | 0 1- 1 0- 6- 2 1 0 | 0 0 1 0 0 0 1 | 0 0 0 1 Add the 2 nd row to the 1 st row: ⇔ 1 0 0 0 | 1 1 0 0 4- 1 0 0 | 0 1- 1 0- 6- 2 1 0 | 0 0 1 0 0 0 1 | 0 0 0 1 Add 6* 1 st row to the 3 rd row and (-4)* 1 st row to the 2 nd row: ⇔ 1 0 0 0 | 1 1 0 0- 1 0 0 | - 4- 3- 1 0- 2 1 0 | 6 6 1 0 0 0 1 | 0 1 Divide the 2 nd row by (-1) and add 2* 2 nd row to the 3 rd row: ⇔ 1 0 0 0 | 1 1 0 0 0 1 0 0 | 4 3 1 0 0 0 1 0 | 14 12 3 0 0 0 0 1 | 0 0 1...
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psLinAlgebra3Ans - Fall 2007 ARE211 Problem Set#08 Answer...

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