FinalExam_AllStudyContent_2010-0315

FinalExam_AllStudyContent_2010-0315 - -1 A= = ; A = | A | c...

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A = = f e x x d c b a 2 1 ; A -1 = - - a c b d A | | 1 ; |A| = ad-bc ; A1 = 4 0 3 2 ; b1= - 4 5 ; B1 = - 7 1 2 3 ; A2 = 4 2 5 13 ; A3 = 4 1 1 2 ; T1 = 4 3 3 2 2 1 b*b T = - 4 5 [ ] 4 5 - = - - 16 20 20 25 ; b T *b = [ ] 4 5 - - 4 5 = 41 Solving linear equations - basic mathematics. o If A1*x=b => = + 2 1 2 * 4 3 1 * 0 2 b b x x => linear combination of columns or columns space o A1*x =b1 => - = 4 5 2 1 4 0 3 2 x x => 4x2 = -4 => x2=-1; 2x1+3(-1) = 5 => x1=4 => x= - 1 4 Properties of determinants and matrices. Solving eigenvalue problems - basic mathematics. o P(λ) = |A- λI| = - - λ 4 0 3 2 = (2-λ)(4-λ) = 0 => λ 1 =4, λ 2 =2 o Eigenvalues of A2: λ 1 =3, λ 2 =14 => 3+14=17; trace(A2) = 13+4=17 o Eigenvectors of λ 1 =3 => - - 3 4 2 5 3 13 => = 0 0 2 1 1 2 5 10 x x => 0 x2 2x1 0 5x2 10x1 = + = + => ... 6 4 2 | 2 3 2 1 | 1 - - - x x Trace and determinant properties of eigenvalues. o |A T | = |A| ; |A B| = |A| |B| ; A = [L | |U| L is unit matrix with ones on diagonals; get U from Gauss Elim ; if get zero on U, then det(A)=0 o (AB) -1 = B -1 A -1 ; (ABC) -1 = C -1 B -1 A -1 ; (AB) T = B T A T ; (ABC) T = C T B T A T ; (A, B, C are square) o Trace(matrix) = sum of diagonals = λ 1 + λ 2 + λ 3 … + λ n ; |A| = λ 1 * λ 2 * λ 3 … * λ n o Trace(A1)=2+4=6 ; Trace(B1)=3+7=10 ; Trace(T1)=1+3=4 Gerschgorin circle theorem for eigenvalue localization (estimate the eigenvalues of A [below]) Starting with row one, we take the element on the diagonal, a ii as the center for the disc. We then take the remaining elements in the row and apply formula: to obtain the 4 discs… D (10,2); D (8,0.6); D (2,3); D ( − 11,3). Every eigenvalue of A lies within one of these four discs. The eigenvalues are −10.86, 1.8995, 9.8218, 8.1478 Properties of special matrices such as the Hilbert matrix . o N by N matrix with elements 1/(i+j-1); famous example of badly conditioned matrix. o Condition number of 1.0 is perfect; cond (A) = K(A) = (largest SV) / (smallest SV) Gaussian elimination with and without partial pivoting. o RegGauss Elim . 0 2 6 5 0 2 0 9 7 3 2 1 3 2 1 3 2 = + + = - + = + x x x x x x x x => - 0 | 2 6 5 0 | 1 1 2 0 | 9 7 0 => - 0 | 0 0 0 0 | 7 / 9 1 0 0 | 2 / 1 2 / 1 1 ; rank =2 Getting 0 = 0 implies infinite number of solutions (lines on top); If 0 = 1 then no solutions (parallel if graphing); o W/o pivoting
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This note was uploaded on 04/26/2010 for the course CEG 616 taught by Professor Taylor during the Winter '10 term at Wright State.

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FinalExam_AllStudyContent_2010-0315 - -1 A= = ; A = | A | c...

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