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ENM 503 Linear Algebra Review  1. u = (2, -4, 3) and v = (-5, 1, 2).

ENM 503 Linear Algebra Review  1. u = (2, -4, 3) and v = (-5, 1, 2). u + v = (2-5, -4+1, 3 +2) = (-3, -3, 5). 2 u = (4, -8, 6); -v = (5, -1, -2); u . v = ( 2*-5 + -4*1+ 3*2) = -8. || u || = 2. x(2, 3) = (y, 6) => x = 2, y = 4.

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ENM 503 Linear Algebra Review  1. u = (2, -4, 3) and v = (-5, 1, 2).

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  1. ENM 503 Linear Algebra Review  1. u = (2, -4, 3) and v = (-5, 1, 2). u + v = (2-5, -4+1, 3 +2) = (-3, -3, 5). 2u = (4, -8, 6); -v = (5, -1, -2); u . v = ( 2*-5 + -4*1+ 3*2) = -8. ||u|| = 2. x(2, 3) = (y, 6) => x = 2, y = 4.

  2. Find k so that u and v and are orthogonal with • = 0 => orthogonal u = (-1, k, -2) and v = (4, -2, 5). • -4 –2k –10 = 0 => k = -7. •  4. Let u =(k, (sqrt 3), 4). Find k so that ||u|| = 10. • (k2 + 3 + 16)1/2 = 10 => k = 9. • Let u = (2, –5, 1) and v = (3, 0, 2). Find . • = 2*3 - 5*0 + 1*2 = 8. • 6. Solve 2x + 3y = 6. Infinite many solutions.

  3. Solve by i) substitution, ii) elimination, iii) Cramer’s Rule, iv) Gaussian reduction, v) elementary row operations. • 2x + 3y = 45x + 4y = 3 • i) x = (4 – 3y)/2 substituted for x in 2nd equation yields • 10 – 7.5y + 4y = 3 => y = 2; x = -1. •  ii) 2x + 3y = 4; multiply by 5  10x + 15y = 20 5x + 4y = 3; multiply by 2  10x + 8y = 6 • Then subtract to get 7y = 14 => y = 2 and x = -1.

  4. iii) Cramer's Rule 2x + 3y = 4 5x + 4y = 3 = 7 / -7 = -1 = -14/-4 = 2

  5. Matrix Inverse • AX= b • A-1AX = A-1b • IX = X = A-1 b • Where X = (x, y) and b = (4, 3) • A-1 A X A-1 = b

  6. v) 2 3 4 1 1.5 2 1 1.5 2 1 0 -1 5 4 3 0 -3.5 -7 0 1 2 0 1 2 2x + 3y = 4 5x + 4y = 3

  7. System of linear equations Consistent Inconsistent Infinite number of solutions No solution Unique solution

  8. Find the inverse of a 2 by 2 matrix.

  9. A matrix may be looked upon as a function. Consider a matrix A which maps all vectors in the plane. For example, A = A: (1, 3)  = Trace (A) = 3 + 2 = 5 = sum of main diagonal elements = a11 + a22 + … + ann

  10. 9. Let A = and B = Find A2, AB, AT, (AB)T. Find A’s characteristic equation and show that A satisfies it. Ax = tx => (tI – A)x = 0 => A2 - 7I = O A2 = AB = = AT = (AB)T =

  11. Let f(t) = 2t2 –5t + 6 and g(t) = t3 –2t2 + t +3. • Find f(A) and g(A) and f(B) and g(B) for matrices • (M+ (K*mat 2 (mmult amat amat)) (k*mat -5 amat) (k*mat 6 (identity-matrix 2))) #2A((-26 -3)(5 -27)) = f(A)#2A((3 6)(0 9)) = f(B)t2 - 3t + 17 = 0 (M+ (mmult amat amat) (K*mat -3 amat) (K*mat 17 (identity-matrix 2))) • A = B = • A2 = A3 =f(A) = 2A2 – 5A + 6I • g(A) = A3-2A2 + A + 3I;

  12. 11. Find the inverse of the following matrices by forming the adjacency matrix, and then by fixing the identity matrix to it and converting the initial matrix to the identity matrix.   B adjB |B|I A = A-1 = B-1 = 1 4 1 0 1 4 1 0 1 0 -3/5 4/5 2 3 0 1 0 -5 -2 1 0 1 2/5 -1/5 

  13. Matrix Inverse e.g., Cross out Row 3 and Column 1, evaluate the remaining determinant as 6, sum the indices 3 + 1 = 4, even => put 6 in transposed indices (1,3); if sum is odd put negated determinant in transposed indices. Do the (3, 2) entry -6. The determinant is -6 but the sum of the indices is 5, odd, thus a – (-6) = 6 is put in transposed indices (2, 3) . Do the (1, 2) entry -3. The determinant is -6 but the sum of the indices is 3, odd, thus a – (-6) = 6 is put in transposed indices (2, 1) . Do the (2, 3) entry 3. The determinant is 12 but the sum of the indices is 5, odd, thus a – 12 is put in transposed indices (3, 2). Continued and finally divide the transposed adjacently matrix by the determinant of B. B = Badj =

  14. Find the eigenvalues and corresponding eigenvectors of • A = (tI – A)u = 0, where u= (x, y) • = (t – 5)(t + 1) => t = 5, -1 • = 0 => x = y or (1, 1); eigenvector = -2x –4y = 0 => x = -2y or (2, -1) • let P = P-1AP is diagonal matrix with the eigenvalues. PAP-1 • (mmult (inverse #2A((1 2)(1 -1))) #2A((1 4)(2 3)) #2A((1 2)(1 -1)))  #2A((5 0)(0 -1)) • Eigenvalues appear along the main diagonal.

  15. 13. (setf am #2A((1 0)(1 2))) (eigenvalues am) (2.0 1.0) (eigenvalues (inverse am)) (1.0 0.5) Observe inverse eigenvalues are reciprocals of original matrix's eigenvalues. 14. Show that AB and BA have same eigenvalues. (setf am #2A((1 0)(1 2)) bm #2A((4 7)(9 2))) (eigenvalues (mmult am bm))  (20.393796 -5.393797) (eigenvalues (mmultbm am))  (20.393796 -5.393797)

  16. 15. Matrices Upper triangular and Det = product of main diagonal 5 * 1* 3* 2 Symmetric since A = AT Det (AB) = [(Det A)*(Det B)]; Note: AB  BA in general.

  17. 16. Matrix Properties A(BC) = (AB)C Associative A(B + C) = AB + AC Distributive (A + B)C = AC + BC Distributive (AB)T = BTAT Transpose AB  BA in general; unless (commutative) matrices. An = AAA…AAA Power of A n A's multiplied

  18. Determinants Compute the determinant of each matrix. A = B = |A| = (det #2A((1 2 3)(4 -2 3)(2 5 -1))) 79 |B| = (det #2A((2 0 1)(4 2 -3)(5 3 1)))  24

  19. |AB| = |A| |B| (setf am #2A((1 0)(1 2)) bm #2A((4 7)(9 2)) (det (mmult am bm))  -110 (* (det am) (det bm)) -110 (eigen (mmult A B)) = (eigen (mmult B A)) (mmult A B)  #2A((12 8 0)(6 4 0)(-7 -2 1)) (mmult B A)  #2A((12 6 1)(8 4 -2)(-2 -1 1)) AB  BA

  20. Eigenvalues: AB = BA (eigen (mmult am bm))((4.0 -8.0 -8.0) (#2A((4.0 -4 4)(-8 8.0 4)(0 0 12.0)) #2A((-8.0 -4 4)(-8 -4.0 4)(0 0 0.0)) #2A((-8.0 -4 4)(-8 -4.0 4)(0 0 0.0)))) (eigen (mmultbm am)) ((4.0 -8.0 -8.0) (#2A((10.0 -10 10)(-2 2.0 10)(0 0 12.0)) #2A((-2.0 -10 10)(-2 -10.0 10)(0 0 0.0)) #2A((-2.0 -10 10)(-2 -10.0 10)(0 0 0.0))))

  21. A *(adj A)= (adj A)*A= |A|I (adj-mat am)  #2A((2 0)(-1 1) (adj-mat #2A((1 -3 3)(3 -5 3)(6 -6 4)))  #2A((-2 -6 6)(6 -14 6)(12 -12 4)) (det #2A((1 -3 3)(3 -5 3)(6 -6 4)))  16 (mmult #2A(( 1 -3 3)(3 -5 3)( 6 -6 4)) #2A((-2 -6 6)(6 -14 6)(12 -12 4))) #2A((16 0 0)(0 16 0)(0 0 16))

  22. Characteristic Equations (setf am #2A(( 1 -3 3)(3 -5 3)(6 -6 4)) bm #2A((-3 1 -1)(-7 5 -1)(-6 6 -2))) (char-e am)  1t3 + 0t2 -12t -16 (characteristic equation) (char-e bm)  1t3 + 0t2 -12t -16 Matrices am amand bm have the same characteristic equations but am has 3 independent eigenvectors and bm has two; and thus the two matrices are not similar.

  23. Powers of Square Matrix (setf A-mat #2A((1 2)(3 4))) (expt-matrix A-mat 5)  #2A((1069 1558)(2337 3406)) (apply #' mmult (list-of 5 a-mat)) #2A((1069 1558)(2337 3406))

  24. AB = AC; but B  C (setf A #2a((4 2 0)(2 1 0)(-2 -1 1)) B #2a((2 3 1)(2 -2 -2)(-1 2 1)) C #2a((3 1 -3)(0 2 6)(-1 2 1))) (mmult a b)  #2A((12 8 0)(6 4 0)(-7 -2 1))) (mmult a c)  #2A((12 8 0)(6 4 0)(-7 -2 1))) Note that (det A) 0; A has no inverse.

  25. Induction Example 1 + 2 + … + n = n(n+1)/2; 1 = 1(1 + 1)/2 1 + 2 + … + n = n(n + 1)/2; assumed true 1 + 2 + … + n + (n + 1) = n(n+1)/2 + (n+1) = [n(n+1) + 2(n+1)]/2 = (n+1)(n+2)/2

  26. Laws of the Algebra of Propositions Idempotent: p + p = p pp = p Associative: (p+q)+r = p+(q+r) (pq)r = p(qr) Commutative: p+q = q+p pq = qp Distributive: p+(qr) = (p+q)(p+r) p(q+r) = (pq) + (pr) Identity: p + 0 = p p1 = p p + 1 = 1 p0 = 0 Complement: p + ~p = 1 p ~p = 0 ~ ~p = p ~1 = 0; ~0 = 1 DeMorgan: ~(p+q) = ~p~q ~(pq) = ~p + ~q 1 = true; 0 = false; ~ = not

  27. Truth Table for De Morgan's Laws ~(p+q) = ~p~q p q ~p ~q p+q~(p+q) ~p~q~(pq) ~p+~q 0 0 1 1 0 1 1 1 10 1 1 0 1 0 0 1 11 0 0 1 1 0 0 1 11 1 0 0 1 0 0 0 0

  28. Conditional Statement p  q converse inverse contrapositive p q p  q q  p ~p  ~q ~q  ~p 0 0 1 1 1 10 1 1 0 0 11 0 0 1 1 01 1 1 1 1 1

  29. Logical Implication p, p  q |- q Law of Detachment p q p  q0 0 10 1 11 0 01 1 1

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