1. 11/6/20151 ENM 503Linear 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. 11/6/20152 3.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. (k 2 + 3 + 16) 1/2 = 10 => k = 9. 5.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. 11/6/20153 7.Solve by i) substitution, ii) elimination, iii) Cramer’s Rule, iv) Gaussian reduction, v) elementary row operations. 2x + 3y = 4 5x + 4y = 3 i) x = (4 – 3y)/2 substituted for x in 2 nd 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. 11/6/20154 = 7 / -7 = -1 = -14/-4 = 2 iii) Cramer's Rule 2x + 3y = 4 5x + 4y = 3

5. 11/6/20155 iv)Matrix Inverse AX = b A -1 AX = A-1b IX = X = A -1 b Where X = (x, y) and b = (4, 3) A -1 A X A -1 = b

6. 11/6/20156 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. 11/6/20157 System of linear equations Inconsistent Consistent No solutionUnique solution Infinite number of solutions

8. 11/6/20158 8.Find the inverse of a 2 by 2 matrix.

9. 11/6/20159 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 = a 11 + a 22 + … + a nn

10. 11/6/201510 9. Let A = and B = Find A 2, AB, A T, (AB) T. Find A’s characteristic equation and show that A satisfies it. Ax = tx => (tI – A)x = 0 => A 2 - 7I = O A 2 = AB = = A T = (AB) T =

11. 11/6/201511 10.Let f(t) = 2t 2 –5t + 6 and g(t) = t 3 –2t 2 + 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) t 2 - 3t + 17 = 0 (M+ (mmult amat amat) (K*mat -3 amat) (K*mat 17 (identity-matrix 2))) A = B = A 2 = A 3 = f(A) = 2A 2 – 5A + 6I g(A) = A 3 -2A 2 + A + 3I;

12. 11/6/201512 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 adj B |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. 11/6/201513 Matrix Inverse B = B adj = 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.

14. 11/6/201514 12.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 -1 AP 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. 11/6/201515 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 (mmult bm am))  (20.393796 -5.393797)

16. 11/6/201516 15. Matrices Upper triangular and Det = product of main diagonal 5 * 1* 3* 2 Symmetric since A = A T Det (AB) = [(Det A)*(Det B)]; Note: AB  BA in general.

17. 11/6/201517 16. Matrix Properties A(BC) = (AB)CAssociative A(B + C) = AB + ACDistributive (A + B)C = AC + BCDistributive (AB) T = B T A T Transpose AB  BA in general; unless (commutative) matrices. A n = 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 11/6/201518

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 11/6/201519

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 (mmult bm 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)))) 11/6/201520

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)) 11/6/201521

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)  1t 3 + 0t 2 -12t -16 (characteristic equation) (char-e bm)  1t 3 + 0t 2 -12t -16 Matrices am am and bm have the same characteristic equations but am has 3 independent eigenvectors and bm has two; and thus the two matrices are not similar. 11/6/201522

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)) 11/6/201523

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. 11/6/201524

25. 11/6/201525 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+ppq = qp Distributive: p+(qr) = (p+q)(p+r)p(q+r) = (pq) + (pr) Identity: p + 0 = pp1 = p p + 1 = 1p0 = 0 Complement: p + ~p = 1p ~p = 0 ~ ~p = p~1 = 0; ~0 = 1 DeMorgan: ~(p+q) = ~p~q~(pq) = ~p + ~q 1 = true; 0 = false; ~ = not 11/6/201526

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 1 0 1 1 0 1 0 0 1 1 1 0 0 1 1 0 0 1 1 1 1 0 0 1 0 0 0 0 11/6/201527

28. Conditional Statement p  q converse inverse contrapositive p q p  q q  p ~p  ~q~q  ~p 0 0 11 1 1 0 1 1 0 0 1 1 0 0 1 1 0 1 1 1 1 1 1 11/6/201528

29. Logical Implication p, p  q |- q Law of Detachment p q p  q 0 0 1 0 1 1 1 0 0 1 1 1 11/6/201529