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* *1+ 3*2) = -8. ||u|| = 2. x(2, 3) = (y, 6) => x = 2, y = 4.

11/6/ 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 = Let u = (k, (sqrt 3), 4). Find k so that ||u|| = 10. (k ) 1/2 = 10 => k = 9. 5.Let u = (2, –5, 1) and v = (3, 0, 2). Find. = 2*3 - 5*0 + 1*2 = Solve 2x + 3y = 6. Infinite many solutions.

11/6/ 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.

11/6/20154 = 7 / -7 = -1 = -14/-4 = 2 iii) Cramer's Rule 2x + 3y = 4 5x + 4y = 3

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

11/6/20156 v) x + 3y = 4 5x + 4y = 3

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

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

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) = = 5 = sum of main diagonal elements = a 11 + a 22 + … + a nn

11/6/ 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 =

11/6/ 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;

11/6/ 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 = /5 4/ /5 -1/5

11/6/ Matrix Inverse B = B adj = e.g., Cross out Row 3 and Column 1, evaluate the remaining determinant as 6, sum the indices = 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.

11/6/ 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.

11/6/ (setf am #2A((1 0)(1 2))) (eigenvalues am)  ( ) (eigenvalues (inverse am))  ( ) 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))  ( ) (eigenvalues (mmult bm am))  ( )

11/6/ 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.

11/6/ 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

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

|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)( )) (mmult B A)  #2A((12 6 1)(8 4 -2)( )) AB  BA 11/6/201519

Eigenvalues: AB = BA (eigen (mmult am bm))  (( ) (#2A(( )( )( )) #2A(( )( )( )) #2A(( )( )( )))) (eigen (mmult bm am))  (( ) (#2A(( )( )( )) #2A(( )( )( )) #2A(( )( )( )))) 11/6/201520

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(( )( )( )) (det #2A((1 -3 3)(3 -5 3)(6 -6 4)))  16 (mmult #2A(( )(3 -5 3)( )) #2A(( )( )( )))  #2A((16 0 0)(0 16 0)(0 0 16)) 11/6/201521

Characteristic Equations (setf am #2A(( )(3 -5 3)(6 -6 4)) bm #2A(( )( )( ))) (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

Powers of Square Matrix (setf A-mat #2A((1 2)(3 4))) (expt-matrix A-mat 5)  #2A(( )( )) (apply #' mmult (list-of 5 a-mat))  #2A(( )( )) 11/6/201523

AB = AC; but B  C (setf A #2a((4 2 0)(2 1 0)( )) B #2a((2 3 1)( )(-1 2 1)) C #2a((3 1 -3)(0 2 6)(-1 2 1))) (mmult a b)  #2A((12 8 0)(6 4 0)( ))) (mmult a c)  #2A((12 8 0)(6 4 0)( ))) Note that (det A)  0; A has no inverse. 11/6/201524

11/6/ Induction Example … + n = n(n+1)/2; 1 = 1(1 + 1)/ … + n = n(n + 1)/2; assumed true … + n + (n + 1) = n(n+1)/2 + (n+1) = [n(n+1) + 2(n+1)]/2 = (n+1)(n+2)/2

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

Truth Table for De Morgan's Laws ~(p+q) = ~p~q p q ~p ~q p+q ~(p+q) ~p~q ~(pq) ~p+~q /6/201527

Conditional Statement p  q converse inverse contrapositive p q p  q q  p ~p  ~q~q  ~p /6/201528

Logical Implication p, p  q |- q Law of Detachment p q p  q /6/201529