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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 2 2u u 2 4 3 29

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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.

(k2 + 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.

u•v

u•vu•v

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7. 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.

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4 3

3 42 3

5 4

x = 7 / -7 = -1

2 4

5 3

7y

= -14/-4 = 2

iii) Cramer's Rule 2x + 3y = 4 5x + 4y = 3

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4 / 7 3 / 7 2 3 4 / 7 3 / 7 4 1

5 / 7 2 / 7 5 4 5 / 7 2 / 7 3 2

x x

y y

iv) Matrix Inverse AX = bA-1AX = A-1bIX = X = A-1 b

Where X = (x, y) and b = (4, 3)

A-1 A X A-1 = b

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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

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System of linear equations

Inconsistent Consistent

No solution Unique solutionInfinite number of solutions

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

1

d -b

a b -c a

c d ad-bc

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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

1

3

15

7

3 4

1 2

3 4

1 2

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9. Let A = 1 3

2 1

and B = 1 2

4 3

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 = 1 2

3 1

1 3

2 1

1 2

4 3

13 7

2 7

13 2 1 4 1 2

7 7 2 3 3 1

7 0

0 7

0 0

0 0

7 0

0 7

7 0

0 7

21 3| | 7 0

2 1

ttI A t

t

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10. 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;

2 3

5 1

1 2

0 3

67 24

40 59

11 9

15 14

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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 

1 4

2 3

1 3 3 2 6 6 16 0 0

3 5 3 6 14 6 0 16 0

6 6 4 12 12 4 0 0 16

3 4

2 1

5

2 6 6

6 14 6

12 12 4

16

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Matrix InverseMatrix Inverse1 3 3

3 5 3

6 6 4

B =

Badj =

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.

2 6 6

6 14 6

12 12 4

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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-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.

1 4

2 3

2t-1 -4=t -4t-5

-2 t-3

5 1 44 4

2 5 3x y

1 2

1 1

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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)

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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.

5 3 2 9

0 1 4 6

0 0 3 7

0 0 0 2

4 1 2

1 1 7

2 7 1

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16. Matrix PropertiesA(BC) = (AB)C AssociativeA(B + C) = AB + AC Distributive (A + B)C = AC + BC Distributive

(AB)T = BTAT TransposeAB BA in general; unless (commutative) matrices.An = AAA…AAA Power of A n A's multiplied

DeterminantsDeterminants

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

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1 2 3

4 2 3

2 5 1

2 0 1

4 2 3

5 3 1

|AB| = |A| |B||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

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Eigenvalues: AB = BAEigenvalues: 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))))

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A *(adj A)= (adj A)*A= |A|IA *(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))

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Characteristic EquationsCharacteristic 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 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.

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Powers of Square MatrixPowers 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))

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AB = AC; but B AB = AC; but B C 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.

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Induction ExampleInduction 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

Laws of the Algebra of PropositionsLaws 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

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Truth Table for De Morgan's LawsTruth 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

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Conditional StatementConditional 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

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Logical ImplicationLogical Implication

p, p q |- q Law of Detachment

p q p q0 0 10 1 11 0 01 1 1

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