1. ENM 500 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); v
u . v = ( 2* *1+ 3*2) = -8.
||u|| = is the length
  d(u,v) = is distance
2. x(2, 3) = (y, 6) => 2x = y, 3x = 6 => x = 2, y = 4.
u + v u 3u
-2u
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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.
(k )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. Infinitely many solutions
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3. i) x = (4 – 3y)/2 substituted for x in 2nd equation yields
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 2nd equation yields
10 – 7.5y + 4y = 3 => y = 2; x = -1.
 ii) 2x + 3y = 4; multiply by 5  10x + 15y = x + 4y = 3; multiply by 2  10x + 8y = 6
Then subtract to get 7y = 14 => y = 2 and x = -1.
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4. = 7 / -7 = -1 = -14/-4 = 2 iii) Cramer's Rule 2x + 3y = 4 5x + 4y = 3
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5. iv) Matrix Inverse: Ax = b => A-1Ax = x = A- b
Where x = (x, y) and b = (4, 3)
(inverse #2A((2 3)(5 4))) 
#2A((-4/7 3/7)(5/7 -2/7)
(print-matrix *) 
-4/ /7
5/ /7
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6. v) Elementary Row Operations
2x + 3y = x + 4y = 3
Find the inverse
/2 ½ /2 ½ /7 3/ /2 -5/ /7 -2/ /7 -2/7
-4/ /7
5/ /7
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7. Find the inverse of a 2 by 2 matrix.
(inverse #2A((1 3)(4 -2)))  #2A((1/7 3/14)(2/7 -1/14))
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8. Trace (A) = 3 + 2 = 5 = sum of main diagonal elements
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
= a11 + a22 + … + ann
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9. Ax = tx => (tI – A)x = 0 =>
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 =
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10. A = and B = f(A) = 2A2 – 5A + 6I g(A) = A3-2A2 + A + 3I
10. Let f(t) = 2t2 –5t + 6 and g(t) = t3 –2t2 + t Find f(A) and g(A) and f(B) and g(B) for matrices
A = and B = f(A) = 2A2 – 5A + 6I g(A) = A3-2A2 + A + 3I
A2 =
A3 =
(expt-mat #2A((2 -3)(5 1)) 2)  #2A((-11 -9)(15 -14))
(m-list-ops (list (expt-matrix #2A((2 -3)(5 1)) 2) (K*mat #2A((2 -3)(5 1)) -3)
(K*mat (identity-matrix 2) 17)) #' M+)
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11. 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 =
/5 4/ /5 -1/5 
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12. Matrix Inverse
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.
B =
Badj =
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13. 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
(m-list-ops (list (inverse #2A((1 2)(1 -1))) #2A((1 4)(2 3)) #2A((1 2)(1 -1))))  #2A((5 0)(0 -1))
(eigenvalues #2A((1 4)(2 3)))
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14. (eigenvalues (inverse am))  (1.0 0.5)
13. (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))  ( )
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15. Matrices
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16. Upper triangular and Det = product of main diagonal 5 * 1* 3* 2
15. Matrices
Upper triangular and Det = product of main diagonal * 1* 3* 2
Symmetric since A = AT
Det (AB) = [(Det A)*(Det B)]; Note: AB  BA in general.
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17. A(BC) = (AB)C Associative A(B + C) = AB + AC Distributive
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 = AAAA…AAAA Power of A n A's multiplied
AmAn = Am+n
(Am)n = Amn
AI = IA = A Commutative with Identity matrix
AB = 0 but A, B  0
Example: (setf ma #2a((0 1 0)(1 0 1)(0 1 0))
mb #2a((1 -2 1)(0 0 0)( )))
(Mmult ma mb)  #2a((0 0 0)(0 0 0)(0 0 0))
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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
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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)( )) (mmult B A)  #2A((12 6 1)(8 4 -2)( )) AB  BA
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20. Eigenvalues: AB = BA
(eigen (mmult am bm))  (( ) (#2A(( )( )( )) #2A(( )( )( )) #2A(( )( )( )))) (eigen (mmult bm am))  (( ) (#2A(( )( )( )) #2A(( )( )( )) #2A(( )( )( ))))
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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(( )( )( )) (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))
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22. Characteristic Equations
(setf am #2A(( )(3 -5 3)(6 -6 4)) bm #2A(( )( )( ))) (char-e am)  1t3 + 0t2 -12t -16 (characteristic equation) (char-e bm)  1t3 + 0t2 -12t -16 Matrices 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. (M+ (mmult am am am) (k*mat -12 am) (k*mat -16 (identity-matrix 3)))  #2A((0 0 0)(0 0 0)(0 0 0)); am satisfies C-E
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23. Powers of Square Matrix
(setf A-mat #2A((1 2)(3 4))) (expt-matrix A-mat 5)  #2A(( )( )) (m-list-ops (list-of 5 a-mat)) 
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24. 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.
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25. Induction Example
Prove for n = 1; Assume true for n = k' prove true for n = k + 1
… + n = n(n+1)/2
… + n + (n + 1) = n(n+1)/2 + (n+1)
= [n(n+1) + 2(n+1)]/2
= (n+1)(n+2)/2
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