ENGR 1405 Engineering Mathematics 1

Faculty of Engineering and Applied Science
2000 Fall

Quiz 2
Solution

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[Note:   if you see symbols like ¹ or Þ in various places, then your browser is not reading the style sheet for this Web page properly (or the Symbol font is not installed on your computer).   The translation is:

        “Þ” as “Þ”   (implies),
        “-” as “-”   (minus),
        “ú” as “ú”   (set of real numbers),
        “¬” as “¬”   (left arrow),
        “«” as “«”   (two-way arrow).]

Given the matrices A and X₀, given as


        [ 0.5  0.7  0.1 ]          [ 0.8 ]
    A = [ 0.4  0.2  0.6 ] ,   X  = [ 0.2 ]
        [ 0.1  0.1  0.3 ]      0   [ 0.0 ]

1. Determine   X₁ = A X₀.

   ---

   X₁ = A X₀

         [ 0.5  0.7  0.1 ] [ 0.8 ]   [ (.5*.8 + .7*.2 + .1*0) ]
       = [ 0.4  0.2  0.6 ] [ 0.2 ] = [ (.4*.8 + .2*.2 + .6*0) ]
         [ 0.1  0.1  0.3 ] [ 0.0 ]   [ (.1*.8 + .1*.2 + .3*0) ]
   
         [ 0.54 ]
       = [ 0.36 ]
         [ 0.10 ]
         ========
   

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2. For the matrix
   

           [ u ]
       X = [ v ]
           [ w ]
   

   determine values for   u, v and w such that (A - I) X = 0 and   u + v + w = 1 .

   ---

   Matrix   A - I =
   

       [-0.5  0.7  0.1 ]
       [ 0.4 -0.8  0.6 ] .
       [ 0.1  0.1 -0.7 ] 
   

   Method 1:

   Solving the homogeneous linear system   [(A - I) | 0]:
   Multiply all rows by 10 and swap rows 1 and 3:

       [ 1  1 -7 | 0 ] 
       [ 4 -8  6 | 0 ] 
       [-5  7  1 | 0 ]
   

   R₂ ¬ R₂ - 4 R₁,
   R₃ ¬ R₃ + 5 R₁:
   

       [ 1   1  -7 | 0 ] 
       [ 0 -12  34 | 0 ] 
       [ 0  12 -34 | 0 ]
   

   R₃ ¬ R₃ + R₂, then
   R₂ ¬ R₂ / -12:
   

       [ 1   1   -7  | 0 ] 
       [ 0   1 -17/6 | 0 ] 
       [ 0   0    0  | 0 ]
   

   R₁ ¬ R₁ - R₂:
   

       [ 1   0 -25/6 | 0 ] 
       [ 0   1 -17/6 | 0 ] 
       [ 0   0    0  | 0 ]
   

   This generates a one-parameter family of solutions
   (u, v, w) = (25, 17, 6) t,   where t is any real number.
   However, we also have the constraint u + v + w = 1 .
   Þ     (25 + 17 + 6) t = 1     Þ     t = 1/48

   Therefore the unique solution is
           (u, v, w) = (25/48, 17/48, 1/8).

   ---

   Method 2:

   Incorporate the constraint   u + v + w = 1   into the linear system, as an extra row.

       [-0.5  0.7  0.1 |  0 ]
       [ 0.4 -0.8  0.6 |  0 ]  
       [ 0.1  0.1 -0.7 |  0 ] 
       [  1    1    1  |  1 ]
   

   Multiply rows 1, 2 and 3 by 10 and swap rows 1 and 3:

       [  1   1  -7 | 0 ] 
       [  4  -8   6 | 0 ]  
       [ -5   7   1 | 0 ]
       [  1   1   1 | 1 ]
   

   R₂ ¬ R₂ - 4 R₁,
   R₃ ¬ R₃ + 5 R₁:
   R₄ ¬ R₄ - R₁:
   

       [  1   1  -7 | 0 ] 
       [  0 -12  34 | 0 ] 
       [  0  12 -34 | 0 ]
       [  0   0   8 | 1 ]
   

   R₃ ¬ R₃ + R₂, then
   R₂ ¬ R₂ / -12 and
   R₄ ¬ R₄ / 8 :
   

       [  1   1   -7  |  0  ] 
       [  0   1 -17/6 |  0  ] 
       [  0   0    0  |  0  ]
       [  0   0    1  | 1/8 ]
   

   R₁ ¬ R₁ - R₂ and
   R₃ « R₄:
   

       [  1   0 -25/6 |  0  ] 
       [  0   1 -17/6 |  0  ] 
       [  0   0    1  | 1/8 ]
       [  0   0    0  |  0  ]
   

   R₁ ¬ R₁ +25/6 R₃,
   R₂ ¬ R₂ +17/6 R₃,
   

       [  1   0    0  | 25/48 ] 
       [  0   1    0  | 17/48 ] 
       [  0   0    1  |  1/8  ]
       [  0   0    0  |   0   ]
   

   Therefore the unique solution is
   

   (u, v, w) = (25/48, 17/48, 1/8).

   [Note:   this is the steady state vector for the Markov chain whose transition matrix is A.]

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Created 2000 10 04 and modified 2000 10 04 by Dr. G.H. George