ENGR 1405 Engineering Mathematics 1

Faculty of Engineering and Applied Science
2000 Fall

Problem Set 1
Solutions

[Note: some browsers do not display symbols correctly. Read
“Þ” as “Þ” (implies),
“«” as “«” (two-way arrow),
“¬” as “¬” (left arrow),
“-” as “-” (minus) and
“ú” as “ú” (set of real numbers).]

In each of the following questions use the three operations, introduced in class, to reduce the given linear system to triangular form as much as possible. Indicate whether or not the system has a solution. If a solution exists, then say what type, and write a summary statement expressing the solution. Be sure that the recommended order of operations is being used.

In this question, the output from the linsys.exe program appears here:

3 x -	y =	4
-5 x +	2 y =	6

Program to handle the Arithmetic in the Row Reduction of a Linear System
VisualBASIC version (c)1999, Glyn George.

Linear System:

Row 1:    3/1       -1/1        4/1    
Row 2:   -5/1        2/1        6/1    

-----------------------------------------------------------------------------
Selected row operation:    Divide row 1 by  3 / 1 .

Linear System:

Row 1:    1/1       -1/3        4/3    
Row 2:   -5/1        2/1        6/1    

-----------------------------------------------------------------------------
Selected row operation:    Subtract -5 / 1 times row 1 from row 2 .

Linear System:

Row 1:    1/1       -1/3        4/3    
Row 2:    0/1        1/3       38/3    

-----------------------------------------------------------------------------
Selected row operation:    Divide row 2 by  1 / 3 .

Linear System:

Row 1:    1/1       -1/3        4/3    
Row 2:    0/1        1/1       38/1    

-----------------------------------------------------------------------------
Selected row operation:    Subtract -1 / 3 times row 2 from row 1 .

Linear System:

Row 1:    1/1        0/1       14/1    
Row 2:    0/1        1/1       38/1    

-----------------------------------------------------------------------------

Program execution terminated.

The linear system therefore has the unique solution (x, y) = (14, 38).

[Note: This linear system can be represented by a pair of lines with distinct slopes in ú² space. The solution set is then the single point of intersection of the two lines.]

x	-	z =	3
x +	y -	z =	5
2 x +	y -	2 z =	8

Program to handle the Arithmetic in the Row Reduction of a Linear System
VisualBASIC version (c)1999, Glyn George.

Linear System:

Row 1:    1/1        0/1       -1/1        3/1    
Row 2:    1/1        1/1       -1/1        5/1    
Row 3:    2/1        1/1       -2/1        8/1    

-----------------------------------------------------------------------------
Selected row operation:    Subtract  1 / 1 times row 1 from row 2 .

Linear System:

Row 1:    1/1        0/1       -1/1        3/1    
Row 2:    0/1        1/1        0/1        2/1    
Row 3:    2/1        1/1       -2/1        8/1    

-----------------------------------------------------------------------------
Selected row operation:    Subtract  2 / 1 times row 1 from row 3 .

Linear System:

Row 1:    1/1        0/1       -1/1        3/1    
Row 2:    0/1        1/1        0/1        2/1    
Row 3:    0/1        1/1        0/1        2/1    

-----------------------------------------------------------------------------
Selected row operation:    Subtract  1 / 1 times row 2 from row 3 .

Linear System:

Row 1:    1/1        0/1       -1/1        3/1    
Row 2:    0/1        1/1        0/1        2/1    
Row 3:    0/1        0/1        0/1        0/1    

-----------------------------------------------------------------------------

Program execution terminated.

The linear system therefore has the one parameter family of solutions (x, y, z) = (3 + z, 2, z), where z can be any real number.

The solution set can also be written in the equivalent form
(x, y, z) = (3, 2, 0) + t (1, 0, 1), (t = any real number),
(which is the vector parametric equation of a straight line).

[Note: One can also see at a glance that equation 3 in the original linear system is simply the sum of equations 1 and 2. There are therefore only two independent equations for the three unknowns, so that a unique solution becomes impossible. As the remaining two equations represent a pair of intersecting planes, the geometrical representation of the solution set must be a straight line (which is a one dimensional object = a one parameter family of solutions)].

r +	s -	8 t =	6
	s -	4 t =	0
r -	s -	t =	5

Program to handle the Arithmetic in the Row Reduction of a Linear System
VisualBASIC version (c)1999, Glyn George.

Linear System:

Row 1:    1/1        1/1       -8/1        6/1    
Row 2:    0/1        1/1       -4/1        0/1    
Row 3:    1/1       -1/1       -1/1        5/1    

-----------------------------------------------------------------------------
Selected row operation:    Subtract  1 / 1 times row 1 from row 3 .

Linear System:

Row 1:    1/1        1/1       -8/1        6/1    
Row 2:    0/1        1/1       -4/1        0/1    
Row 3:    0/1       -2/1        7/1       -1/1    

-----------------------------------------------------------------------------
Selected row operation:    Subtract -2 / 1 times row 2 from row 3 .

Linear System:

Row 1:    1/1        1/1       -8/1        6/1    
Row 2:    0/1        1/1       -4/1        0/1    
Row 3:    0/1        0/1       -1/1       -1/1    

-----------------------------------------------------------------------------
Selected row operation:    Divide row 3 by -1 / 1 .

Linear System:

Row 1:    1/1        1/1       -8/1        6/1    
Row 2:    0/1        1/1       -4/1        0/1    
Row 3:    0/1        0/1        1/1        1/1    

-----------------------------------------------------------------------------
Selected row operation:    Subtract  1 / 1 times row 2 from row 1 .

Linear System:

Row 1:    1/1        0/1       -4/1        6/1    
Row 2:    0/1        1/1       -4/1        0/1    
Row 3:    0/1        0/1        1/1        1/1    

-----------------------------------------------------------------------------
Selected row operation:    Subtract -4 / 1 times row 3 from row 1 .

Linear System:

Row 1:    1/1        0/1        0/1       10/1    
Row 2:    0/1        1/1       -4/1        0/1    
Row 3:    0/1        0/1        1/1        1/1    

-----------------------------------------------------------------------------
Selected row operation:    Subtract -4 / 1 times row 3 from row 2 .

Linear System:

Row 1:    1/1        0/1        0/1       10/1    
Row 2:    0/1        1/1        0/1        4/1    
Row 3:    0/1        0/1        1/1        1/1    

-----------------------------------------------------------------------------

Program execution terminated.

The linear system therefore has the unique solution (r, s, t) = (10, 4, 1).

[Note: This linear system can be represented by three planes with non-coplanar normal directions in ú³ space. The solution set is then the unique point of intersection of the three planes.]

	x₂ -	x₃ +	2 x₄ =	-2
x₁	+	x₃ -	x₄ =	3
x₁ +	x₂	+	x₄ =	1
2 x₁ -	3 x₂ +	5 x₃ -	8 x₄ =	12

Program to handle the Arithmetic in the Row Reduction of a Linear System
VisualBASIC version (c)1999, Glyn George.

Linear System:

Row 1:    0/1        1/1       -1/1        2/1       -2/1    
Row 2:    1/1        0/1        1/1       -1/1        3/1    
Row 3:    1/1        1/1        0/1        1/1        1/1    
Row 4:    2/1       -3/1        5/1       -8/1       12/1    

-----------------------------------------------------------------------------
Selected row operation:    Swap rows 2 and 1 .

Linear System:

Row 1:    1/1        0/1        1/1       -1/1        3/1    
Row 2:    0/1        1/1       -1/1        2/1       -2/1    
Row 3:    1/1        1/1        0/1        1/1        1/1    
Row 4:    2/1       -3/1        5/1       -8/1       12/1    

-----------------------------------------------------------------------------
Selected row operation:    Subtract  1 / 1 times row 1 from row 3 .

Linear System:

Row 1:    1/1        0/1        1/1       -1/1        3/1    
Row 2:    0/1        1/1       -1/1        2/1       -2/1    
Row 3:    0/1        1/1       -1/1        2/1       -2/1    
Row 4:    2/1       -3/1        5/1       -8/1       12/1    

-----------------------------------------------------------------------------
Selected row operation:    Subtract  2 / 1 times row 1 from row 4 .

Linear System:

Row 1:    1/1        0/1        1/1       -1/1        3/1    
Row 2:    0/1        1/1       -1/1        2/1       -2/1    
Row 3:    0/1        1/1       -1/1        2/1       -2/1    
Row 4:    0/1       -3/1        3/1       -6/1        6/1    

-----------------------------------------------------------------------------
Selected row operation:    Subtract  1 / 1 times row 2 from row 3 .

Linear System:

Row 1:    1/1        0/1        1/1       -1/1        3/1    
Row 2:    0/1        1/1       -1/1        2/1       -2/1    
Row 3:    0/1        0/1        0/1        0/1        0/1    
Row 4:    0/1       -3/1        3/1       -6/1        6/1    

-----------------------------------------------------------------------------
Selected row operation:    Subtract -3 / 1 times row 2 from row 4 .

Linear System:

Row 1:    1/1        0/1        1/1       -1/1        3/1    
Row 2:    0/1        1/1       -1/1        2/1       -2/1    
Row 3:    0/1        0/1        0/1        0/1        0/1    
Row 4:    0/1        0/1        0/1        0/1        0/1    

-----------------------------------------------------------------------------

Program execution terminated.

The linear system therefore has the two parameter family of solutions (x₁, x₂, x₃, x₄) = (3 - s + t, -2 + s - 2 t, s, t), where s and t can be any real numbers.

The solution set can also be written in the equivalent form
(x₁, x₂, x₃, x₄) = (3, -2, 0, 0) + s (-1, 1, 1, 0) + t (1, -2, 0, 1),
(which is the vector parametric equation of a plane in ú⁴ space).

[Note: Visual inspection of the original linear system leads quickly to the observation that equation 3 is simply the sum of equations 1 and 2 and that equation 4 = two times (equation 2) minus three times (equation 1). There are therefore only two independent equations for the four unknowns, so that a unique solution becomes impossible.

The geometrical interpretation is difficult to visualize because a linear system in four unknowns requires the four spatial dimensions of ú⁴ space to represent it. As the remaining two equations represent a pair of intersecting three-dimensional hyperplanes in ú⁴ space, the geometrical representation of the solution set must be a plane (which is a two dimensional object = a two parameter family of solutions)].

Determine the value of the matrix if it exists or explain, briefly, why it does not exist.

Both matrices have dimensions (2×3).
Their sum is therefore defined.

A =  [  1 -3  7 ] + [  3 -1  4 ] = [  4 -4 11 ]
     [  4  5 -6 ]   [ -7  3 -2 ]   [ -3  8 -8 ]
                                   ============

Both matrices have dimensions (2×2).
The linear combination is therefore defined.

B = [  2  4 ] + [ -15   6 ]  =  [ -13  10 ] 
    [ 12 -6 ]   [ -12 -21 ]     [   0 -27 ]
                                ===========

The first matrix has dimensions (2×2) but the second has dimensions (2×3).
These incompatible dimensions Þ C does not exist.
The first matrix has dimensions (2×1) and the second has dimensions (1×2).
The product D is therefore defined as a (2×2) matrix.
```
D  =  [ (5x8)  (5x-3) ]  =  [ 40 -15 ]
      [ (4x8)  (4x-3) ]     [ 32 -12 ] 
                            ==========
```

E = [11 4 / -2 5 / 3 -6] * [2 -6 8 / 1 -2 6]

The first matrix has dimensions (3×2) and the second has dimensions (2×3).
The product E is therefore defined as a (3×3) matrix.

      [ (11x2 +  4x1)  (11x-6 +  4x-2)  (11x8 +  4x6) ] 
E  =  [ (-2x2 +  5x1)  (-2x-6 +  5x-2)  (-2x8 +  5x6) ]
      [ ( 3x2 + -6x1)  ( 3x-6 + -6x-2)  ( 3x8 + -6x6) ]

      [  26  -74  112 ] 
   =  [   1    2   14 ] 
      [   0   -6  -12 ] 
      =================

Find, if possible, values for “k” such that the system of linear equations below will have

(i) a unique solution

(ii) infinitely many solutions

(iii) no solution

(2k + 1) x -	y +	2 z =	2
-2 x +	y +	z =	1
k² x	+	3 z =	4

    [  2k+1   -1    2   |  2  ]
    [    -2    1    1   |  1  ]
    [   k^2    0    3   |  4  ]

Try to avoid a pivot on an element containing k.
The first row operation is therefore R₁ « R₂:

    [    -2    1    1   |  1  ]
    [  2k+1   -1    2   |  2  ]
    [   k^2    0    3   |  4  ]

R₁ ¬ R₁ × (-1/2):

    [     1   -1/2 -1/2 | -1/2  ]
    [  2k+1   -1    2   |  2    ]
    [   k^2    0    3   |  4    ]

R₂ ¬ R₂ + (- (2k + 1)) R₁ and
R₃ ¬ R₃ + (- k²) R₁:

    [   1   -1/2   -1/2     | -1/2     ]
    [   0   k-1/2  k+5/2    | k+5/2    ]
    [   0   k^2/2  k^2/2+3  | k^2/2+4  ]

If k = 0 then the linear system is

    [   1   -1/2   -1/2  | -1/2 ]
    [   0   -1/2    5/2  |  5/2 ]
    [   0     0     3    |  4   ]

which, after row divisions in rows 2 and 3, will leave a row echelon form with a leading “1” in every column on the left. The linear system therefore has a unique solution when k = 0.

For all other values of k,
R₃ « R₂:

    [   1   -1/2   -1/2     | -1/2     ]
    [   0   k^2/2  k^2/2+3  | k^2/2+4  ] 
    [   0   k-1/2  k+5/2    | k+5/2    ]

R₂ ¬ R₂ × (2 / k² ):

    [   1   -1/2   -1/2     | -1/2     ]
    [   0    1    1+6/k^2   | 1+8/k^2  ] 
    [   0   k-1/2  k+5/2    | k+5/2    ]

R₃ ¬ R₃ + ( (1/2)-k) R₂:

    [ 1   -1/2         -1/2              |           -1/2             ]
    [ 0    1          1+6/k^2            |          1+8/k^2           ] 
    [ 0    0  (k+5/2 - (k-1/2)(1+6/k^2)) | (k+5/2 - (k-1/2)(1+8/k^2)) ]

Rank(A|b) = 3 for all non-zero values of k.
Rank(A) = 3 unless ((2k + 5)×(k²) = (2k - 1)×(k² + 6))
Þ 2 k³ + 5 k² = 2 k³ - k² + 12 k - 6
Þ 6 k² - 12 k + 6 = 0
Þ (k - 1)² = 0
Þ k = 1 only.

Therefore the linear system has
no solutions when k = 1; and
a unique solution for all other values of k.

5 r +	4 s +	4 t +	5 u +	k ⁴ v =
4 r +	3 s +	3 t +	4 u +	k ³ v =
3 r +	2 s +	2 t +	3 u +	k ² v =
2 r +	1 s +	1 t +	2 u +	k ¹ v =

This is a homogeneous linear system.
Therefore the trivial solution (r, s, t, u, v) = (0, 0, 0, 0, 0) is guaranteed to exist, regardless of the value of k.
There are fewer equations than unknowns. Therefore there are infinitely many solutions regardless of the value of k.
The row reduction below is not necessary to answer this question!

    [ 5   4   4   5   k^4 |  0 ]
    [ 4   3   3   4   k^3 |  0 ] 
    [ 3   2   2   3   k^2 |  0 ] 
    [ 2   1   1   2    k  |  0 ]

R₁ ¬ R₁ × (1/5):

    [ 1  4/5 4/5  1  k^4/5 |  0 ]
    [ 4   3   3   4   k^3  |  0 ] 
    [ 3   2   2   3   k^2  |  0 ] 
    [ 2   1   1   2    k   |  0 ]

R₂ ¬ R₂ + (-4) R₁,
R₃ ¬ R₃ + (-3) R₁,
R₄ ¬ R₄ + (-2) R₁:

    [ 1  4/5  4/5  1     k^4/5      |  0 ]
    [ 0 -1/5 -1/5  0  k^3 - 4*k^4/5 |  0 ] 
    [ 0 -2/5 -2/5  0  k^2 - 3*k^4/5 |  0 ] 
    [ 0 -3/5 -3/5  0   k  - 2*k^4/5 |  0 ]

R₂ ¬ R₂ × (-5):

    [ 1  4/5  4/5  1     k^4/5      |  0 ]
    [ 0   1    1   0  4*k^4 - 5*k^3 |  0 ] 
    [ 0 -2/5 -2/5  0  k^2 - 3*k^4/5 |  0 ] 
    [ 0 -3/5 -3/5  0   k  - 2*k^4/5 |  0 ]

R₃ ¬ R₃ + (2/5) R₂,
R₄ ¬ R₄ + (3/5) R₁:

    [ 1  4/5  4/5  1         k^4/5      |  0 ]
    [ 0   1    1   0    4*k^4 - 5*k^3   |  0 ] 
    [ 0   0    0   0  k^2 + k^4 - 2*k^3 |  0 ] 
    [ 0   0    0   0  k + 2*k^4 - 3*k^3 |  0 ]
=
    [ 1  4/5  4/5  1            k^4/5           |  0 ]
    [ 0   1    1   0       k^3 * (4*k - 5)      |  0 ] 
    [ 0   0    0   0        k^2 * (k-1)^2       |  0 ] 
    [ 0   0    0   0  k*(k-1)*(2*k^2 + 2*k - 1) |  0 ]

If k = 0 or k = 1, then in this partially reduced matrix, elements a₃₅ = a₄₅ = 0 and rank(A) = 2. This generates a three parameter family of solutions.

For all other values of k,

R₃ ¬ R₃ × (1 / [k² × (k - 1)²]):

    [ 1  4/5  4/5  1           k^4/5            |  0 ]
    [ 0   1    1   0      k^3 * (4*k - 5)       |  0 ] 
    [ 0   0    0   0             1              |  0 ] 
    [ 0   0    0   0  k*(k-1)*(2*k^2 + 2*k - 1) |  0 ]

R₄ ¬ R₄ + (- [k × (k - 1) × (2k² + 2k - 1)]) R₃:

    [ 1  4/5  4/5  1           k^4/5          |  0 ]
    [ 0   1    1   0     k^3 * (4*k - 5)      |  0 ] 
    [ 0   0    0   0            1             |  0 ] 
    [ 0   0    0   0            0             |  0 ]

which leaves rank(A) = 3 and a two parameter family of solutions.

Answer:
The linear system has infinitely many solutions for all values of k.
It has a three parameter family of solutions for k = 0 and k = 1.
It has a two parameter family of solutions for all other values of k .

A fish farm “grows” two different species of fish, trout and perch. The total number of fish at the farm on 6 September, 2000 is “N”. The ratio of perch to trout at this time is “m/n”. Based on previous studies it has been determined that each perch will eat “k” grammes of food per day and that each trout consumes twice as much food as a perch does. On 6 September, 2000, it is necessary to supply “W” kg of food.

Determine the number of each type of fish present, for each of the following sets of values.
1. N = 10,000 , m = 2 , n = 3 , W = 32 kg , k = 2 g .
2. N = 10,000 , m = 3 , n = 2 , W = 20 kg , k = 2 g .
3. N = 20,000 , m = 3 , n = -1 , W = 20 kg , k = 2 g .
Let P = number of perch
and T = number of trout, then

Number of fish: P + T = N

Food eaten: Pk + T(2k) = W Þ P + 2T = W/k

Ratio of fish: P/T = m/n Þ nP - mT = 0

[The middle equation assumes that the masses W and k are measured in the same units.]

This generates the linear system for unknowns P, T:
```
        [  1   1 |  N  ]
        [  1   2 | W/k ]
        [  n  -m |  0  ]
```
Rather than solve each case in turn, we can solve the general case:
R₂ ¬ R₂ - R₁;
R₃ ¬ R₃ - n × R₁:
```
        [  1     1   |     N   ]
        [  0     1   | W/k - N ]
        [  0  -(m+n) |    -nN  ]
```
R₃ ¬ -R₃:
```
        [  1    1   |     N   ]
        [  0    1   | W/k - N ]
        [  0   m+n  |    nN   ]
```
R₁ ¬ R₁ - R₂;
R₃ ¬ R₃ - (m+n)×R₂:
```
        [  1    0   |      2N - W/k         ]
        [  0    1   |     W/k - N           ]
        [  0    0   | (m+2n)N - (m+n)(W/k)  ]
```
This system is inconsistent unless (m + 2n)×N = (m + n)×(W/k)

If this condition is satisfied, then the unique solution to the linear system is
P = 2N - W/k ,
T = W/k - N.
1. (m + 2n)×N = (2+6)×10000 = 80000
  (m + n)×(W/k) = (2+3)×16000 = 80000
  [Note that W/k = 32 kg / 2 g = 32,000 g / 2 g = 16,000 , (not 16)].
  The linear system therefore has a unique solution.
  P = 2N - W/k = 20000 - 16000 = 4,000
  T = W/k - N = 16000 - 10000 = 6,000
  As a final check, one can quickly verify that the numbers of perch and trout are indeed in the ratio 2:3.
  
  Answer: There are 4,000 perch and 6,000 trout.
2. (m + 2n)×N = (3+4)×10000 = 70000
  (m + n)×(W/k) = (3+2)×10000 = 50000
  The linear system is therefore inconsistent and there is no solution.
3. At first sight, this linear system does have a unique solution, because
  (m + 2n)×N = (3-2)×20000 = 20000
  (m + n)×(W/k) = (3-1)×10000 = 20000
  But look at the application!
  The given data cannot be from a realistic population because the given ratio “m:n” is negative. This therefore leads to a negative number of one of the two types of fish, which is clearly impossible.
  There is therefore no solution.
  [The linear system leads to the nonsense solution
  P = 30000 ,
  T = -10000,
  which is, indeed, in the ratio 3:-1.]

(i)		a unique solution
(ii)		infinitely many solutions
(iii)		no solution

Number of fish:		P + T = N
Food eaten:	Pk + T(2k) = W Þ	P + 2T = W/k
Ratio of fish:	P/T = m/n Þ	nP - mT = 0

ENGR 1405 Engineering Mathematics 1

Problem Set 1 Solutions

Problem Set 1
Solutions