ENGR 1405 Engineering Mathematics 1

Faculty of Engineering and Applied Science
2000 Fall

Quiz 1
Solution

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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

x	+	y	-	2 z	=	1
-x	+	2 y	+	(2k + 5) z	=	2
		3 y	+	k2 z	=	k

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

R₂ ¬ R₂ + R₁:

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

R₂ ¬ R₂ / 3:

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

R₃ ¬ R₃ - 3 R₂:

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

But   k² - 2k - 3 = (k + 1) (k - 3)

Case k = -1:

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

Clearly   rank(A) < rank(A|b).
Therefore the linear system is inconsistent and has no solutions when k = -1.

Case k = 3:

    [  1   1  -2  |  1 ]
    [  0   1   3  |  1 ]
    [  0   0   0  |  0 ]

Clearly   rank(A) = rank(A|b) = 2 (< n = 3).
Therefore the linear system has infinitely many solutions when k = 3.

Case else:


R₃ ¬ R₃ / ((k + 1) (k - 3)):

    [  1   1     -2     |   1     ]
    [  0   1 (2k/3) + 1 |   1     ]
    [  0   0      1     | 1/(k+1) ]

Clearly   rank(A) = rank(A|b) = n = 3.
Therefore the linear system has a unique solution for all values of k except -1 and 3.

Answer:
The linear system has


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• infinitely many solutions for k = 3 only;
• no solutions for k = -1 only;
• a unique solution for all other values of k.

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Note:
There are other valid ways of reaching echelon form, such as

R₂ « R₃; then
R₂ ¬ R₂ / 3; then
R₃ ¬ R₃ + R₁; then
R₃ ¬ R₃ - 3 R₂; leading to

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

The same conclusions then follow.

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Created 2000 09 21 and modified 2000 09 22 by Dr. G.H. George

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