Linear Algebra problem (linear equations)

  • Thread starter kirab
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Homework Statement



Given augmented matrix

[tex]\left(\begin{array}{ccc}a&1&-1\\2&1&b\end{array}\right)[/tex]

list conditions on a and b such that there is:

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

The Attempt at a Solution



I row-reduced the matrix to

[tex]\left(\begin{array}{ccc}1&0&\frac{b+1}{2-a}\\0&1&\frac{-2-a}{2-a}\end{array}\right)[/tex]

and ended up with

i) for no solution, a = 0, 2 (since on the steps to row reducing, there was a [tex]\frac{1}{a}[/tex] in one of the entries). My textbook, however, says that a = 2, b != -1 are the conditions for no solution. Why b != -1 and why not a = 0?

Now I didn't get a result for unique solution and infinitely many solutions separately.

I only got that [tex]x1 = \frac{b+1}{2-a}[/tex] and [tex]x2 = \frac{-2 - ab}{2-a}[/tex]
which is the correct unique solution except that the book places restrictions on a and b, namely that they must be a = 2, b = -1. Where are these restrictions coming from and wouldn't a = 2 make it have no solution? Also how would one get an infinite amount of solutions in this case? Thanks.
 
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Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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Look at the result for x1 and x2. If a=2 there is no solution if either numerator doesn't vanish. But if they both have the form 0/0 then you have an infinite number of solutions. To verify this in a clearer way just put a=2 and b=-1 into the original equation. Then the first row and second rows of the matrix are the same, so you really only have one equation in two unknowns.
 

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