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This is my first time posting a question and I'm really stuck on this one. Help would be appreciated. I'm given this matrix

$$\left[ \begin{array}{ccc|c} 2&3&0&9\\0&1&\lambda+6&4\\ 0&0&\lambda^2-5\lambda+6&9-3\lambda \end{array}\right], $$

and I need to find for which λ ∈ ℝ this matrix has

  • one solution
  • no solutions
  • infinite solutions
  • 0
    Hello and welcome. What have you tried so far? – 2017-02-16
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    Well I do know how to solve it for one solution? I can just solve the last row and find λ right? But I'm afraid I don't know how to go past that. – 2017-02-16
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    Perhaps the two questions should be posted as separate questions. – 2017-02-16
  • 0
    Oh okay sorry about that. – 2017-02-16
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    @T.J.Dom No worries. You're doing fine. – 2017-02-16
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    do you mean this is an augmented matrix of the for $[A|b]$ corresponding to $Ax = b?$ – 2017-02-16
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    @abel Yes, I copied a link to here and didn't include the vertical bar. Fixed now. – 2017-02-16

3 Answers 3

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There are some conditions dealing with pivots that will help you decide these three cases. The conditions are:

  1. No solution: A pivot in the constant column. Since the first two rows have pivots, such a pivot must occur in the third row. Therefore, the constant entry in the third row must be nonzero and this is the leading entry of the row. In other words, $\lambda^2-5\lambda+6=0$ and $9-3\lambda\not=0$.

  2. A unique solution: There are no pivots in the constant column and every variable is a pivot variable. The first two variables are pivot variables, to make the third variable a pivot variable, you need the leading entry in the third row to be in the third column. In other words, $\lambda^2-5\lambda+6\not=0$

  3. Infinitely many solutions: There are no pivots in the constant column and there is a free variable. Since the first two variables are pivot variables, the third variable is the only choice to be free. To be free, you need the leading entry of the third row to not be in the third column. Once you've made that choice, you may worry that there is a pivot in the constant column. You'll need to make sure that the $\lambda$ you pick does not give a pivot in the constant column of the third row. In other words, $\lambda^2-5\lambda+6=0$ and $9-3\lambda=0$.

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Use the determinant formula for the first three columns of the matrix (that compose an upper triangular matrix, so you just need to get the product of the diagonal elements).

There is one solution if and only if the determinant you got is not equal to zero (Determinant theorem). Then, check the values for which the determinant is equal to 0 and write the new matrices using that values.

For every new matrix, you have to study how many dependent/independent rows or columns you have (you can use Gauss algorithm, too).

Knowing that, you can have a conclusion using the Rouché-Capelli theorem.

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see if there is a pivot in column three. you have one if $\lambda \neq 2$ and $\lambda \neq 3$ implying in these case you have a unique solution. now all you are left with are the cases $\lambda = 2$ and $\lambda = 3.$ i will let you handle those.