\begin{bmatrix} a & 0 & b & 2 \\ a & a & 4 & 4 \\ 0 & a & 2 & b \\ \end{bmatrix}

Find the values of a and b such that the system has:

a) A unique solution

So I think I got this although I'm not sure.

$|A| \not= 0$

reduced it down to:

\begin{bmatrix} 1 & 0 & b/a \\ 0 & 1 & 4-1/a \\ 0 & 0 & b-2 \\ \end{bmatrix}

Therefore $b\not=2$ and $a\not=0$ Do I mention anything else here?

b) A two-parameter solution

This is the one I'm having trouble with. I don't know how to answer it and can't find anything telling me how to.

I got this from simplifying everything down and assuming $y=v$ and $z=w$


Therefore $a\not=0$ and $b$=all real


1 Answer 1


Here is one way to approach this problem:

Reduce $\begin{bmatrix}a&0&b&2\\a&a&4&4\\0&a&2&b\end{bmatrix}\longrightarrow\begin{bmatrix}a&0&b&2\\0&a&4-b&2\\0&a&2&b\end{bmatrix}\longrightarrow\begin{bmatrix}a&0&b&2\\0&a&4-b&2\\0&0&b-2&b-2\end{bmatrix}$.

Now continue reduction to row echelon form in each of the following cases:

1) $a\ne0, b\ne 2$

2) $a\ne0, b=2$

3) $a=0, b=2$

4) $a=0, b\ne2$

  • $\begingroup$ This helps with the first part, but the second part is where I'm having the most trouble $\endgroup$
    – Rick
    Apr 25, 2015 at 1:18
  • $\begingroup$ To get a two-parameter solution, you need to have two free variables (so there are two columns with no leading ones), and the system has to be consistent; so now you just have to determine which of the last 3 cases will give you this. $\endgroup$
    – user84413
    Apr 25, 2015 at 17:59

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