Thread: matrix row reduction solution question..

1. matrix row reduction solution question..

$\displaystyle \begin{pmatrix} 1 & 1& \lambda &| 1\\ 0& \lambda -1 &1-\lambda &| 0\\ 0&1 &1-\lambda &|1-\lambda ^2 \end{pmatrix}$

cant we say that for lambda=1 we have endless solutions?

but also we can divide equation 2 and 3 by lambda-1 and after that so for lambda=1 we get that there is no solutions
so what is the correct solution

2. You know, when $\displaystyle \lambda=1$, we have that $\displaystyle \lambda-1=0$. So in this case, it doesn't make sense to divide by $\displaystyle \lambda-1$.

3. Originally Posted by transgalactic
$\displaystyle \begin{pmatrix} 1 & 1& \lambda &| 1\\ 0& \lambda -1 &1-\lambda &| 0\\ 0&1 &1-\lambda &|1-\lambda ^2 \end{pmatrix}$

cant we say that for lambda=1 we have endless solutions?

but also we can divide equation 2 and 3 by lambda-1 and after that so for lambda=1 we get that there is no solutions
so what is the correct solution
With $\displaystyle \lambda= 1$ the matrix is
$\displaystyle \begin{pmatrix} 1 & 1& 1 &| 1\\ 0& 0 & 0 &| 0\\ 0&1 & 0 &| 0\end{pmatrix}$

Which corresponds to the equations x+ y+ z= 1, 0= 0, 0= 0. It should be obvious that you can choose, say, y and z to be any numbers you want and solve for the corresponding x. That is why there are an infinite number of solutions.

As HappyJoe said (may he long be happy), you can only divide by $\displaystyle \lambda- 1$ as long as $\displaystyle \lambda- 1\ne 0$ so that it is not valid to do that and then let $\displaystyle \lambda= 1$.