0
$\begingroup$

Hello I have two $3\times3$ matrices and need to prove that they are not row equivalent for any values of $a,b,c$ where they are all real numbers

$$ \begin{pmatrix} 2 & 0 & 0 \\ a & -1 & 0 \\ b & c & 3 \\ \end{pmatrix} $$ $$ \begin{pmatrix} 1 & 1 & 2 \\ -2 & 0 & -1 \\ 1 & 3 & 5 \\ \end{pmatrix} $$

so what is did was I did a bunch of row operation on the second matrix to try and get the first one but since they are supposed to be not row equivalent I couldn't get it. this is the final matrix I got. $$ \begin{pmatrix} 2 & 0 & 1 \\ 3 & -1 & 0 \\ 0 & 2 & 3 \\ \end{pmatrix} $$ Since my knowledge on matrices is very limited I have no idea if I am on the right track or not or how I would prove this. Can you help me please?.

$\endgroup$
1
$\begingroup$

The first matrix has full row rank. Notice it is already lower triangular with nonzero diagonal elements. If you are more comfortable with RREF, you can do that too and arrive at the identity matrix.

On the other hand, if you perform RREF on the second matrix, you will find that it has row rank $2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.