1
$\begingroup$

As the title stated, what is the most efficient method for finding the variable matrix?

For example, in the linear system $Ax = b$, given $$A = \begin{pmatrix} 1 & 1& 5 \\ 2& 3& 4 \\ 0& 1& -1\\ 1 & 1& 2 \\ \end{pmatrix}$$ and $$b = \begin{pmatrix} 16 \\ 13 \\ -4\\ 7 \\ \end{pmatrix}$$ what would then be the variable matrix $x$?

For my attempt, i've thought of reducing $(A|b)$ into its reduced row-echelon form:

I've got the $I_{4}$ instead, which probably doesn't tell me much.

Please advise, thanks.

$\endgroup$
3
$\begingroup$

Why don't you think it tells you much? If you've reduced the equation to $$I_{4\times 4} x = \tilde b$$ for some new right-hand side $\tilde b$, what is $x$?

$\endgroup$
  • $\begingroup$ Does it mean that $x = (0, 0, 0)$? However, apparently i've got the answer that $x = (2, -1, 3)$ as the solution to the linear system. How does one derive that variable matrix? $\endgroup$ – idolo Oct 28 '17 at 10:25

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.