0
$\begingroup$

I'm having a hard time understanding this question:

Determine the reduced echelon form of the homogeneous linear system of three equations in variables $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ such that $x_{1}, x_{2}, x_{4}$ are leading variables; $x_{3}, x_{5}$ are free variables and which has solutions

$$\begin{pmatrix}2 \\-1\\1\\0\\0 \end{pmatrix} \quad\text{and}\quad \begin{pmatrix}-3 \\2\\0\\-4\\1 \end{pmatrix} $$

$\endgroup$
0
$\begingroup$

You have \begin{cases} x_1=2x_3-3x_5\\ x_2=-x_3+2x_5\\ x_4=-4x_5 \end{cases} which means \begin{cases} x_1-2x_3+3x_5=0\\ x_2+x_3-2x_5=0\\ x_4+4x_5=0 \end{cases} This system's matrix is what you're looking for.

If you can't see why, here's a different strategy: the matrix must be of the form \begin{bmatrix} 1 & 0 & a & 0 & b \\ 0 & 1 & c & 0 & d \\ 0 & 0 & 0 & 1 & e \end{bmatrix} and the two given vectors should be in its null space.

$\endgroup$
  • $\begingroup$ that makes sense, thanks!! $\endgroup$ – 01000101 Jan 20 '17 at 9:33

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.