MHB How to find a non-zero vector in the column space of M

shamieh
Messages
538
Reaction score
0
Let the matrix $M = \begin{bmatrix}-12&-12&16&-15\\-6&-8&-8&-10\\0&20&0&25\end{bmatrix}$

Find a non zero vector in the column space of $M$

Is it not true that $\begin{bmatrix}-12\\-8\\20\end{bmatrix}$ is a non zero vector in the column space of $M$ ? For some reason it keeps telling me "that is incorrect your answer doesn't seem to be a Vector"
 
Physics news on Phys.org
shamieh said:
Let the matrix $M = \begin{bmatrix}-12&-12&16&-15\\-6&-8&-8&-10\\0&20&0&25\end{bmatrix}$

Find a non zero vector in the column space of $M$

Is it not true that $\begin{bmatrix}-12\\-8\\20\end{bmatrix}$ is a non zero vector in the column space of $M$ ? For some reason it keeps telling me "that is incorrect your answer doesn't seem to be a Vector"

Hi shameih,

It is true that $\begin{bmatrix}-12\\-8\\20\end{bmatrix}$ is a non-zero vector that is in the column space of $M$. I see nothing wrong with your answer. Is it possible that the way you inserted it into the computer might be incorrect?
 
Thread 'Determine whether ##125## is a unit in ##\mathbb{Z_471}##'
This is the question, I understand the concept, in ##\mathbb{Z_n}## an element is a is a unit if and only if gcd( a,n) =1. My understanding of backwards substitution, ... i have using Euclidean algorithm, ##471 = 3⋅121 + 108## ##121 = 1⋅108 + 13## ##108 =8⋅13+4## ##13=3⋅4+1## ##4=4⋅1+0## using back-substitution, ##1=13-3⋅4## ##=(121-1⋅108)-3(108-8⋅13)## ... ##= 121-(471-3⋅121)-3⋅471+9⋅121+24⋅121-24(471-3⋅121## ##=121-471+3⋅121-3⋅471+9⋅121+24⋅121-24⋅471+72⋅121##...

Similar threads

Back
Top