0
$\begingroup$

I need help on how to do part a) and b). I know for part c), I just have to row reduce and get my vector X.

enter image description here

1 Answers 1

1

a) Any linear combination of the columns (distinct from the columns).

b) Find first the equation(s) of the image, i.e. solve for $A\cdot X=0$. Then take any vector which does not satisfy at least one of these equations. The number of linearly independent equations is $4-\operatorname{rank}A$.

  • 0
    Could you elaborate more on how to find a linear combination from just that A matrix? I know how to do it if I have, say a B, vector. Also, I don't quite understand how to find a vector that doesn't satisfy one of the equations. When finding the equations of the image, from your method, is that the X vector? – 2017-01-23
  • 0
    You don't have to ‘find’ a linear combination: just choose a set of 4 coefficients (not all equal to $0$ but one, of course), and compute $\lambda_1C_1+\lambda_2C_2+\dotsm$. Yes, the $X$ vector gives the coefficients of the equation(s). Suppose there's only one equation, say $x-2y-z+t=0$; just choose $x,y,z,t $ so that this equation is *not* satisfied. It shouldn't be really hard… – 2017-01-23