What are the image and kernel of matrices A and B?

[toothpaste666]

I tried an example with three x's and made up the value for 2 and solved for the last but i ended up with 2 different values for that x one for each equation, but they didnt work for both

 

[Mark44]

so a nontrivial relation would be
for the 2x2 matrix B would be
$5\begin{bmatrix}2\\6\end{bmatrix} + \frac{-10}{3}\begin{bmatrix}3\\9\end{bmatrix} = 0$

Sure, that works. A simpler example is
$1\begin{bmatrix}2\\6\end{bmatrix} + \frac{-2}{3}\begin{bmatrix}3\\9\end{bmatrix} = 0$

 

[Mark44]

I tried an example with three x's and made up the value for 2 and solved for the last but i ended up with 2 different values for that x one for each equation, but they didnt work for both

Which matrix are you referring to here?

 

[toothpaste666]

I wanted to find a nontrivial relation between the column vectors
$\begin{bmatrix}1\\2\end{bmatrix}, \begin{bmatrix}2\\3\end{bmatrix}, \begin{bmatrix}3\\4\end{bmatrix}$
so i picked
$x_1 \begin{bmatrix}1\\2\end{bmatrix} + 2 \begin{bmatrix}2\\3\end{bmatrix} + 3 \begin{bmatrix}3\\4\end{bmatrix}$
$= \begin{bmatrix}x_1\\2x_1\end{bmatrix} + \begin{bmatrix}4\\6\end{bmatrix} + \begin{bmatrix}9\\12\end{bmatrix}$
$= \begin{bmatrix}x_1 + 4 + 9\\2x_1 + 6 + 12\end{bmatrix}$
$= \begin{bmatrix}x_1 + 13\\2x_1 + 18\end{bmatrix}$
so for the top row x1 = -13
and for the bottom row x1 = -9 but if i plug either of this into x1 in the first step neither of them work

 

[Mark44]

I wanted to find a nontrivial relation between the column vectors
$\begin{bmatrix}1\\2\end{bmatrix}, \begin{bmatrix}2\\3\end{bmatrix}, \begin{bmatrix}3\\4\end{bmatrix}$

This is a different problem from the two you posted at the start of this thread (so you should have started a new thread...)
The advice I gave about picking two of the constants applied only to matrix A. It doesn't apply to this matrix.

What you're doing here is to solve this matrix equation for the constants:
$$\begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4\end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3\end{bmatrix} = \begin{bmatrix} 0 \\ 0\end{bmatrix}$$
To do this, row reduce the matrix on the left. Once the matrix is reduced, you can read off the constants.

so i picked
$x_1 \begin{bmatrix}1\\2\end{bmatrix} + 2 \begin{bmatrix}2\\3\end{bmatrix} + 3 \begin{bmatrix}3\\4\end{bmatrix}$
$= \begin{bmatrix}x_1\\2x_1\end{bmatrix} + \begin{bmatrix}4\\6\end{bmatrix} + \begin{bmatrix}9\\12\end{bmatrix}$
$= \begin{bmatrix}x_1 + 4 + 9\\2x_1 + 6 + 12\end{bmatrix}$
$= \begin{bmatrix}x_1 + 13\\2x_1 + 18\end{bmatrix}$
so for the top row x1 = -13
and for the bottom row x1 = -9 but if i plug either of this into x1 in the first step neither of them work

 

[toothpaste666]

oops sorry! that makes sense. is the reason that only applies to A because it is a square matrix?

 

[Mark44]

oops sorry! that makes sense. is the reason that only applies to A because it is a square matrix?

A is not a square matrix: it is 3 X 1. B is a square matrix. The vectors in post #34 are unrelated to the A and B matrices.

Please start a new thread.