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For the following matrix:

$$ \begin{bmatrix} 1 & 2 & 1 & 3 \\ 2 & 5 & 5 & 6 \\ 3 & 7 & 6 & 11 \\ 1 & 5 & 10 & 8 \\ \end{bmatrix} $$

I want to find a basis for the Row space, Column space, Kernel (if it represents a Linear Transformation), and Image(if it represents a Linear Transformation).

The row echelon form of this matrix is: $$ \begin{bmatrix} 1 & 2 & 1 & 3 \\ 0 & 1 & 3 & 0 \\ 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix} $$

Row space basis:

$$\{(1,2,1,3),(0,1,3,0),(0,0,0,2)\}$$

Column space basis:

$$\{(1,2,3,1),(2,5,7,5),(3,6,11,8)\}$$

Kernel basis:

$x+2y+z+3w=0$

$y+3z=0$

$2w=0$

$(x,y,z,w)=z(5,-3,1,0)$

$$\{(5,-3,1,0)\}$$

Image basis (same as column space):

$$\{(1,2,3,1),(2,5,7,5),(3,6,11,8)\}$$


Are my above solutions correct? Of not then why?

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    $\begingroup$ Without checking the numerical calculations, it looks good. $\endgroup$ – A.Γ. Apr 20 at 20:55
  • $\begingroup$ You can check that you’ve gotten the correct basis for the kernel yourself by multiplying that vector by the matrix. $\endgroup$ – amd Apr 20 at 23:33

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