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All entries left blank in the determinant below are zeros.
Click on the image for a step by step proof.
If RREF of $A$ is $I_3$, then $A$ is invertible.
This implication is proved in Theorem 7 in Section 2.2 . This proof is important!Step  the row operation  the elementary matrix  the inverse of ele mat 

1st  The third row is replaced by the the sum of the first row and the third row  $E_1 = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right]$  $E_1^{1} = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right]$ 
2nd  The third row and the second row are interchanged  $E_2 = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right]$  $E_2^{1} = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right]$ 
3rd  The third row is replaced by the the sum of the third row and the second row multiplied by $(2)$  $E_3 = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end{array}\right]$  $E_3^{1} = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end{array}\right]$ 
4th  The first row is replaced by the the sum of the first row and the third row  $E_4 = \left[\!\begin{array}{rrr} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$  $E_4^{1} = \left[\!\begin{array}{rrr} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$ 
$\displaystyle \begin{bmatrix}0 \\ 0 \\0\end{bmatrix}$ Black  $\displaystyle \begin{bmatrix}1/2 \\ 1/2 \\ 1/2\end{bmatrix}$ Gray  $\displaystyle \begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}$ White  

$\displaystyle \begin{bmatrix}1 \\ 0 \\0\end{bmatrix}$ Red  $\displaystyle \begin{bmatrix}1/2 \\ 0 \\0\end{bmatrix}$ Maroon  $\displaystyle \begin{bmatrix}1/2 \\ 1/2 \\0\end{bmatrix}$ Olive  $\displaystyle \begin{bmatrix}1 \\ 1/2 \\0\end{bmatrix}$ Orange  $\displaystyle \begin{bmatrix}1 \\ 1 \\0\end{bmatrix}$ Yellow  
$\displaystyle \begin{bmatrix}0 \\ 1 \\0\end{bmatrix}$ Green 

$\displaystyle \begin{bmatrix}1/2 \\ 1 \\0\end{bmatrix}$ Chartruse  $\displaystyle \begin{bmatrix}0 \\ 1/2 \\ 1/2\end{bmatrix}$ Teal 


$\displaystyle \begin{bmatrix}0 \\ 0 \\1\end{bmatrix}$ Blue  $\displaystyle \begin{bmatrix}0 \\ 0 \\ 1/2\end{bmatrix}$ Navy  $\displaystyle \begin{bmatrix}1/2 \\ 0 \\ 1/2\end{bmatrix}$ Purple  $\displaystyle \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}$ Magenta  $\displaystyle \begin{bmatrix}0 \\ 1 \\ 1\end{bmatrix}$ Cyan  







$\displaystyle \begin{bmatrix} 1 \\ 1/2 \\ 1/2 \end{bmatrix}$ Salmon 


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Since Teal and Yellow are the heads of particular vectors in the Color Cube, to construct a transition I connected the heads with a line segment. Points on this line segment are the heads of special linear combinations of the vectors representing Teal and Yellow. As an exercise write the linear combinations which are used in the above transition.
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In the above animation I used the colors from the line segment connecting Teal and Yellow to color the rectangles in the middle of the square.
Above is the unit circle colored using colors from the line segment connecting Teal and Yellow.