311.1.5.17 geometric description

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The discussion focuses on the geometric description of the solution set for the linear system represented by the equations given in matrix form. The reduced row echelon form (RREF) of the matrix indicates that the solution is the trivial solution (0,0,0). The determinant of the matrix is calculated to be -48, confirming that the transformation is one-to-one, meaning the only vector that maps to the zero vector is the zero vector itself.

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karush
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$\tiny{311.1.5.17}$
Give a geometric description of the solution set.

$\begin{array}{rrrrr}
-2x_1&+2x_2&+4x_3&=0\\
-4x_1&-4x_2&-8x_3&=0\\
&-3x_2&-3x_3&=0
\end{array}$
this can be written as
$\left[\begin{array}{rrr|rr}-2&2&4&0\\-4&-4&-8&0\\&-3&-3&0\end{array}\right]$

$\text{RREF}=\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array} \right]$
ok I am not sure how you get a geometric description with everything going to zero
 
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the point (0,0,0)
 
Equivalently the matrix [math]\begin{bmatrix}-2 & 2 & 4 \\ -4 & -4 & 8 \\ 0 & -3 & -3\end{bmatrix}[/math] has determinant [math]\left|\begin{array}{ccc}-2 & 2 & 4 \\ -4 & -4 & 8 \\ 0 & -3 & -3 \end{array}\right|= 3\left|\begin{array}{cc}-2 & 4 \\ -4 & 8\end{array}\right|- 3\left|\begin{array}{cc}-2 & 2 \\ -4 & -4\end{array}\right|= 3(-16+ 16)- 3(8+ 8)= -3(16)= -48 [/math] which is not 0 so is a "one to one" transformation. The only vector that is mapped to the 0 vector is the 0 vector itself.
 

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