# 311.1.5.17 geometric description

• MHB
• karush
In summary, the given system of equations can be written as a matrix and its reduced row echelon form is an identity matrix, indicating that the solution set is a single point at the origin (0,0,0). This can also be seen from the fact that the determinant of the matrix is non-zero, indicating a one-to-one transformation.
karush
Gold Member
MHB
$\tiny{311.1.5.17}$
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

the point (0,0,0)

Equivalently the matrix $$\displaystyle \begin{bmatrix}-2 & 2 & 4 \\ -4 & -4 & 8 \\ 0 & -3 & -3\end{bmatrix}$$ has determinant $$\displaystyle \left|\begin{array}{ccc}-2 & 2 & 4 \\ -4 & -4 & 8 \\ 0 & -3 & -3 \end{array}\right|= 3\left|\begin{array}{cc}-2 & 4 \\ -4 & 8\end{array}\right|- 3\left|\begin{array}{cc}-2 & 2 \\ -4 & -4\end{array}\right|= 3(-16+ 16)- 3(8+ 8)= -3(16)= -48$$ 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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