1. ## Matrices

$\displaystyle \begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;1 & 0 & 0 & 14\;\\\;0 & 0 & 1 & -13\end{vmatrix}$

If the variables in the equations of this matrix were x, y, and z:

x = 14
z = -13

I'm just not sure what to answer for y. Does having no y value in the augmented matrix mean the system has infinitely many solutions or a unique solution?

2. Originally Posted by Thomas $\displaystyle \begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;1 & 0 & 0 & 14\;\\\;0 & 0 & 1 & -13\end{vmatrix}$

If the variables in the equations of this matrix were x, y, and z:

x = 14
z = -13

I'm just not sure what to answer for y. Does having no y value in the augmented matrix mean the system has infinitely many solutions or a unique solution?
there is no leading 1 in the y column, it means we can equate y to a parameter (thus there are infinitely many solutions, because the parameter can be essentially anything, and each new parameter gives rise to a new solution set)... you know this

3. Thank you for the help.

For some reason, when doing questions that actually mean something, I start doubting my instincts.

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