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?
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