# Simultaneous equation matrices, how do my answers look here?

How many solutions does each system have

• A. Unique solution
• B. No solutions
• C. Infintely many solutions
• D. None of the above.

$$\left(\begin{array}{cc|r} 1 & 0 & 6\\ 0 & 1 & -4\\ \end{array}\right)$$

My answer: A - Unique solution

x = 6, y= -4

$$\left(\begin{array}{cc|r} 1 & 0 & 8\\ 0 & 1 & -11\\ 0 & 0 & 0 \end{array}\right)$$

My answer: A - Unique solution

x = 8, y= -11 Although, I think I recall hearing that if the bottom row is all zeroes, has infintely many solutions?

$$\left(\begin{array}{ccc|r} 1 & 0 & 14 & 0\\ 0 & 1 & 15 & 0\\ 0 & 0 & 0 & 1\\ \end{array}\right)$$

My answer: B - No solutions

As the bottom row is invalid. Or am I allowed to 'overlook that'? And say it has infinitely many solutions?

$$\left(\begin{array}{ccc|r} 0 & 1 & 0 & -4\\ 0 & 0 & 1 & 2\\ \end{array}\right)$$

My answer: A - Unique solution y = -4, z = 2

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For the second one, the third row puts no further constraints on $x$ and $y$. So unique solution. For next to last, can't "overlook." For last, not unique, there are $3$ variables, so infinitely many solutions, $x=$ anything, $y=-4$, $z=2$. By the way, it's not matrices that have or don't have solutions, it is systems of equations. – André Nicolas Feb 9 '12 at 2:07
Note that "D" could never be an answer for this type of question. – David Mitra Feb 9 '12 at 2:09