Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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


share|improve this question
3  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.