Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$$ \left\{ \begin{array}{l} x + y + az + at = a \\ 2x + y - a^2z + 2at = 1 \\ 4x + 3y + a^2 z + 4a^2 t = 1 \end{array} \right. $$

Is there a way to predict the number of solutions $(x, y, z, t)$, depending on the value of the parameter $a$, without solving the system explicitely?

share|cite|improve this question
The rank of the system of linear equations can be at most 3, which means there are either no solutions or infinitely many solutions. – Daryl Aug 17 '12 at 10:44
up vote 1 down vote accepted

Yes, by reducing the augmented matrix of the system to row echelon form. For your system, the augmented matrix is $$ \left[ \begin{array}{cccc|c} 1 & 1 & a & a & a \\ 2 & 1 & -a^2 & 2a & 1 \\ 4 & 3 & a^2 & 4a^2 & 1 \end{array}\right] $$ and the row echelon form $$ \left[ \begin{array}{cccc|c} 1 & 1 & a & a & a \\ 0 & 1 & a^2+2a & 0 & 2a-1 \\ 0 & 0 & 2(a^2-a) & 4(a^2-a) & -2a \end{array}\right]. $$ First thing to look for is consistency, so that the last row does not correspond to $0=$non-zero. This guarantees existence of at least one solution. After that, the dimension of the solution space = the number of free variables = $n - \rm{rank}(A)$, where $n$ is the number of variables of the system, $A$ is the augmented matrix, whose rank is given by the number of non-zero rows in the row echelon form. (Of course if you've calculated the row echelon form, you're only 30 seconds from the explicit solution. If you need to know the dimension of the solution space, I don't think there's any alternative: you need the rank of $A$.)

share|cite|improve this answer
If $a=1$ though, the system is inconsistent. – Daryl Aug 17 '12 at 11:05

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.