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.

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

I have a problem that says to find $h$ such that the matrix is the augmented matrix of a consistent linear system. Here's the matrix: $$\left[\begin{matrix}1&-1&4\\-2&3&h\end{matrix}\right]$$ Now why is it that any $h$ makes this consistent? I was able to reduce the matrix down to $$\left[\begin{matrix}1&-1&4\\0&1&h+8\end{matrix}\right]$$But I'm not sure where to go from here. How would one realize that $h$ can be any number?

share|cite|improve this question
If $h=-8$, then the matrix will be inconsistent. – Calvin Lin Feb 5 '13 at 6:43
@CalvinLin Wouldn't $(4,0)$ be a solution? – Ben West Feb 5 '13 at 6:44
The matrix isn't consistent, but the linear system. Any $h$ is applicable because $rank(A)=2$ and $rank(A|b)$ must be two also. Then you can conclude that you have either a unique or infinite number of solutions. – Daryl Feb 5 '13 at 6:45
@BenW. Haha, yes ... My bad. I was thinking that consistent meant unique solution. – Calvin Lin Feb 5 '13 at 6:46
up vote 1 down vote accepted

Concretely, your reduced matrix is essentially asking you to find solutions to $$ x-y=4\qquad\text{and}\qquad y=h+8. $$

Say you let $h=h_0$ be any value. Then a possible solution to your system is $(12+h_0,h_0+8)$. So the system is consistent regardless of your choice for $h$. The key here is that $h+8$ is not a pivot point in your augmented matrix, since the $(2,2)$ entry is nonzero.

share|cite|improve this answer

For any $h$ the system will be \begin{split} x&-y&=4\\ -2x&+3y&=h \end{split} or equivalently $$Ax=b$$ with $$A=\begin{pmatrix}1 &-1 \\-2 & 3\end{pmatrix}, \ \ b=\begin{pmatrix}4\\h\end{pmatrix}.$$

Since $A$ is invertible the system has a solution for any $h$ (that is $A^{-1}b$).

share|cite|improve this answer

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.