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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?

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If $h=-8$, then the matrix will be inconsistent. –  Calvin Lin Feb 5 '13 at 6:43
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@CalvinLin Wouldn't $(4,0)$ be a solution? –  000 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

2 Answers 2

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.

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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 $A^{-1}b$).

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