# Thread: Reduce Augmented Matrix to find value to consistent equations.

1. ## Reduce Augmented Matrix to find value to consistent equations.

I'm not sure how to do this one, I row reduced a bit but don't know where to go with it.
$
\begin{bmatrix}
1 & 2 & 2 & | & 1 \\
2 & 1 & 3 & | & 3 \\
4 & 5 & 7 & | & \lambda \\
\end{bmatrix}
$

How do I find the value of $\lambda$ for which the equations are consistent?

Thanks

2. Originally Posted by alexgeek
I'm not sure how to do this one, I row reduced a bit but don't know where to go with it.
$
\begin{bmatrix}
1 & 2 & 2 & | & 1 \\
2 & 1 & 3 & | & 3 \\
4 & 5 & 7 & | & \lambda \\
\end{bmatrix}
$

How do I find the value of $\lambda$ for which the equations are consistent?

Thanks
Reduce it to upper triangular and post your results.

I just did that myself and I don't think it can be consistent.

3. What did you get when you "row reduced a bit"? Yes, there is a single value of $\lambda$ that makes this system consistent. What it is should be clear after you have "done" the first column.

4. Originally Posted by HallsofIvy
What did you get when you "row reduced a bit"? Yes, there is a single value of $\lambda$ that makes this system consistent. What it is should be clear after you have "done" the first column.
I was lazy so I let my calculator do the ref. By doing it by hand, I see the answer is 1.

5. Oh I reduced it a bit last night then didn't know where to go, tried it this morning and got 5 but if you did it with a calculator I doubt I'm right.

$R_2 - 2R_1$
$
\begin{bmatrix}
1 & 2 & 2 & | & 1 \\
0 & -3 & -1 & | & 1 \\
4 & 5 & 7 & | \lambda \\
\end{bmatrix}
$

$R_1 + 2R_2$
$
\begin{bmatrix}
1 & -4 & 0 & | & 3 \\
0 & -3 & -1 & | & 1 \\
4 & 5 & 7 & | \lambda \\
\end{bmatrix}
$

$R_3 -4 R_1$
$
\begin{bmatrix}
1 & -4 & 0 & | & 3 \\
0 & -3 & -1 & | & 1 \\
0 & 21 & 7 & | \lambda -12 \\
\end{bmatrix}
$

$R_3 + 7 R_2$
$
\begin{bmatrix}
1 & -4 & 0 & | & 3 \\
0 & -3 & -1 & | & 1 \\
0 & 0 & 0 & | \lambda -5 \\
\end{bmatrix}
$

So I think if $\lambda = 5$ then the equations are linearly dependent as there's a row if zeros yes?

Thanks

6. After row reduction, an inconsistent system looks like a row of zeros before the augmented matrix sign, but a nonzero entry after it. Such an equation can't hold, can it? So you have two choices here: either the system is consistent and has infinitely many solutions (but only one choice for $\lambda$), or the system is inconsistent and has no solutions (but any other choice for $\lambda$ is fine).

7. Yes, that is correct. But you are doing too much work. Your second step, adding a multiple of the second row to the first, is unecessary. Always work by columns, from left to right. Since there is already a "1" in the upper left, you don't need to do anything to the first row. Subtract twice the first row from the second and 4 times the first row from the third to get
$\left[\begin{array}{ccc}1 & 2 & 2 \\ 0 & -3 & -1 \\ 0 & -3 & -1\end{array}\left|\begin{array}{c}1 \\ 1\\ \lambda- 4\end{array}\right]$

Now since the first three numbers in the last two rows are the same, subracting the second row from the third will give "0 0 0" so the matrix is not "invertible"- there is not a unique solution, which is what dwsmith really meant to say. If $\lambda= 5$, then $\lambda- 4= 1$ so that the two last rows are identical. In that case, there are an infinite number of solutions so the system certainly is "consistent". If $\lambda$ is any other number, there is NO solution which is what "inconsistent" means.