# Thread: Use Cramer's Rule to find x_2 when Ax=...

1. ## Use Cramer's Rule to find x_2 when Ax=...

I think I was handed a trick question.

For the matrix $\displaystyle \left(\begin{array}{ccc}1 & 2 & 3 \\2 & 1 & 3 \\1 & -1 & -1\end{array}\right)$, use Cramer's Rule to find $\displaystyle x_2$ when $\displaystyle Ax=\left(\begin{array}{c}1 \\-2\\1\end{array}\right)$.

My current answer is $\displaystyle x_2=\frac{16}{3}$. Is this correct, or do I have to determine what $\displaystyle x$ is by inverting $\displaystyle A$?

2. Originally Posted by Runty
I think I was handed a trick question.

For the matrix $\displaystyle \left(\begin{array}{ccc}1 & 2 & 3 \\2 & 1 & 3 \\1 & -1 & -1\end{array}\right)$, use Cramer's Rule to find $\displaystyle x_2$ when $\displaystyle Ax=\left(\begin{array}{c}1 \\-2\\1\end{array}\right)$.

My current answer is $\displaystyle x_2=\frac{16}{3}$. Is this correct, or do I have to determine what $\displaystyle x$ is by inverting $\displaystyle A$?
You are correct as you are only asked for $\displaystyle x_2$ .

Solving for $\displaystyle x$ means solving for $\displaystyle (x_1,x_2,x_3)$

as $\displaystyle x = \left(\begin{array}{c}x_1 \\ x_2 \\ x_3 \end{array}\right)$

This requires inverting $\displaystyle A$

3. Originally Posted by pickslides
You are correct as you are only asked for $\displaystyle x_2$ .

Solving for $\displaystyle x$ means solving for $\displaystyle (x_1,x_2,x_3)$

as $\displaystyle x = \left(\begin{array}{c}x_1 \\ x_2 \\ x_3 \end{array}\right)$

This requires inverting $\displaystyle A$
Alright, doing that, my new answer is $\displaystyle x_2=-\frac{65}{9}$, with $\displaystyle x=\left(\begin{array}{c}7/3 \\ 16/3 \\ -4 \end{array}\right)$. Is that correct, or have I made a mistake somewhere?

Remember, the question says to use Cramer's Rule to find $\displaystyle x_2$.

4. Have a look at Online Matrix Calculator

5. The problem told you to use "Cramer's rule":

$\displaystyle x_2= \frac{\left|\begin{array}{ccc}1 & 1 & 3 \\ 2 & -2 & 3 \\ 1 & 1 & -1\end{array}\right|}{\left|\begin{array}{ccc}1 & 2 & 3 \\2 & 1 & 3 \\ 1 & -1 & 3\end{array}\right|}$.

I have no idea why you would consider that a "trick question".

6. Originally Posted by HallsofIvy
The problem told you to use "Cramer's rule":

$\displaystyle x_2= \frac{\left|\begin{array}{ccc}1 & 1 & 3 \\ 2 & -2 & 3 \\ 1 & 1 & -1\end{array}\right|}{\left|\begin{array}{ccc}1 & 2 & 3 \\2 & 1 & 3 \\ 1 & -1 & 3\end{array}\right|}$.

I have no idea why you would consider that a "trick question".
I say it because I believe that by stating $\displaystyle Ax=\left(\begin{array}{c}1 \\-2\\1\end{array}\right)$, it could mean that you need to do this: $\displaystyle x=A^{-1}\left(\begin{array}{c}1 \\-2\\1\end{array}\right)$, before you apply Cramer's Rule. This would come out to $\displaystyle x=\left(\begin{array}{c}7/3 \\ 16/3 \\ -4 \end{array}\right)$, and this $\displaystyle x$ would be used in Cramer's Rule, not $\displaystyle Ax=\left(\begin{array}{c}1 \\-2\\1\end{array}\right)$.

Of course, I could be wrong. That's why I thought this might be a trick question.

7. Nevermind, I just realized the mistake I made. It was a notation mistake. The correct answer is $\displaystyle x_2=\frac{16}{3}$.