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

**Runty** · April 12th 2010, 05:11 PM

I think I was handed a trick question.

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

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

**pickslides** · April 12th 2010, 05:22 PM

Originally Posted by Runty

I think I was handed a trick question.

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

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

You are correct as you are only asked for $x_2$ .

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

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

This requires inverting $A$

**Runty** · April 12th 2010, 05:58 PM

Originally Posted by pickslides

You are correct as you are only asked for $x_2$ .

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

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

This requires inverting $A$

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

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

**pickslides** · April 12th 2010, 07:13 PM

Have a look at Online Matrix Calculator

**HallsofIvy** · April 13th 2010, 12:36 AM

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

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

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

**Runty** · April 13th 2010, 11:26 AM

Originally Posted by HallsofIvy

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

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

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

I say it because I believe that by stating $Ax=\left(\begin{array}{c}1 \\-2\\1\end{array}\right)$ , it could mean that you need to do this: $x=A^{-1}\left(\begin{array}{c}1 \\-2\\1\end{array}\right)$ , before you apply Cramer's Rule. This would come out to $x=\left(\begin{array}{c}7/3 \\ 16/3 \\ -4 \end{array}\right)$ , and this $x$ would be used in Cramer's Rule, not $Ax=\left(\begin{array}{c}1 \\-2\\1\end{array}\right)$ .

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

**Runty** · April 13th 2010, 01:37 PM

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

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