Imaginary roots factoring help

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July 29th 2009, 01:46 PM
diddledabble

$r^{2}-6r+6=0$
This might result in an $i$ root. I am not sure just stuck.

This is the determinant of my matrix and I need to find the eigenvalues. No square roots or fractions allowed. Must be a whole number or something like $2i$
July 29th 2009, 01:59 PM
pickslides

Quote:

Originally Posted by diddledabble

$r^{2}-6r+6=0$

This might result in an $i$ root. I am not sure just stuck.

$r^{2}-6r+6=0$

I would complete the square here

$(r^{2}-6r+9)+6-9=0$

$(r^{2}-6r+9)-3=0$

$(r-3)^{2}-3=0$

$(r-3)^{2}=3$

Now solving for r will determine the nature of the roots.

$(r-3)^{2}=3$

$r-3=\sqrt{3}$

$r=3\pm\sqrt{3}$

These look real to me. (Rofl)
July 29th 2009, 02:01 PM
Chris L T521

Quote:

Originally Posted by diddledabble

$r^{2}-6r+6=0$
This might result in an $i$ root. I am not sure just stuck.

[I take this to be the characteristic equation from a DE]

To find r, use the quadratic formula. You should get $r=3\pm\sqrt{3}$ .
July 29th 2009, 02:08 PM
Chris L T521

Quote:

Originally Posted by diddledabble

This is the determinant of my matrix and I need to find the eigenvalues. No square roots or fractions allowed. Must be a whole number or something like $2i$

What is the matrix you're working with?
July 29th 2009, 02:29 PM
Chris L T521

Quote:

Originally Posted by diddledabble

$\mathbf{x}^{\prime}(t)=\begin{bmatrix}6&-3\\2&1\end{bmatrix}{\color{red}\mathbf{x}(t)}$ with initial value of $\mathbf{x}(0)=\begin{bmatrix}-10\\-6\end{bmatrix}$

I think you left out an important part (in red)

Let $\mathbf{A}=\begin{bmatrix}6&-3\\2&1\end{bmatrix}$ .

Then to find eigenvalues, we set $\det\left(\mathbf{A}-\lambda \mathbf{I}\right)=0$ .

In this case, it turns out that $\mathbf{A}-\lambda \mathbf{I}=\begin{bmatrix}6-\lambda & -3\\2&1-\lambda\end{bmatrix}$ .

It now follows that $\det\left(\mathbf{A}-\lambda \mathbf{I}\right)=\left(6-\lambda\right)\left(1-\lambda\right)-\left(2\right)\left(-3\right)=6-7\lambda+\lambda^2+6=\lambda^2-7\lambda+12$

Thus, $\det\left(\mathbf{A}-\lambda \mathbf{I}\right)=0\implies\lambda^2-7\lambda+12=0\implies\left(\lambda-4\right)\left(\lambda-3\right)=0\implies$ $\lambda=4$ or $\lambda=3$ .