Determinates

[jsbeckton]

Any one see the stratagy here?

Let A=\(\displaystyle \[
\begin{array}
2 & 1 & 0 \\
0 & 1 & { - 1} \\
0 & 2 & 4 \\
\end{array}
\]\). Find all real numbers X such that det\(\displaystyle (xI_3 - A) = 0\)

Should I start out with something like det\(\displaystyle (xI_3)\) - det(A) = 0, then bring the x out front? And if so where should I go from there? Thanks

 

[pka]

\(\displaystyle \L
\left| {\matrix{
{x - 2} & { - 1} & 0 \cr
0 & {x - 1} & 1 \cr
0 & { - 2} & {x - 4} \cr

} } \right| = x^3 - 7x^2 + 16x - 12\)

 

[stapel]

Should I start out with something like det\(\displaystyle (xI_3)\) - det(A) = 0...?

Whoa! Do determinants really work like that? :shock:

You've got det(xI₃ - A), so work with that:

. . . . .\(\displaystyle \large{xI_3\mbox{ }-\mbox{ }A\mbox{ }=\mbox{ }\left[\begin{array}{ccc}x-2&-1&0\\0&x-1&1\\0&-2&x-4\end{array}\right]}\)

Then, doing the "tack on copies of the first two columns at the end, and criss-cross multiply" thingy, you get:

. . . . .(x - 2)(x - 1)(x - 4) + 0 + 0 - 0 - (-2)(1)(x - 2) - 0

. . . . . . .= (x - 2)(x² - 5x + 4) + 2(x - 2)

. . . . . . .= (x - 2)(x² - 5x + 6)

You can finish the factoring. Then set the factored form equal to zero, and solve for the three values of x.

Eliz.

 

[jsbeckton]

thanks a lot, I knew that I was wrong.