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Find a general solution of the given system using the method (A - [tex]\lambda[/tex]I)V2 = V1.

[tex]x'_1 = 2x_1 - 5x_2, x'_2 = 4x_1 - 2x_2[/tex]

[tex]x' =

\left(\begin{array}{cc}2&-5\\4&-2\end{array}\right)[/tex]

characteristic equation:

(2 - [tex]\lambda[/tex])((-2) - [tex]\lambda[/tex]) + 20 = 0

[tex]\lambda[/tex]^2 + 16 = 0

[tex]\lambda[/tex] = 4i

Using this method:

(A - 4i[tex]\lambda[/tex])V_2 = V_1

[tex]\left(\begin{array}{cc}2-4i&-5\\4&-2-4i\end{array}\right) *[/tex] [tex]\left(\begin{array}{cc}a\\b\end{array}\right) = [/tex] [tex]\left(\begin{array}{cc}0\\0\end{array}\right) [/tex]

(2 - 4i)a - 5b = 0

4a - (-2 - 4i)b = 0

When there are no complex roots, I can set a or b = 1 to find the value of the other. And when I row reduce this in my ti89, I get

[tex]\left(\begin{array}{cc}1&i 0\\0&0 0\end{array}\right) [/tex] with a space between the i and 0 in the top row, and the two 0's in the 2nd row.

How do I find a and b in this problem? When I find their values I will have V_1.

[tex]x'_1 = 2x_1 - 5x_2, x'_2 = 4x_1 - 2x_2[/tex]

[tex]x' =

\left(\begin{array}{cc}2&-5\\4&-2\end{array}\right)[/tex]

characteristic equation:

(2 - [tex]\lambda[/tex])((-2) - [tex]\lambda[/tex]) + 20 = 0

[tex]\lambda[/tex]^2 + 16 = 0

[tex]\lambda[/tex] = 4i

Using this method:

(A - 4i[tex]\lambda[/tex])V_2 = V_1

[tex]\left(\begin{array}{cc}2-4i&-5\\4&-2-4i\end{array}\right) *[/tex] [tex]\left(\begin{array}{cc}a\\b\end{array}\right) = [/tex] [tex]\left(\begin{array}{cc}0\\0\end{array}\right) [/tex]

(2 - 4i)a - 5b = 0

4a - (-2 - 4i)b = 0

When there are no complex roots, I can set a or b = 1 to find the value of the other. And when I row reduce this in my ti89, I get

[tex]\left(\begin{array}{cc}1&i 0\\0&0 0\end{array}\right) [/tex] with a space between the i and 0 in the top row, and the two 0's in the 2nd row.

How do I find a and b in this problem? When I find their values I will have V_1.

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