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Determine which of x1, x2, and x3 are eigenvectors for the matrix
a. for those that are, determine the associated eigenvalue. (for each vector, enter the associated eigenvalue, if it exists. if an eigenvalue does not exist, enter dne.) a = −1 2 0 3 , x1 = 0 2 , x2 = 1 5 , x3 = 1 2

Respuesta :

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For a given matrix [tex]\mathbf A[/tex] with eigenvalues [tex]\lambda[/tex] and eigenvectors [tex]\mathbf v[/tex], the following relation holds (by definition):

[tex]\mathbf{Av}=\lambda\mathbf v[/tex]

In other words, if [tex]\mathbf v[/tex] is an eigenvector, then the multiplication of [tex]\mathbf v[/tex] by [tex]\mathbf A[/tex] results in a scaled version of the vector [tex]\mathbf v[/tex].

Given the matrix

[tex]\mathbf A=\begin{bmatrix}-1&2\\0&3\end{bmatrix}[/tex]

we want to decide which of the vectors [tex]\mathbf x_i[/tex] are eigenvectors. Taking [tex]\mathbf x_1[/tex], we have

[tex]\mathbf{Ax}_1=\begin{bmatrix}-1&2\\0&3\end{bmatrix}\begin{bmatrix}0\\2\end{bmatrix}=\begin{bmatrix}4\\6\end{bmatrix}[/tex]

but there is no scalar we can remove from the resulting vector to make it a scalar multiple of [tex]\mathbf x_1[/tex]. So [tex]\mathbf x_1[/tex] is not an eigenvector of [tex]\mathbf A[/tex]. We do the same sort of check for the remaining two vectors.

[tex]\mathbf{Ax}_2=\begin{bmatrix}-1&2\\0&3\end{bmatrix}\begin{bmatrix}1\\5\end{bmatrix}=\begin{bmatrix}9\\15\end{bmatrix}[/tex]

Same as before; [tex]\mathbf x_2[/tex] is not an eigenvector of [tex]\mathbf A[/tex].

[tex]\mathbf{Ax}_3=\begin{bmatrix}-1&2\\0&3\end{bmatrix}\begin{bmatrix}1\\2\end{bmatrix}=\begin{bmatrix}3\\6\end{bmatrix}=3\begin{bmatrix}1\\2\end{bmatrix}[/tex]

So [tex]\mathbf x_3[/tex] is an eigenvector of [tex]\mathbf A[/tex] with associated eigenvalue [tex]\lambda=3[/tex].

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