2, 14; x1, x2
2, 14; x1 + x2, x1 - x2
2, 0; x1, x2
2, 0; x1 + x2, x1 - x2
Answer is Right!
Answer is Wrong!
The correct answer is $\boxed{\text{(B)}}$.
Let $A$ be a 2×2 matrix with eigenvalues $\lambda_1=1$ and $\lambda_2=-2$, and corresponding eigenvectors $x_1$ and $x_2$. Then, we know that $Ax_1=\lambda_1x_1$ and $Ax_2=\lambda_2x_2$.
We can also write this as $A(x_1+x_2)=(\lambda_1+0)x_1+(\lambda_2+0)x_2=x_1+x_2$ and $A(x_1-x_2)=(\lambda_1-0)x_1+(\lambda_2-0)x_2=x_1-x_2$.
Therefore, the eigenvalues of $A^2-3A+4I$ are $1+0+4=5$ and $-2+0+4=2$, and the corresponding eigenvectors are $x_1+x_2$ and $x_1-x_2$.
Here is a more detailed explanation of each option:
- Option A: The eigenvalues of $A^2-3A+4I$ are $2$ and $14$, which is not correct.
- Option B: The eigenvalues of $A^2-3A+4I$ are $2$ and $14$, and the corresponding eigenvectors are $x_1$ and $x_2$. This is correct, since $Ax_1=\lambda_1x_1$ and $Ax_2=\lambda_2x_2$.
- Option C: The eigenvalues of $A^2-3A+4I$ are $2$ and $0$, and the corresponding eigenvectors are $x_1$ and $x_2$. This is not correct, since the eigenvectors of $A^2-3A+4I$ are not necessarily the same as the eigenvectors of $A$.
- Option D: The eigenvalues of $A^2-3A+4I$ are $2$ and $0$, and the corresponding eigenvectors are $x_1+x_2$ and $x_1-x_2$. This is correct, since $A(x_1+x_2)=(\lambda_1+0)x_1+(\lambda_2+0)x_2=x_1+x_2$ and $A(x_1-x_2)=(\lambda_1-0)x_1+(\lambda_2-0)x_2=x_1-x_2$.