3 and -3
-3 and -5
3 and 5
5 and 0
Answer is Right!
Answer is Wrong!
The correct answer is $\boxed{\text{C. }3 \text{ and }5}$.
To find the eigenvalues of a matrix, we can use the following formula:
$$\lambda = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$
where $a$, $b$, and $c$ are the coefficients of the characteristic polynomial of the matrix.
The characteristic polynomial of the matrix $\left[ {\begin{array}{*{20}{c}} 4&1 \ 1&4 \end{array}} \right]$ is $p(x) = x^2 – 8x + 16$.
The discriminant of $p(x)$ is $b^2 – 4ac = 64 – 64 = 0$.
Therefore, the eigenvalues of the matrix $\left[ {\begin{array}{*{20}{c}} 4&1 \ 1&4 \end{array}} \right]$ are $\boxed{\text{C. }3 \text{ and }5}$.
Here is a brief explanation of each option:
- Option A: $3$ and $-3$. These are not eigenvalues of the matrix $\left[ {\begin{array}{*{20}{c}} 4&1 \ 1&4 \end{array}} \right]$. To see this, we can substitute $x = 3$ and $x = -3$ into the characteristic polynomial $p(x)$ and see that we do not get $0$.
- Option B: $-3$ and $-5$. These are not eigenvalues of the matrix $\left[ {\begin{array}{*{20}{c}} 4&1 \ 1&4 \end{array}} \right]$. To see this, we can substitute $x = -3$ and $x = -5$ into the characteristic polynomial $p(x)$ and see that we do not get $0$.
- Option C: $3$ and $5$. These are eigenvalues of the matrix $\left[ {\begin{array}{*{20}{c}} 4&1 \ 1&4 \end{array}} \right]$. To see this, we can substitute $x = 3$ and $x = 5$ into the characteristic polynomial $p(x)$ and see that we get $0$.