If the column vector $\left[\begin{array}{l}-2 \\ -2\end{array}\right]$ is an eigenvector of the $2 \times 2$ matrix $A=\left[\begin{array}{ll}a & 1 \\ 1 & 2\end{array}\right]$ corresponding to the eigenvalue $\lambda$, then $a+\lambda=$ ?
(A) 7
(B) 4
(C) 8
(D) 5
(E) 6

Answer

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Answer

Solving the equation, we get $a = 2$ and $\lambda = 3$. Therefore, $a + \lambda = 2 + 3 = \boxed{5}$.

Steps

Step 1 ：Given eigenvector $v = \begin{bmatrix} -2 \\ -2 \end{bmatrix}$ and matrix $A = \begin{bmatrix} a & 1 \\ 1 & 2 \end{bmatrix}$, we need to find $a$ and $\lambda$ that satisfy the equation $Av = \lambda v$.

Step 2 ：Solving the equation, we get $a = 2$ and $\lambda = 3$. Therefore, $a + \lambda = 2 + 3 = \boxed{5}$.

If the column vector $\left[\begin{array}{l}-2 \\ -2\end{array}\right]$ is an eigenvector of the $2 \times 2$ matrix $A=\left[\begin{array}{ll}a & 1 \\ 1 & 2\end{array}\right]$ corresponding to the eigenvalue $\lambda$, then $a+\lambda=$ ?(A) 7(B) 4(C) 8(D) 5(E) 6

Answer

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If the column vector $\left[\begin{array}{l}-2 \\ -2\end{array}\right]$ is an eigenvector of the $2 \times 2$ matrix $A=\left[\begin{array}{ll}a & 1 \\ 1 & 2\end{array}\right]$ corresponding to the eigenvalue $\lambda$, then $a+\lambda=$ ?
(A) 7
(B) 4
(C) 8
(D) 5
(E) 6