GATE Overflow Test Series | Linear Algebra | Test 1 | Question: 26

Question

The value of $k^{2} + 2k + 1$ such that the vector $\begin{bmatrix} 2\\ 4\\ 6 \end{bmatrix}$ is an eigenvector of the matrix  $\begin{bmatrix} 4 & 1 & 2\\ k & 2 & 1\\14 & -4 & 10\\ \end{bmatrix}$ is __________

Answer 1

We know that $AX = \lambda X$

$\implies \begin{bmatrix} 4 & 1 & 2 \\ k & 2 & 1\\ 14 & -4 & 10 \end{bmatrix} \begin{bmatrix} 2  \\ 4 \\ 6 \end{bmatrix} = \lambda  \begin{bmatrix} 2 \\ 4 \\ 6  \end{bmatrix}$

$\implies \begin{bmatrix} 24 \\ 2k + 14 \\ 72  \end{bmatrix} = \begin{bmatrix} 2\lambda \\ 4\lambda \\ 6\lambda \end{bmatrix}$

Here, $2 \lambda = 24 \implies \lambda = 12$

$2k + 14 = 4\lambda \implies 2k + 14 = 48 \implies k = 17$

Now, the value of $k^{2} + 2k + 1 = 17^{2} + 2(17) + 1 = 289 + 34 + 1 = 324.$
    
So, the correct answer is $324.$