What do you mean by generalized eigenvector?

Hint: Since a matrix is a rectangular array filled with numbers or variables. Row elements are horizontal lines and columns are vertical elements like $(2,1)$ is the row one element.
If we need to find the eigenvector, first we need to solve the given matrix and we need to find the eigenvalues, and then only we can find the eigenvectors using the eigenvalues of the given matrix.
Eigenvectors are defined as the reference of a square (same order) matrix.

Complete step-by-step solution:
Eigenvectors of the given matrix generally represent the system of linear equations.
Method to find the eigenvectors:
The method of determining the eigenvector of a matrix is,
Let A be a \[n \times n\] matrix. A scalar $\lambda $ is an eigenvalue of a matrix A if there is a non-zero column vector $v$ , such that $Av = \lambda v$. Which is equation $(1)$
The vector $v$ is called an eigenvector of a matrix A corresponding to the
eigenvalue $\lambda $.
from the equation $(1)$ we have $Av = \lambda v \Rightarrow Av - \lambda v = 0$
since $I$ is an identity matrix, then we get \[(A - \lambda {I_n})v = {\text{ }}0\] which is an equation $(2)$
the equation $(2)$ has non-trivial solutions if and only if \[\det (A - \lambda {I_n}) = {\text{ }}0\] which is an equation $(3)$
The referring equation $(3)$ \[p(x) = \det (A - \lambda {I_n})\] is a polynomial of degree n called the characteristic polynomial of the n x n matrix A. The characteristic equation is given by $p(x) = 0$
Generalized eigenvector,
Eigenvector is not very different from generalized eigenvectors; it is defined as follows:
A generalized eigenvector is associated with the eigenvalue $\lambda $ of n-times and \[n \times n\] matrix denoted by the non-zero vector of X and defined as \[{(A - \lambda {I_n})^k} = {\text{ }}0\] where K is some positive integer.
For $k = 1$ then we get \[(A - \lambda {I_n}) = {\text{ }}0\]
Therefore, if $k = 1$ then the eigenvector of the matrix A is its generalized eigenvector.

Note: To find the determinant of the matrix
If for example take a $2 \times 2$ matrix which is $A = \left( {\begin{array}{*{20}{c}}
  2&1 \\
  4&3
\end{array}} \right)$ then we need to find its determinant
If we see $\left| A \right| = \left| {\begin{array}{*{20}{c}}
  2&1 \\
  4&3
\end{array}} \right| = (2 \times 3) - (1 \times 4)$
$ \Rightarrow 6 - 4$
$ \Rightarrow 2$
Thus, the determinant of a matrix $A$is non zero, that is $\left| A \right| = 2$
A matrix is nonsingular ($\left| A \right| \ne 0$) then we are able to find its inverse form.
If it singular ($\left| A \right| = 0$) then we cannot find its inverse form.
Also, in a matrix nonsingular matrix = invertible matrix.
If the order of the elements is not equal (not the same size), then it is called a non-square matrix.
$1 \times 2,5 \times 7$