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A zero has a multiplicity, which refers to the number of times that its associated factor appears in the polynomial. For instance, the quadratic (x + 3)(x  2) has the zeroes x = 3 and x = 2, each occuring once.
The characteristic polynomial of the matrix is $p_A(x) = \det (xI-A)$. In your case, $A = \begin{bmatrix} 1 & 4 \\ 2 & 3\end{bmatrix}$, so $p_A(x) = (x+1)(x-5)$. Hence it has two distinct eigenvalues and each occurs only once, so the algebraic multiplicity of both is one. If $B=\begin{bmatrix} 5 & 0 \\ 0 & 5\end{bmatrix}$, then $p_B(x) = (x-5)^2$, hence the eigenvalue $5$ has algebraic multiplicity $2$. Since $\dim \ker (5I-B) = 2$, the geometric multiplicity is also $2$. If $C=\begin{bmatrix} 5 &1 \\ 0 & 5\end{bmatrix}$, then $p_C(x) = (x-5)^2$ (same as $p_C$), hence the eigenvalue $5$ has algebraic multiplicity $2$.