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 $2$. However, $\dim \ker (5I-C) = 1$, the geometric multiplicity is $1$.
Very loosely speaking, the matrix is 'deficient' in some sense when the two multiplicities do not match.
The algebraic multiplicity of an eigenvalue $\lambda$ is the power $m$ of the term $(x-\lambda)^m$ in the characteristic polynomial.
The geometric multiplicity is the number of linearly independent eigenvectors you can find for an eigenvalue.