Find all $2 \times 2$ matrices $\bold{A}$ that have the property that for any $2 \times 2$ matrix $\bold{B}$,
\[\bold{A} \bold{B} = \bold{B} \bold{A}.\]
Let $\bold{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and $\bold{B} = \begin{pmatrix} e & f \\ g & h \end{pmatrix}$ be arbitrary $2 \times 2$ matrices. Then,
\[\bold{A} \bold{B} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} e & f \\ g & h \end{pmatrix} = \begin{pmatrix} ae + bg & af + bh \\ ce + dg & cf + dh \end{pmatrix}\]and
\[\bold{B} \bold{A} = \begin{pmatrix} e & f \\ g & h \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} ea + fc & eb + fd \\ ga + hc & gb + hd \end{pmatrix}.\]Setting the two matrices equal, we get
\begin{align*}
ae + bg &= ea + fc, \\
af + bh &= eb + fd, \\
ce + dg &= ga + hc, \\
cf + dh &= gb + hd.
\end{align*}From the first equation, $ae + bg = ea + fc,$ so $bg = fc.$ Similarly, $af + bh = eb + fd$ implies that $af = eb.$ Subtracting these equations, we get $bg - af = fc - eb,$ so $a(f - e) = b(f - e).$ If $f \neq e,$ then we can cancel $f - e,$ and we get $a = b.$ But if $f = e,$ then $a$ and $b$ can be anything.
The remaining equations are $ce + dg = ga + hc$ and $cf + dh = gb + hd.$ Substituting $c = a,$ $e = f,$ $d = e,$ and $g = h,$ respectively, we get $ae + deg = ga + ha$ and $af + e^2 = gb + he.$ Thus,
\begin{align*}
(a - h) e + deg - ga &= 0, \\
(a - g) f + e^2 - gb &= 0.
\end{align*}If $a \neq g,$ then we can divide the first equation by $a - g,$ and we get $e = 0.$ But if $a = g,$ then $e$ can be anything.
Similarly, if $a \neq h,$ then we can divide the second equation by $a - h,$ and we get $f = 0.$ Otherwise, $f$ can be anything.
Therefore, the matrices $\bold{A} = \boxed{\begin{pmatrix} a & b \\ c & d \end{pmatrix}}$ that commute with every $2 \times 2$ matrix are those where $a$ and $d$ are arbitrary, and the entries of $b$ and $c$ are freely chosen, as long as $b(f - e) = 0$ for all $e$ and $f,$ and $c(a - h) = 0$ for all $a$ and $h.$
To find all $2 \times 2$ matrices $\bold{A}$ that satisfy $\bold{A} \bold{B} = \bold{B} \bold{A}$ for any $2 \times 2$ matrix $\bold{B}$, we can use the properties of matrix multiplication to determine the conditions on the elements of $\bold{A}$.
Let's write $\bold{A}$ and $\bold{B}$ as follows:
\[\bold{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, \quad \bold{B} = \begin{pmatrix} e & f \\ g & h \end{pmatrix}.\]
Using matrix multiplication, we have:
\[\bold{A} \bold{B} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} e & f \\ g & h \end{pmatrix} = \begin{pmatrix} ae + bg & af + bh \\ ce + dg & cf + dh \end{pmatrix}.\]
Similarly,
\[\bold{B} \bold{A} = \begin{pmatrix} e & f \\ g & h \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} ea + fc & eb + fd \\ ga + hc & gb + hd \end{pmatrix}.\]
Since $\bold{A} \bold{B} = \bold{B} \bold{A}$ for any $\bold{B}$, we have:
\[\begin{pmatrix} ae + bg & af + bh \\ ce + dg & cf + dh \end{pmatrix} = \begin{pmatrix} ea + fc & eb + fd \\ ga + hc & gb + hd \end{pmatrix}.\]
Comparing the corresponding entries, we get the following conditions:
1. $ae + bg = ea + fc$
2. $af + bh = eb + fd$
3. $ce + dg = ga + hc$
4. $cf + dh = gb + hd$
Now, let's solve these four equations to find the possible values of $a$, $b$, $c$, and $d$.
From equation 1, we get $ag = fc$, and from equation 3, we have $cg = ha$. Comparing these two equations, we see that $ag = fc = cg = ha$. Since $g$ and $h$ can take any value, we conclude that $a = c = 0$.
Similarly, from equation 2, we get $bf = ed$, and from equation 4, we have $hf = ge$. Comparing these two equations, we see that $bf = ed = hf = ge$. Since $e$ and $f$ can take any value, we conclude that $b = d = 0$.
Therefore, the matrices $\bold{A}$ that satisfy $\bold{A} \bold{B} = \bold{B} \bold{A}$ for any $2 \times 2$ matrix $\bold{B}$ are of the form:
\[\bold{A} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}.\]
Recall that Matrix multiplication is always commutative if ...
... one matrix is the Identity matrix.
... one matrix is the Zero matrix.
... both matrices are 2×2 rotation matrices. (basically case #2)
... both matrices are Diagonal matrices.