In An Operation on Matrices, we saw that matrix multiplication was a non-commutative operation on the set of \(2 \times 2\) matrices with real entries. Here we will continue to work with this set and show that the usual matrix addition is also an operation on this set. Furthermore, the set with these two operations will form a ring.

**A Ring**

Let \(M(2, \mathbb{R})\) be the set of all \(2 \times 2\) matrices with real entries. Then, \(M(2,\mathbb{R})\) is a ring with identity under the usual matrix addition and multiplication.

**Addition**

In order to show that this set forms a ring, we will need to begin by showing that it is a commutative group under addition. That is, we need to show that addition is an operation, it is commutative, there is an identity and additive inverses exist for all elements in the set.

**Operation**

To show that addition is an operation, we need to show that it is a mapping from \((M(2,\mathbb{R}))^{2}\) to \(M(2,\mathbb{R})\). That is, if we choose two elements of \(M(2,\mathbb{R})\) and add them together, we will indeed have a unique element of \(M(2,\mathbb{R})\). Therefore, we let \(a,b \in M(2,\mathbb{R})\). Then we can note that

\begin{align*}

a&=\begin{bmatrix}

a_{11} & a_{12} \\

a_{21} & a_{22}

\end{bmatrix} \text{ and } \\

b&=\begin{bmatrix}

b_{11} & b_{12} \\

b_{21} & b_{22}

\end{bmatrix},

\end{align*}

for some \(a_{11},a_{12},a_{21},a_{22},b_{11},b_{12},b_{21},b_{22} \in \mathbb{R}\). If we then add these, we will get

\begin{align*}

a+b&=\begin{bmatrix}

a_{11} & a_{12} \\

a_{21} & a_{22}

\end{bmatrix} +

\begin{bmatrix}

b_{11} & b_{12} \\

b_{21} & b_{22}

\end{bmatrix} \\

&=\begin{bmatrix}

a_{11}+b_{11} & a_{12}+b_{12} \\

a_{21}+b_{21} & a_{22}+b_{22}

\end{bmatrix}.

\end{align*}

Now, since the real numbers are closed under addition, we note that each of the entries are again well-defined real numbers. Furthermore, the result is a \(2 \times 2\) matrix, so the result is in \(M(2,\mathbb{R})\). Hence, addition will be an operation. We will save the formal proof until we are finished with are requirements for addition.

**Associative**

We will also need to show that \(+\) is associative. That is, if \(a,b,c \in M(2,\mathbb{R})\), then we need to show that \(a+(b+c)=(a+b)+c\). We will, therefore, work through each of these using the same definition for \(a\) and \(b\) as above, with the further assumption that

\begin{align*}

c=\begin{bmatrix}

c_{11} & c_{12} \\

c_{21} & c_{22}

\end{bmatrix}.

\end{align*}

We now get

\begin{align*}

a+(b+c)&=\begin{bmatrix}

a_{11} & a_{12} \\

a_{21} & a_{22}

\end{bmatrix} +

\left(\begin{bmatrix}

b_{11} & b_{12} \\

b_{21} & b_{22}

\end{bmatrix}+

\begin{bmatrix}

c_{11} & c_{12} \\

c_{21} & c_{22}

\end{bmatrix} \right) \\

&=\begin{bmatrix}

a_{11} & a_{12} \\

a_{21} & a_{22}

\end{bmatrix} +

\begin{bmatrix}

b_{11}+c_{11} & b_{12}+c_{12} \\

b_{21}+c_{21} & b_{22}+c_{22}

\end{bmatrix} \\

&=\begin{bmatrix}

a_{11}+(b_{11}+c_{11}) & a_{12}+(b_{12}+c_{12}) \\

a_{21}+(b_{21}+c_{21}) & a_{22}+(b_{22}+c_{22})

\end{bmatrix} \\

&=\begin{bmatrix}

(a_{11}+b_{11})+c_{11} & (a_{12}+b_{12})+c_{12} \\

(a_{21}+b_{21})+c_{21} & (a_{22}+b_{22})+c_{22}

\end{bmatrix},

\end{align*}

where the last step follows from the associativity of real numbers. If we then follow the same outline, going backwards from this point, we will indeed get out \((a+b)+c\).

**Commutative**

We now need to show that addition is commutative. Using the same definitions for \(a,b \in M(2, \mathbb{R})\), we then have

\begin{align*}

b+a&=\begin{bmatrix}

b_{11} & b_{12} \\

b_{21} & b_{22}

\end{bmatrix}+\begin{bmatrix}

a_{11} & a_{12} \\

a_{21} & a_{22}

\end{bmatrix} \\

&=\begin{bmatrix}

b_{11}+a_{11} & b_{12}+a_{12} \\

b_{21}+a_{21} & b_{22}+a_{22}

\end{bmatrix}.

\end{align*}

Note that, since the real numbers are commutative, we get that \(b_{11}+a_{11}=a_{11}+b_{11}\). This would also apply to the other entries as well. We would then get

\begin{align*}

\begin{bmatrix}

b_{11}+a_{11} & b_{12}+a_{12} \\

b_{21}+a_{21} & b_{22}+a_{22}

\end{bmatrix}=

\begin{bmatrix}

a_{11}+b_{11} & a_{12}+b_{12} \\

a_{21}+b_{21} & a_{22}+b_{22}

\end{bmatrix},

\end{align*}

which is the same as \(a+b\) from the work we have above. Therefore, \(a+b=b+a\), so addition will be commutative.

**Identity and Inverses**

In order to show that an additive identity and inverses exists, we will need to find them. Now when finding these, we should recall that the additive identity of real numbers is \(0\). Therefore, we would expect the additive identity for matrices to be the matrix analog of \(0\). In this way, we should guess

\begin{align*}

0&=\begin{bmatrix}

0 & 0 \\

0 & 0

\end{bmatrix}.

\end{align*}

Now, if we test this by adding it to any other matrix, we quickly see that we get the other matrix out. Hence, this will be our identity.

For inverses we will follow a similar though pattern. The additive inverse of a real number \(a\) is \(-a\), so the additive inverse of a matrix \(a\) should be

\begin{align*}

-a&=\begin{bmatrix}

-a_{11} & -a_{12} \\

-a_{21} & -a_{22}

\end{bmatrix}.

\end{align*}

Now, to show that this is indeed an inverses, we should add \(a\) and \(-a\). In this case, we will indeed get out the zero matrix, so this is the additive inverse.

We are now ready to show that \(\langle M(2,\mathbb{R}),+ \rangle\) as a commutative group.

**Proof**

Let \(a,b,c \in M(2, \mathbb{R})\). Then, we know that

\begin{align*}

a&=\begin{bmatrix}

a_{11} & a_{12} \\

a_{21} & a_{22}

\end{bmatrix}, \\

b&=\begin{bmatrix}

b_{11} & b_{12} \\

b_{21} & b_{22}

\end{bmatrix}\text{ and } \\

c&=\begin{bmatrix}

c_{11} & c_{12} \\

c_{21} & c_{22}

\end{bmatrix}

\end{align*}

for some \(a_{11},a_{12},a_{21},a_{22},\)\(b_{11},b_{12},b_{21},b_{22},\)\(c_{11},c_{12},c_{21},c_{22} \in \mathbb{R}\).

Now, we can see that

\begin{align*}

a+b&=\begin{bmatrix}

a_{11} & a_{12} \\

a_{21} & a_{22}

\end{bmatrix} + \begin{bmatrix}

b_{11} & b_{12} \\

b_{21} & b_{22}

\end{bmatrix} \\

&=\begin{bmatrix}

a_{11}+b_{11} & a_{12}+b_{12} \\

a_{21}+b_{21} & a_{22}+b_{22}

\end{bmatrix}.

\end{align*}

Each of these entries are real since the real numbers are closed under addition, and, since this is a \(2 \times 2\) matrix, we have that addition is an operation on \(M(2,\mathbb{R})\).

Next, we note that

\begin{align*}

a+(b+c)&=\begin{bmatrix}

a_{11} & a_{12} \\

a_{21} & a_{22}

\end{bmatrix} +

\left(\begin{bmatrix}

b_{11} & b_{12} \\

b_{21} & b_{22}

\end{bmatrix}+

\begin{bmatrix}

c_{11} & c_{12} \\

c_{21} & c_{22}

\end{bmatrix} \right) \\

&=\begin{bmatrix}

a_{11} & a_{12} \\

a_{21} & a_{22}

\end{bmatrix} +

\begin{bmatrix}

b_{11}+c_{11} & b_{12}+c_{12} \\

b_{21}+c_{21} & b_{22}+c_{22}

\end{bmatrix} \\

&=\begin{bmatrix}

a_{11}+(b_{11}+c_{11}) & a_{12}+(b_{12}+c_{12}) \\

a_{21}+(b_{21}+c_{21}) & a_{22}+(b_{22}+c_{22})

\end{bmatrix} \\

&=\begin{bmatrix}

(a_{11}+b_{11})+c_{11} & (a_{12}+b_{12})+c_{12} \\

(a_{21}+b_{21})+c_{21} & (a_{22}+b_{22})+c_{22}

\end{bmatrix},

\end{align*}

since addition is associative for real numbers. Therefore,

\begin{align*}

a+(b+c)&=\begin{bmatrix}

(a_{11}+b_{11})+c_{11} & (a_{12}+b_{12})+c_{12} \\

(a_{21}+b_{21})+c_{21} & (a_{22}+b_{22})+c_{22}

\end{bmatrix} \\

&=\begin{bmatrix}

a_{11}+b_{11} & a_{12}+b_{12} \\

a_{21}+b_{21} & a_{22}+b_{22}

\end{bmatrix} +

\begin{bmatrix}

c_{11} & c_{12} \\

c_{21} & c_{22}

\end{bmatrix} \\

&=\left(\begin{bmatrix}

a_{11} & a_{12} \\

a_{21} & a_{22}

\end{bmatrix}+

\begin{bmatrix}

b_{11} & b_{12} \\

b_{21} & b_{22}

\end{bmatrix}\right)+

\begin{bmatrix}

c_{11} & c_{12} \\

c_{21} & c_{22}

\end{bmatrix} \\

&=(a+b)+c.

\end{align*}

Hence, the addition is an associative operation.

Now, using the work we have above, we can see that

\begin{align*}

a+b&=\begin{bmatrix}

a_{11}+b_{11} & a_{12}+b_{12} \\

a_{21}+b_{21} & a_{22}+b_{22}

\end{bmatrix}\\

&=\begin{bmatrix}

b_{11}+a_{11} & b_{12}+a_{12} \\

b_{21}+a_{21} & b_{22}+a_{22}

\end{bmatrix},

\end{align*}

since addition is commutative for real numbers. Hence,

\begin{align*}

a+b&=\begin{bmatrix}

b_{11}+a_{11} & b_{12}+a_{12} \\

b_{21}+a_{21} & b_{22}+a_{22}

\end{bmatrix} \\

&=\begin{bmatrix}

b_{11} & b_{12} \\

b_{21} & b_{22}

\end{bmatrix}+

\begin{bmatrix}

a_{11} & a_{12} \\

a_{21} & a_{22}

\end{bmatrix} \\

&=b+a.

\end{align*}

Therefore, the addition of matrices is commutative.

Next, let

\begin{align*}

0=\begin{bmatrix}

0 & 0 \\

0 & 0

\end{bmatrix}.

\end{align*}

We can now see that

\begin{align*}

a+0&=\begin{bmatrix}

a_{11} & a_{12} \\

a_{21} & a_{22}

\end{bmatrix}+

\begin{bmatrix}

0 & 0 \\

0 & 0

\end{bmatrix} \\

&=\begin{bmatrix}

a_{11}+ 0 & a_{12}+0 \\

a_{21}+ 0 & a_{22} +0

\end{bmatrix} \\

&=\begin{bmatrix}

a_{11} & a_{12} \\

a_{21} & a_{22}

\end{bmatrix}\\

&=a.

\end{align*}

Furthermore, since we have shown commutativity, \(0+a=a+0=a\). Hence, \(0\) is the identity for addition.

We now define \(-a\) as the matrix

\begin{align*}

-a&=\begin{bmatrix}

-a_{11} & -a_{12} \\

-a_{21} & -a_{22}

\end{bmatrix}.

\end{align*}

Then, taking the sum of \(a\) and \(-a\), we find that

\begin{align*}

a+(-a)&=\begin{bmatrix}

a_{11} & a_{12} \\

a_{21} & a_{22}

\end{bmatrix}+

\begin{bmatrix}

-a_{11} & -a_{12} \\

-a_{21} & -a_{22}

\end{bmatrix} \\

&=\begin{bmatrix}

a_{11}+(-a_{11}) & a_{12}+(-a_{12}) \\

a_{21}+(-a_{21}) & a_{22}+(-a_{22})

\end{bmatrix} \\

&=\begin{bmatrix}

0 & 0 \\

0 & 0

\end{bmatrix} \\

&=0.

\end{align*}

Furthermore, since we have shown commutativity, we have that \((-a)+a=a+(-a)=0\), so \(-a\) is the additive identity of \(a\).

We now have that on \(M(2,\mathbb{R})\), addition is an associative and commutative operation. Furthermore, an additive identity exists and additive inverses exist for all elements of \(M(2,\mathbb{R})\). Hence, \(\langle M(2,\mathbb{R}),+\rangle\) is an Abelian (commutative) group.

**Multiplication**

We now know that \(M(2,\mathbb{R})\) is a commutative group under addition, so we are now left to show that \(M(2,\mathbb{R})\) is a monoid. That is, we need to show that the multiplication is an associative operation and that an identity exists.

**Operation**

Note that we have already shown that matrix multiplication is an operation on \(M(2,\mathbb{R})\) in the post An Operation on Matrices. We will, therefore, reference this post in the proof.

**Associative**

In order to show associativity, we let \(a,b,c \in M(2,\mathbb{R})\). Then

#####
\begin{align*}

&(a\cdot b)\cdot c \\

&=\left(\begin{bmatrix}

a_{11} & a_{12} \\

a_{21} & a_{22}

\end{bmatrix} \cdot

\begin{bmatrix}

b_{11} & b_{12} \\

b_{21} & b_{22}

\end{bmatrix}\right) \cdot

\begin{bmatrix}

c_{11} & c_{12} \\

c_{21} & c_{22}

\end{bmatrix} \\

&=\begin{bmatrix}

a_{11}b_{11}+a_{12}b_{21} & a_{11}b_{12}+a_{12}b_{22} \\

a_{21}b_{11}+a_{22}b_{12} & a_{21}b_{12}+a_{22}b_{22}

\end{bmatrix} \cdot

\begin{bmatrix}

c_{11} & c_{12} \\

c_{21} & c_{22}

\end{bmatrix} \\

&=\begin{bmatrix}

(a_{11}b_{11}+a_{12}b_{21})c_{11}+(a_{11}b_{12}+a_{12}b_{22})c_{21} & (a_{11}b_{11}+a_{12}b_{21})c_{12}+(a_{11}b_{12}+a_{12}b_{22})c_{22} \\

(a_{21}b_{11}+a_{22}b_{12})c_{11} + (a_{21}b_{12}+a_{22}b_{22})c_{21} & (a_{21}b_{11}+a_{22}b_{12})c_{12} + (a_{21}b_{12}+a_{22}b_{22})c_{22}

\end{bmatrix}\\

&=\begin{bmatrix}

a_{11}b_{11}c_{11}+a_{12}b_{21}c_{11}+a_{11}b_{12}c_{21}+a_{12}b_{22}c_{21} & a_{11}b_{11}c_{12}+a_{12}b_{21}c_{12}+a_{11}b_{12}c_{22}+a_{12}b_{22}c_{22} \\

a_{21}b_{11}c_{11}+a_{22}b_{12}c_{11}+a_{21}b_{12}c_{21}+a_{22}b_{22}c_{21} & a_{21}b_{11}c_{12}+a_{22}b_{12}c_{12}+a_{21}b_{12}c_{22}+a_{22}b_{22}c_{22}

\end{bmatrix}.

\end{align*}

We will now start with \(a \cdot (b \cdot c)\) and see if we can simplify from both ends to get an equality. When we write up the proof, we will keep this as one statement, but, for now, it will be easier to work on both ends. Therefore, we see that

#####
\begin{align*}

&a\cdot (b\cdot c)\\

&=\begin{bmatrix}

a_{11} & a_{12} \\

a_{21} & a_{22}

\end{bmatrix} \cdot

\left(\begin{bmatrix}

b_{11} & b_{12} \\

b_{21} & b_{22}

\end{bmatrix} \cdot

\begin{bmatrix}

c_{11} & c_{12} \\

c_{21} & c_{22}

\end{bmatrix} \right) \\

&=\begin{bmatrix}

a_{11} & a_{12} \\

a_{21} & a_{22}

\end{bmatrix}\cdot

\begin{bmatrix}

b_{11}c_{11}+b_{12}c_{21} & b_{11}c_{12}+b_{12}c_{22} \\

b_{21}c_{11}+b_{22}c_{12} & b_{21}c_{12}+b_{22}c_{22}

\end{bmatrix} \\

&=\begin{bmatrix}

a_{11}(b_{11}c_{11}+b_{12}c_{21})+a_{12}(b_{21}c_{11}+b_{22}c_{12}) & a_{11}(b_{11}c_{12}+b_{12}c_{22})+a_{12}(b_{21}c_{12}+b_{22}c_{22}) \\

a_{21}(b_{11}c_{11}+b_{12}c_{21})+a_{22}(b_{21}c_{11}+b_{22}c_{12}) & a_{21}(b_{11}c_{12}+b_{12}c_{22})+a_{22}(b_{21}c_{12}+b_{22}c_{22})

\end{bmatrix}\\

&=\begin{bmatrix}

a_{11}b_{11}c_{11}+a_{11}b_{12}c_{21}+a_{12}b_{21}c_{11}+a_{12}b_{22}c_{12} & a_{11}b_{11}c_{12}+a_{11}b_{12}c_{22}+a_{12}b_{21}c_{12}+a_{12}b_{22}c_{22} \\

a_{21}b_{11}c_{11}+a_{21}b_{12}c_{21}+a_{22}b_{21}c_{11}+a_{22}b_{22}c_{12} & a_{21}b_{11}c_{12}+a_{21}b_{12}c_{22}+a_{22}b_{21}c_{12}+a_{22}b_{22}c_{22}

\end{bmatrix}.

\end{align*}

By looking at the two results, we realize that a reordering of the sums within the entries will indeed give us an equality between the two. Hence, we will have associativity.

**Identity**

We recall that for the real numbers, the identity for multiplication is \(1\), so we will want to find a matrix that behaves like \(1\). We know that, for any \(a,b \in M(2,\mathbb{R}\)

\begin{align*}

a \cdot b&=\begin{bmatrix}

a_{11} & a_{12} \\

a_{21} & a_{22}

\end{bmatrix} \cdot

\begin{bmatrix}

b_{11} & b_{12} \\

b_{21} & b_{22}

\end{bmatrix}\\

&=\begin{bmatrix}

a_{11}b_{11}+a_{12}b_{21} & a_{11}b_{12}+a_{12}b_{22} \\

a_{21}b_{11}+a_{22}b_{12} & a_{21}b_{12}+a_{22}b_{22}

\end{bmatrix}

\end{align*}

Upon looking at this product, if we want the resulting matrix to be \(b\), we note that we should let \(a_{11}=1, a_{12}=0, a_{21}=0,\) and \(a_{22}=1\). That is, the multiplicative identity for matrices would be

\begin{align*}

1 &=\begin{bmatrix}

1& 0 \\

0 & 1

\end{bmatrix}.

\end{align*}

We are now ready to give a proof that matrix multiplication is an associative operation with an identity.

**Proof**

Let \(a,b,c \in M(2, \mathbb{R})\). Then, we know that

\begin{align*}

a&=\begin{bmatrix}

a_{11} & a_{12} \\

a_{21} & a_{22}

\end{bmatrix}, \\

b&=\begin{bmatrix}

b_{11} & b_{12} \\

b_{21} & b_{22}

\end{bmatrix}\text{ and } \\

c&=\begin{bmatrix}

c_{11} & c_{12} \\

c_{21} & c_{22}

\end{bmatrix}

\end{align*}

for some \(a_{11},a_{12},a_{21},a_{22},\)\(b_{11},b_{12},b_{21},b_{22},\)\(c_{11},c_{12},c_{21},c_{22} \in \mathbb{R}\).

We note that matrix multiplication is an operation on the set \(M(2,\mathbb{R})\) by the post An Operation on Matrices.

Next, we note that

#####
\begin{align*}

&(a\cdot b)\cdot c\\

&=\left(\begin{bmatrix}

a_{11} & a_{12} \\

a_{21} & a_{22}

\end{bmatrix} \cdot

\begin{bmatrix}

b_{11} & b_{12} \\

b_{21} & b_{22}

\end{bmatrix}\right) \cdot

\begin{bmatrix}

c_{11} & c_{12} \\

c_{21} & c_{22}

\end{bmatrix} \\

&=\begin{bmatrix}

a_{11}b_{11}+a_{12}b_{21} & a_{11}b_{12}+a_{12}b_{22} \\

a_{21}b_{11}+a_{22}b_{12} & a_{21}b_{12}+a_{22}b_{22}

\end{bmatrix} \cdot

\begin{bmatrix}

c_{11} & c_{12} \\

c_{21} & c_{22}

\end{bmatrix} \\

&=\begin{bmatrix}

(a_{11}b_{11}+a_{12}b_{21})c_{11}+(a_{11}b_{12}+a_{12}b_{22})c_{21} & (a_{11}b_{11}+a_{12}b_{21})c_{12}+(a_{11}b_{12}+a_{12}b_{22})c_{22} \\

(a_{21}b_{11}+a_{22}b_{12})c_{11} + (a_{21}b_{12}+a_{22}b_{22})c_{21} & (a_{21}b_{11}+a_{22}b_{12})c_{12} + (a_{21}b_{12}+a_{22}b_{22})c_{22}

\end{bmatrix}\\

&=\begin{bmatrix}

a_{11}b_{11}c_{11}+a_{12}b_{21}c_{11}+a_{11}b_{12}c_{21}+a_{12}b_{22}c_{21} & a_{11}b_{11}c_{12}+a_{12}b_{21}c_{12}+a_{11}b_{12}c_{22}+a_{12}b_{22}c_{22} \\

a_{21}b_{11}c_{11}+a_{22}b_{12}c_{11}+a_{21}b_{12}c_{21}+a_{22}b_{22}c_{21} & a_{21}b_{11}c_{12}+a_{22}b_{12}c_{12}+a_{21}b_{12}c_{22}+a_{22}b_{22}c_{22}

\end{bmatrix} \\

&=\begin{bmatrix}

a_{11}b_{11}c_{11}+a_{11}b_{12}c_{21}+a_{12}b_{21}c_{11}+a_{12}b_{22}c_{12} & a_{11}b_{11}c_{12}+a_{11}b_{12}c_{22}+a_{12}b_{21}c_{12}+a_{12}b_{22}c_{22} \\

a_{21}b_{11}c_{11}+a_{21}b_{12}c_{21}+a_{22}b_{21}c_{11}+a_{22}b_{22}c_{12} & a_{21}b_{11}c_{12}+a_{21}b_{12}c_{22}+a_{22}b_{21}c_{12}+a_{22}b_{22}c_{22}

\end{bmatrix} \\

&=\begin{bmatrix}

a_{11}(b_{11}c_{11}+b_{12}c_{21})+a_{12}(b_{21}c_{11}+b_{22}c_{12}) & a_{11}(b_{11}c_{12}+b_{12}c_{22})+a_{12}(b_{21}c_{12}+b_{22}c_{22}) \\

a_{21}(b_{11}c_{11}+b_{12}c_{21})+a_{22}(b_{21}c_{11}+b_{22}c_{12}) & a_{21}(b_{11}c_{12}+b_{12}c_{22})+a_{22}(b_{21}c_{12}+b_{22}c_{22})

\end{bmatrix} \\

&=\begin{bmatrix}

a_{11} & a_{12} \\

a_{21} & a_{22}

\end{bmatrix}\cdot

\begin{bmatrix}

b_{11}c_{11}+b_{12}c_{21} & b_{11}c_{12}+b_{12}c_{22} \\

b_{21}c_{11}+b_{22}c_{12} & b_{21}c_{12}+b_{22}c_{22}

\end{bmatrix} \\

&=\begin{bmatrix}

a_{11} & a_{12} \\

a_{21} & a_{22}

\end{bmatrix} \cdot

\left(\begin{bmatrix}

b_{11} & b_{12} \\

b_{21} & b_{22}

\end{bmatrix} \cdot

\begin{bmatrix}

c_{11} & c_{12} \\

c_{21} & c_{22}

\end{bmatrix} \right) \\

&=a \cdot (b \cdot c).

\end{align*}

Hence, matrix multiplication is an associative operation.

Finally, we let

\begin{align*}

1&=\begin{bmatrix}

1& 0 \\

0 & 1

\end{bmatrix}

\end{align*}

We then note that

\begin{align*}

1 \cdot a & = \begin{bmatrix}

1 & 0 \\

0 & 1

\end{bmatrix} \cdot

\begin{bmatrix}

a_{11} & a_{12} \\

a_{21} & a_{22}

\end{bmatrix} \\

&=\begin{bmatrix}

a_{11} & a_{12} \\

a_{21} & a_{22}

\end{bmatrix} \\

&=a \\

&=\begin{bmatrix}

a_{11} & a_{12} \\

a_{21} & a_{22}

\end{bmatrix} \\

&=\begin{bmatrix}

a_{11} & a_{12} \\

a_{21} & a_{22}

\end{bmatrix} \cdot

\begin{bmatrix}

1 & 0 \\

0 & 1

\end{bmatrix} \\

&=a \cdot 1.

\end{align*}

Hence, \(a \cdot 1=1 \cdot a=a\) for all \(a \in M(2,\mathbb{R})\). Therefore, \(1\) is the multiplicative identity on \(M(2,\mathbb{R}\).

We have now shown that matrix multiplication is an associative operation with identity on \(M(2,\mathbb{R})\). Hence, \(\langle M(2,\mathbb{R}), \cdot \rangle\) is a monoid.

**Distributivity**

We now need to show that when we combine multiplication and addition, we have both left and right distributivity. That is, if we have \(a,b,c \in M(2, \mathbb{R}\), we get that \(a\cdot (b+c)=a\cdot b+a \cdot c\) and \((b+c) \cdot a=(b \cdot a)+(c \cdot a\). As we work through both of these, we will get the desired result by working through the required arithmetic, so we will proceed to the proof.

**Proof of Distributivity**

Let \(a,b,c \in M(2, \mathbb{R})\). Then, we know that

\begin{align*}

a&=\begin{bmatrix}

a_{11} & a_{12} \\

a_{21} & a_{22}

\end{bmatrix}, \\

b&=\begin{bmatrix}

b_{11} & b_{12} \\

b_{21} & b_{22}

\end{bmatrix}\text{ and } \\

c&=\begin{bmatrix}

c_{11} & c_{12} \\

c_{21} & c_{22}

\end{bmatrix}

\end{align*}

for some \(a_{11},a_{12},a_{21},a_{22},\)\(b_{11},b_{12},b_{21},b_{22},\)\(c_{11},c_{12},c_{21},c_{22} \in \mathbb{R}\).

Then, note that

####
\begin{align*}

a& \cdot (b+c) \\

&=\begin{bmatrix}

a_{11} & a_{12} \\

a_{21} & a_{22}

\end{bmatrix} \cdot \left(

\begin{bmatrix}

b_{11} & b_{12} \\

b_{21} & b_{22}

\end{bmatrix}+

\begin{bmatrix}

c_{11} & c_{12} \\

c_{21} & c_{22}

\end{bmatrix}\right) \\

&=\begin{bmatrix}

a_{11} & a_{12} \\

a_{21} & a_{22}

\end{bmatrix} \cdot

\begin{bmatrix}

b_{11}+c_{11} & b_{12}+c_{12} \\

b_{21}+c_{21} & b_{22}+c_{22}

\end{bmatrix} \\

&=\begin{bmatrix}

a_{11}(b_{11}+c_{11})+a_{12}(b_{21}+c_{21}) & a_{11}(b_{12}+c_{12})+a_{12}(b_{22}+c_{22}) \\

a_{21}(b_{11}+c_{11})+a_{22}(b_{21}+c_{21}) & a_{21}(b_{12}+c_{12})+a_{22}(b_{22}+c_{22})

\end{bmatrix} \\

&=\begin{bmatrix}

(a_{11}b_{11}+a_{12}b_{21})+(a_{11}c_{11})+a_{12}c_{22}) & (a_{11}b_{12}+a_{12}b_{22})+(a_{11}c_{12}+a_{12}c_{22}) \\

(a_{21}b_{11}+a_{22}b_{21})+(a_{21}c_{11})+a_{22}c_{22}) & (a_{21}b_{12}+a_{22}b_{22})+(a_{21}c_{12}+a_{22}c_{22})

\end{bmatrix} \\

&=\begin{bmatrix}

(a_{11}b_{11}+a_{12}b_{21}) & (a_{11}b_{12}+a_{12}b_{22}) \\

(a_{21}b_{11}+a_{22}b_{21}) & (a_{21}b_{12}+a_{22}b_{22})

\end{bmatrix}+\begin{bmatrix}

(a_{11}c_{11}+a_{12}c_{22}) & (a_{11}c_{12}+a_{12}c_{22}) \\

(a_{21}c_{11}+a_{22}c_{22}) & (a_{21}c_{12}+a_{22}c_{22})

\end{bmatrix} \\

&=a \cdot b + a \cdot c.

\end{align*}

Hence, the multiplication and addition are left distributive.

Furthermore,

####
\begin{align*}

&(b+c) \cdot a \\

&\left(

\begin{bmatrix}

b_{11} & b_{12} \\

b_{21} & b_{22}

\end{bmatrix}+

\begin{bmatrix}

c_{11} & c_{12} \\

c_{21} & c_{22}

\end{bmatrix}\right) \cdot \begin{bmatrix}

a_{11} & a_{12} \\

a_{21} & a_{22}

\end{bmatrix} \\

& =\begin{bmatrix}

b_{11}+c_{11} & b_{12}+c_{12} \\

b_{21}+c_{21} & b_{22}+c_{22}

\end{bmatrix} \cdot

\begin{bmatrix}

a_{11} & a_{12} \\

a_{21} & a_{22}

\end{bmatrix}\\

&=\begin{bmatrix}

(b_{11}+c_{11})a_{11}+(b_{12}+c_{12})a_{21} & (b_{11}+c_{11})a_{12}+(b_{12}+c_{12})a_{22} \\

(b_{21}+c_{21})a_{11}+(b_{22}+c_{22})a_{21} & (b_{21}+c_{21})a_{12}+(b_{22}+c_{22})a_{22}

\end{bmatrix} \\

&=\begin{bmatrix}

b_{11}a_{11}+b_{12}a_{21} & b_{11}a_{12}+b_{12}a_{22} \\

b_{21}a_{11}+b_{22}a_{21} & b_{21}a_{12}+b_{22}a_{22}

\end{bmatrix}+

\begin{bmatrix}

c_{11}a_{11}+c_{12}a_{21} & c_{11}a_{12}+c_{12}a_{22} \\

c_{21}a_{11}+c_{22}a_{21} & c_{21}a_{12}+c_{22}a_{22}

\end{bmatrix} \\

&=b \cdot a + c \dot a

\end{align*}

Hence, the addition and multiplication satisfy right distributivity.

**Combining**

We have now show that

- \(\langle M(2,\mathbb{R}),+\rangle\) is an Abelian group.
- \(\langle M(2,\mathbb{R}), \cdot \rangle\) is a monoid.
- Distributivity holds.

Hence, \(\langle M(2,\mathbb{R}), +, \cdot\rangle\) is a ring with identity.

**Conclusion**

As a note on this proof, it is quite long. However, by focusing on one step at a time, we can keep it from becoming overwhelming. I hope that this helps you with your study of groups, monoids and rings. If it did, be sure to share the post with anyone else you know that may find it useful.