Solution. The product $\boldsymbol{AB}$ is:

The product $\boldsymbol{BA}$ is:

Since $\boldsymbol{AB}=\boldsymbol{BA}=\boldsymbol{I}_2$, we conclude that $\boldsymbol{A}$ is an inverse of $\boldsymbol{B}$ and vice versa. This means that both $\boldsymbol{A}$ and $\boldsymbol{B}$ are invertible.

Properties of matrix inverses

Proof. Let $\boldsymbol{A}$ be an $m\times{n}$ invertible matrix. Since $\boldsymbol{A}$ is invertible, there exists an inverse matrix $\boldsymbol{B}$ such that:

Since $\boldsymbol{AB}=\boldsymbol{I}_m$, the shape of $\boldsymbol{B}$ must be (in green):

However, $\boldsymbol{BA}=\boldsymbol{I}_m$, which means that the shape of $\boldsymbol{B}$ is (in green):

The only way this is true is when $m=n$, that is, $\boldsymbol{A}$ and $\boldsymbol{B}$ are both square matrices. This completes the proof.

Proof. Suppose $\boldsymbol{A}$ is a matrix that has a row filled with zeros. For $\boldsymbol{A}$ to be invertible, there must exist some other matrix $\boldsymbol{B}$ such that:

Suppose the $i$-th row of $\boldsymbol{A}$ is all zeros. By the nature of matrix multiplication, this means that the $i$-th row of $\boldsymbol{AB}$ will also be all zeros regardless of what $\boldsymbol{B}$ is. Therefore, $\boldsymbol{AB}$ cannot be an identity matrix and so by the definition of invertibility, $\boldsymbol{A}$ is not invertible.

Now, suppose the $j$-th column of $\boldsymbol{A}$ is all zeros. For any matrix $\boldsymbol{B}$, the $j$-th column of the product $\boldsymbol{BA}$ will also be all zeros. Again, this means that $\boldsymbol{BA}\ne\boldsymbol{I}_n$ and thus $\boldsymbol{A}$ is not invertible.

This completes the proof.

Proof. Suppose $\boldsymbol{B}$ and $\boldsymbol{C}$ are inverses of matrix $\boldsymbol{A}$. Our goal is to show that $\boldsymbol{B}$ and $\boldsymbol{C}$ must be equal:

This means that the inverses $\boldsymbol{B}$ and $\boldsymbol{C}$ must be equivalent. Therefore, the inverse of a matrix, given that it exists, is unique. This completes the proof.

Proof. From the definition of invertible matrices, we have that:

Here, we can think of this as the first matrix ($\boldsymbol{A}^{-1}$) being the inverse matrix of the second matrix ($\boldsymbol{A}$). Again, by the definition of invertible matrices, we also have that:

Similarly, we can conclude that the first matrix $\boldsymbol{A}$ is the inverse matrix of the second matrix $\boldsymbol{A}^{-1}$. In other words:

This completes the proof.

Proof. We show that $\boldsymbol{B}^{-1}\boldsymbol{A}^{-1}$ is the inverse of $\boldsymbol{AB}$ like so:

We have that the product of matrix $\boldsymbol{AB}$ and matrix $\boldsymbol{B}^{-1}\boldsymbol{A}^{-1}$ is the identity matrix. This means that $\boldsymbol{B}^{-1}\boldsymbol{A}^{-1}$ is the inverse matrix of $\boldsymbol{AB}$ by definition, that is:

This completes the proof.

Proof. We follow the same logic used in the previous proof:

By definitionlink of an invertible matrix, the inverse of $\boldsymbol{ABC}$ must be $\boldsymbol{C}^{-1}\boldsymbol{B}^{-1}\boldsymbol{A}^{-1}$, that is:

This completes the proof.

We shall now generalize the previous theorem to the case of $n$ matrices.

Proof. We will prove this by induction on the number of matrices. For the base case, we have already shownlink that the theorem holds for $2$ matrices, that is, $(\boldsymbol{A}_1 \boldsymbol{A}_2)^{-1}= \boldsymbol{A}^{-1}_2 \boldsymbol{A}^{-1}_1$. We now assume that the theorem holds for $k-1$ matrices, that is:

Our goal is to show that the theorem holds for $k$ matrices. Let's first use the associative property of matrices to get:

We now use propertylink to get:

Using our inductive assumption \eqref{eq:ZztqL26tiXGoEs0oSrw}, we get:

This completes the proof.

Proof. We must show that the product of matrix $\boldsymbol{A}^T$ and $\big(\boldsymbol{A}^{-1}\big)^T$ is the identity matrix, that is:

Let's take the transpose of the left-hand side:

Here, we used the property $(\boldsymbol{AB})^T=\boldsymbol{B}^T\boldsymbol{A}^T$ for the first step. Now, we take the transpose of both sides:

This means that $(\boldsymbol{A}^{-1})^T$ is the inverse of $\boldsymbol{A}^T$ by definition, that is:

This completes the proof.

Proof. Starting with the left-hand side:

Here, the 2nd equality holds by theoremlink. This completes the proof.

Proof. We know from theoremlink that if matrix $\boldsymbol{A}$ is invertible, then:

Because we can find the inverse of $\boldsymbol{A}^T$, we have that $\boldsymbol{A}^T$ is also invertible by definition:

Next, let's prove the following:

Let $\boldsymbol{B}^T$ be an invertible matrix. Substituting $\boldsymbol{B}^T$ into $\boldsymbol{A}$ in \eqref{eq:oK1Iql8RPMvmpM8pPI6} gives:

Since $\boldsymbol{B}^T$ is invertible, we know that $(\boldsymbol{B}^T)^{-1}$ exists. The inverse of $\boldsymbol{B}$ is the transpose of $(\boldsymbol{B}^T)^{-1}$. Because we can find the inverse of $\boldsymbol{B}$, we have that $\boldsymbol{B}$ is also invertible. Now, instead of using the letter $\boldsymbol{B}$, we can use the letter $\boldsymbol{A}$ for consistency to conclude:

Now, recall from our guide on proof by contraposition that this is logically equivalent to the following contrapositive statement:

This completes the proof.

Proof. We use the matrix property of bringing the scalar values to the front:

Therefore, $k^{-1}\boldsymbol{A}^{-1}$ is the inverse of $k\boldsymbol{A}$ by definition, that is:

This completes the proof.

Proof. Because $\boldsymbol{AB}$ is invertible, there exists a matrix $\boldsymbol{C}$ such that:

Using the associativity of matrix multiplication:

By definition, this means that $\boldsymbol{CA}$ is the inverse of $\boldsymbol{B}$, that is, $\boldsymbol{B}^{-1}=\boldsymbol{CA}$. Because there exists an inverse of $\boldsymbol{B}$, then $\boldsymbol{B}$ must be invertible by definition.

We can use the same logic to show that $\boldsymbol{A}$ is also invertible. Because $\boldsymbol{AB}$ is invertible, there exists a matrix $\boldsymbol{D}$ such that:

This means that $\boldsymbol{A}^{-1}=\boldsymbol{BD}$, and thus $\boldsymbol{A}$ is invertible by definition. This completes the proof.

Recall that to prove a matrix $\boldsymbol{B}$ is an inverse of matrix $\boldsymbol{A}$, we had to show $\boldsymbol{BA}=\boldsymbol{I}$ as well as $\boldsymbol{AB}=\boldsymbol{I}$. Fortunately, it turns out that if $\boldsymbol{BA}=\boldsymbol{I}$, then $\boldsymbol{AB}=\boldsymbol{I}$ and vice versa. This means that we don't have to show both $\boldsymbol{BA}=\boldsymbol{I}$ and $\boldsymbol{AB}=\boldsymbol{I}$ - we just need to show either of the two and the other will also hold.

Proof. We will prove the first statement - the proof for the second statement is identical. To follow this proof, we will use a theoremlink covered in the next section - please come back to this proof after reaching and completing that section.

Let $\boldsymbol{A}$ and $\boldsymbol{B}$ be square matrices of shape $n\times{n}$. Assume $\boldsymbol{BA}=\boldsymbol{I}_n$. To prove that $\boldsymbol{B}=\boldsymbol{A}^{-1}$, we must first show that $\boldsymbol{A}$ is invertible, that is, $\boldsymbol{A}^{-1}$ exists. This allows us to manipulate our assumption to obtain the desired results:

Multiplying $\boldsymbol{A}$ to both sides gives us $\boldsymbol{AB}=\boldsymbol{I}_n$ as well. Therefore, the only task remaining is to show that $\boldsymbol{A}$ is invertible.

We know from theoremlink that if the homogeneous system $\boldsymbol{Ax}=\boldsymbol{0}$ only has the trivial solution $\boldsymbol{x}=\boldsymbol{0}$, then $\boldsymbol{A}$ is invertible. Let $\boldsymbol{x}_0$ be any solution to $\boldsymbol{Ax}=\boldsymbol{0}$, that is:

Multiplying both sides by $\boldsymbol{B}$ gives:

Therefore, the only solution to the homogeneous system is the zero vector, which is the trivial solution. Therefore, $\boldsymbol{A}$ must be invertible - performing steps \eqref{eq:CCAaKX8jLF61MiDz8RJ} completes the proof.

Practice problems