Proof. Let $\boldsymbol{A}$ be an invertible matrix. Let's first prove the following:

From theoremlink, we know that $\boldsymbol{A}$ can be expressed as a product of elementary matrices:

Taking the determinant of both sides gives:

Repeatedly applying theoremlink to the right-hand side gives:

We know from theoremlink that the determinant of an elementary matrix cannot be $0$. Therefore, we have that:

Next, let's prove the converse, that is:

Recall from our guide on proof by contraposition that we can equivalently prove the following contrapositive statement instead:

We know from theoremlink that if $\boldsymbol{A}$ is not invertible, then the reduced row echelon form of $\boldsymbol{A}$ is not an identity matrix and instead has a row with all zeros. This reduced row echelon form is obtained by applying a series of elementary row operations on $\boldsymbol{A}$ like so:

Taking the determinant of both sides gives:

Because $\text{rref}(\boldsymbol{A})$ contains a row with all zeros, we know from theoremlink that its determinant is zero:

After repeated application of theoremlink on the right-hand side, we get:

From theoremlink, the determinant of elementary matrices cannot be equal to zero. Therefore, for \eqref{eq:uh15d3QwVMp4dBC2l9I} to hold, $\det(\boldsymbol{A})=0$. This completes the proof.

Proof. First consider the case when $\boldsymbol{A}$ is invertible. By theoremlink, we know that $\boldsymbol{A}$ can be expressed as a product of elementary matrices:

For reference later, let's take the determinant of both sides:

Now, let's multiply both sides of \eqref{eq:JH87eddXjoAVbCsnHhO} by some matrix $\boldsymbol{B}$ to get:

Taking the determinant of both sides and repeatedly apply theoremlink to get:

Substituting \eqref{eq:A3qN2BOhPSZhDTxp4FJ} into \eqref{eq:EQDIFXv7fQvWqMq5jb7} gives:

Now, suppose $\boldsymbol{A}$ is not invertible. By theoremlink, this means that $\det(\boldsymbol{A})=0$. Therefore, for any matrix $\boldsymbol{B}$, we have that:

We must thus show that $\det(\boldsymbol{AB})=0$. The reduced row echelon form of $\boldsymbol{A}$ can be obtained by applying a series of elementary row operations:

Now, let's multiply both sides by $\boldsymbol{B}$ to get:

Taking the determinant of both sides gives:

Because $\boldsymbol{A}$ is not invertible, the reduced row echelon form of $\boldsymbol{A}$ contains a row with all zeros. The matrix product $\mathrm{rref}(\boldsymbol{A})\boldsymbol{B}$ will therefore result in another matrix containing a row with all zeros. By theoremlink, this means that $\det\big(\text{rref}(\boldsymbol{A})\boldsymbol{B}\big)=0$. Therefore, \eqref{eq:DM0GgDYlP5mWd3hXtvz} becomes:

We now repeatedly apply theoremlink to get:

By theoremlink, we know that the determinant of elementary matrices cannot be zero, and so $\det(\boldsymbol{AB})=0$. This completes the proof.

Proof. By the multiplicative propertylink of determinants, we have that:

The determinant of the identity matrix is $1$ by theoremlink. Therefore, we get:

This completes the proof.

Proof. Consider an $n\times{n}$ matrix $\boldsymbol{A}$. The determinant of $k\boldsymbol{A}$ is:

Here, $\boldsymbol{I}_n$ is an $n\times{n}$ identity matrix. By the multiplicative property of determinantslink, we have that:

Finally, by theoremlink, we have that $\det(k\boldsymbol{I}_n)=k^n$. Therefore, we conclude that:

This completes the proof.

Proof. Suppose $\boldsymbol{A}$ is not invertible. From theoremlink, we have that $\det(\boldsymbol{A})=0$. From theoremlink, if $\boldsymbol{A}$ is not invertible, then $\boldsymbol{A}^T$ is also not invertible, which means $\det(\boldsymbol{A}^T)=0$. Therefore, we have that:

Now, suppose $\boldsymbol{A}$ is invertible. From theoremlink, $\boldsymbol{A}$ can be expressed as a product of elementary matrices:

For reference later, let's take the determinant of both sides and apply theoremlink repeatedly:

Now, taking the transpose of both sides of \eqref{eq:VeE5NBxXWHXYDI9Q3Y5} and applying theoremlink gives:

Taking the determinant of both sides and applying theoremlink repeatedly gives:

From theoremlink, we know that the determinant of an elementary matrix is equal to the determinant of its transpose. Therefore, \eqref{eq:QvFCrdH1upnLdwJWLTK} becomes:

This is the same as \eqref{eq:JUKQmcM56Yj39s7D7al}. Therefore, we conclude that:

This completes the proof.