Proof. Consider the following matrix:

Similarly, consider the identity matrix $\boldsymbol{I}$ below:

Here, the columns of $\boldsymbol{I}$ are the standard unit vectors.

Now, by definition, $\boldsymbol{A}_i(\boldsymbol{b})$ is the same as matrix $\boldsymbol{A}$ except that the $i$-th column is replaced by $\boldsymbol{b}$, that is:

Also, by definition, $\boldsymbol{I}_i(\boldsymbol{x})$ is:

Notice how the $i$-th row contains all zeros except for $x_i$. By Laplace expansion theoremlink, we can compute the determinant of $\boldsymbol{I}_i(\boldsymbol{x})$ by cofactor expansion along the $i$-th row:

Here, we used the fact that $\det(\boldsymbol{I})=1$ by theoremlink.

Now, the matrix product $\boldsymbol{A}\Big(\boldsymbol{I}_i(\boldsymbol{x})\Big)$ is:

Note that the second equality holds because of theoremlink. Now, we take the determinant of both sides of \eqref{eq:KlruqkGMOuMi0fTgm1X} to get:

From the multiplicative property of determinantslink, we have that:

Substituting \eqref{eq:mc88MydaJ22kwUgV9Ox} gives:

This completes the proof.

Solution. The system of linear equations can be expressed as:

Let $\boldsymbol{A}$ represent the matrix on the left and $\boldsymbol{b}$ represent the vector on the right. To use Cramer's rule, we first need to compute the determinant of $\boldsymbol{A}$ like so:

Next, $\boldsymbol{A}_1(\boldsymbol{b})$ and $\boldsymbol{A}_2(\boldsymbol{b})$ are:

We now compute the determinant of each:

Now, we use Cramer's rule to solve the system:

Therefore, the solution is:

Finally, just to confirm that this is indeed the solution to our system of linear equations, let's substitute $x_1$ and $x_2$ into the system:

These are indeed the solutions to the system 🎉!

Solution. The adjugate of $\boldsymbol{A}$ is:

We now need to find the cofactors:

Therefore, the adjugate of $\boldsymbol{A}$ is:

Proof. By definitionlink, a matrix $\boldsymbol{A}$ is invertible if and only if there exists another matrix $\boldsymbol{B}$ such that:

Once again, we treat matrices as a collection of columns:

Where the column vectors in $\boldsymbol{I}$ represent the standard unit vectors. From theoremlink, the matrix product $\boldsymbol{AB}$ can be expressed as:

Aligning the columns of $\boldsymbol{AB}$ and $\boldsymbol{I}$ gives:

Let's consider the $j$-th equality:

This can be thought of as a system of linear equations. The components of $\boldsymbol{b}_j$ can be expressed as:

We can find each component using Cramer's rulelink like so:

The numerator of \eqref{eq:VhAat4egrXF78WCsobx} is:

By definition, column $\boldsymbol{e}_j$ has a $1$ in the $j$-th entry and zero in all other entries. We now perform cofactor expansion along the $i$-th column to obtain the determinant of $\boldsymbol{A}_i(\boldsymbol{e}_j)$ like so:

Where $\boldsymbol{A}_{ji}$ represents the sub-matrix in which the $j$-th row and the $i$-th column of $\boldsymbol{A}$ are removed. For notational convenience, we write the right-hand side as the cofactor $C_{ji}$ of entry $a_{ji}$ to get:

Substituting \eqref{eq:DZT5wYfD8XE7rMz7g0f} into \eqref{eq:VhAat4egrXF78WCsobx} gives:

Using \eqref{eq:YSZHqoEIbE2fcJGaKue}, we now express \eqref{eq:QHoHxlY9MaHnXYENUgz} as:

Remember, $\boldsymbol{b}_j$ represents the $j$-th column of $\boldsymbol{B}$. We can now express $\boldsymbol{B}$ in its entirety:

The matrix in \eqref{eq:vrmyga8REQfbqxH3cqg} is called the adjugatelink of $\boldsymbol{A}$, that is:

Finally, $\boldsymbol{B}$ is the inverse of $\boldsymbol{A}$, so let's write it as $\boldsymbol{A}^{-1}$ instead. Therefore, we end up with the following result:

This completes the proof.

Proof. By theoremlink, we have that:

By theoremlink, we know that the determinant of a $2\times2$ matrix is:

The adjugate of $\boldsymbol{A}$ is:

The cofactors are:

The adjugate of $\boldsymbol{A}$ is therefore:

Substituting \eqref{eq:S6Nx8o9E4W7RCn1bWUG} and \eqref{eq:IprqJxAw30U05ioX3DD} into \eqref{eq:gZWXPHh86uoVaBPL25D} gives:

This completes the proof.

Solution. The inverse of matrix $\boldsymbol{A}$ is:

Let's confirm that this is actually the inverse of $\boldsymbol{A}$ by computing $\boldsymbol{AA}^{-1}$ like so:

Indeed, the $\boldsymbol{A}^{-1}$ that we found is the inverse of $\boldsymbol{A}$.