search
Search
Unlock 100+ guides
search toc
close
Cancel
Post
account_circle
Profile
exit_to_app
Sign out
search
keyboard_voice
close
Searching Tips
Search for a recipe:
"Creating a table in MySQL"
Search for an API documentation: "@append"
Search for code: "!dataframe"
Apply a tag filter: "#python"
Useful Shortcuts
/ to open search panel
Esc to close search panel
to navigate between search results
d to clear all current filters
Enter to expand content preview Doc Search Code Search Beta SORRY NOTHING FOUND!
mic
Start speaking... Voice search is only supported in Safari and Chrome.
Shrink
Navigate to

# Comprehensive Guide on Laplace Expansion Theorem

schedule Aug 12, 2023
Last updated
local_offer
Linear Algebra
Tags
mode_heat
Master the mathematics behind data science with 100+ top-tier guides
Start your free 7-days trial now!

The determinant was initially defined as the cofactor expansion along the first row. In theoremlink, we proved that the determinant can also be computed using the cofactor expansion along the first column. Interestingly, the determinant can be computed as the cofactor expansion along any row or column 🤯.

Theorem.

# Laplace expansion theorem

The determinant of a matrix can be computed by the cofactor expansion along any row or column.

Proof. For simplicity, consider the following $3\times3$ matrix:

$$\boldsymbol{A}=\begin{pmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33} \end{pmatrix}$$

The general proof follows the same logic. By definition, the determinant of $\boldsymbol{A}$ is equal to the cofactor expansion along the first row:

$$\det(\boldsymbol{A}) =a_{11}\det(\boldsymbol{A}_{11}) -a_{12}\det(\boldsymbol{A}_{12}) +a_{13}\det(\boldsymbol{A}_{13})$$

Here, $\boldsymbol{A}_{12}$ represents the sub-matrix after removing the $1$st row $2$nd column from the original matrix $\boldsymbol{A}$.

Our first goal is to show that the determinant of $\boldsymbol{A}$ is also equal to the cofactor expansion along the second row:

$$\begin{equation}\label{eq:Odp4e5yKFsso1Sc6h50} \det(\boldsymbol{A}) =-a_{21}\det(\boldsymbol{A}_{21}) +a_{22}\det(\boldsymbol{A}_{22}) -a_{23}\det(\boldsymbol{A}_{23}) \end{equation}$$

Consider the following matrix $\boldsymbol{B}$ obtained by interchanging the first two rows of $\boldsymbol{A}$ like so:

$$\boldsymbol{B}=\begin{pmatrix} a_{21}&a_{22}&a_{23}\\ a_{11}&a_{12}&a_{13}\\ a_{31}&a_{32}&a_{33} \end{pmatrix}$$

We obtain the determinant of $\boldsymbol{B}$ using the cofactor expansion along the first row:

$$\begin{equation}\label{eq:lC8KOqXT4QxuWblzcJv} \det(\boldsymbol{B}) =a_{21}\det(\boldsymbol{A}_{21}) -a_{22}\det(\boldsymbol{A}_{22}) +a_{23}\det(\boldsymbol{A}_{23}) \end{equation}$$

Now, by theoremlink, interchanging an adjacent pair of rows will flip the sign of the determinant. Therefore, we have that:

$$\begin{equation}\label{eq:gJLennJq7uOx6KlRg2g} \det(\boldsymbol{A})= -\det(\boldsymbol{B}) \end{equation}$$

Substituting \eqref{eq:lC8KOqXT4QxuWblzcJv} into \eqref{eq:gJLennJq7uOx6KlRg2g} gives:

$$\det(\boldsymbol{A}) =-a_{21}\det(\boldsymbol{A}_{21}) +a_{22}\det(\boldsymbol{A}_{22}) -a_{23}\det(\boldsymbol{A}_{23})$$

This is exactly what we wanted to prove \eqref{eq:Odp4e5yKFsso1Sc6h50}.

* * *

Let's now consider the case of computing the determinant using cofactor expansion along any row after the $2$nd row. Let's say that we want to show that the determinant can also be computed by cofactor expansion along the $3$rd row:

$$\begin{equation}\label{eq:CYrxgMWzQseT5HeUgpS} \det(\boldsymbol{A})=a_{31}\det(\boldsymbol{A}_{31}) -a_{32}\det(\boldsymbol{A}_{32}) +a_{33}\det(\boldsymbol{A}_{33}) \end{equation}$$

Now, consider the following matrix where the $3$rd row of $\boldsymbol{A}$ is at the top:

$$\boldsymbol{B}=\begin{pmatrix} a_{31}&a_{32}&a_{33}\\ a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23} \end{pmatrix}$$

To bring the $3$rd row to the top, we performed $2$ elementary row operations of interchanging rows on $\boldsymbol{A}$ like so:

$$\begin{pmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33} \end{pmatrix}\;\;\;\;\rightarrow\;\;\;\; \begin{pmatrix} a_{11}&a_{12}&a_{13}\\ a_{31}&a_{32}&a_{33}\\ a_{21}&a_{22}&a_{23} \end{pmatrix}\;\;\;\;\rightarrow\;\;\;\; \begin{pmatrix} a_{31}&a_{32}&a_{33}\\ a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23} \end{pmatrix}$$

Every row-swapping operation flips the sign of the determinant. Since we row-swapped twice, the sign of the determinant does not change. Therefore, we have that:

$$\begin{equation}\label{eq:nsUGfDpyFjcyyjqa4Rk} \det(\boldsymbol{A})= \det(\boldsymbol{B}) \end{equation}$$

By definition, the determinant of $\boldsymbol{B}$ is the cofactor expansion along the first row:

$$\begin{equation}\label{eq:xpUrDl1GHI4CcpqPWvk} \det(\boldsymbol{B})= a_{31}\det(\boldsymbol{A}_{31}) -a_{32}\det(\boldsymbol{A}_{32}) +a_{33}\det(\boldsymbol{A}_{33}) \end{equation}$$

Equating \eqref{eq:nsUGfDpyFjcyyjqa4Rk} and \eqref{eq:xpUrDl1GHI4CcpqPWvk} gives:

$$\det(\boldsymbol{A})= a_{31}\det(\boldsymbol{A}_{31}) -a_{32}\det(\boldsymbol{A}_{32}) +a_{33}\det(\boldsymbol{A}_{33})$$

This is exactly what we wanted to prove \eqref{eq:CYrxgMWzQseT5HeUgpS}.

Great, we have managed to show that the determinant can be computed using cofactor expansion along any row. We now have to show that we can do so using cofactor expansion along any column as well. Fortunately, this is easy because we have already proven the following theoremlink:

$$\begin{equation}\label{eq:sqv4NTJTaXGonBhAo20} \det(\boldsymbol{A}^T)=\det(\boldsymbol{A}) \end{equation}$$

For example, suppose we have the following matrix:

$$\boldsymbol{A}=\begin{pmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33} \end{pmatrix}$$

Our goal is to show that the determinant can be computed by cofactor expansion along any column. Let's say we choose the $3$rd column - we want to show the following:

$$\begin{equation}\label{eq:iFSxbmTqAYneLDEEPzu} \det(\boldsymbol{A})= a_{13}\det(\boldsymbol{A}_{13}) -a_{23}\det(\boldsymbol{A}_{23}) +a_{33}\det(\boldsymbol{A}_{33}) \end{equation}$$

Now, the transpose of $\boldsymbol{A}$ is:

$$\boldsymbol{A}^T =\begin{pmatrix} a_{11}&a_{21}&a_{31}\\ a_{12}&a_{22}&a_{32}\\ a_{13}&a_{23}&a_{33} \end{pmatrix}$$

We have already shown that we can perform cofactor expansion along any row to obtain the determinant. Let's pick the $3$rd row:

$$\det(\boldsymbol{A}^T)= a_{13}\det(\boldsymbol{A}_{31}^T) -a_{23}\det(\boldsymbol{A}_{32}^T) +a_{33}\det(\boldsymbol{A}_{33}^T)$$

Because $\det(\boldsymbol{A}^T)=\det(\boldsymbol{A})$, we have that:

$$\begin{equation}\label{eq:IhG1QMl5an7mHRK4RE5} \det(\boldsymbol{A})= a_{13}\det(\boldsymbol{A}_{31}^T) -a_{23}\det(\boldsymbol{A}_{32}^T) +a_{33}\det(\boldsymbol{A}_{33}^T) \end{equation}$$

Finally, notice how $\boldsymbol{A}^T_{ij}= \boldsymbol{A}_{ji}$. For instance, consider $\boldsymbol{A}^T_{31}= \boldsymbol{A}_{13}$ below:

$$\boldsymbol{A}^T_{31} =\begin{pmatrix} \color{green}a_{11}&a_{21}&a_{31}\\ \color{green}a_{12}&a_{22}&a_{32}\\ \color{green}a_{13}&\color{green}a_{23}&\color{green}a_{33} \end{pmatrix}= \begin{pmatrix} \color{green}a_{11}&\color{green}a_{12}&\color{green}a_{13}\\ a_{21}&a_{22}&\color{green}a_{23}\\ a_{31}&a_{32}&\color{green}a_{33} \end{pmatrix} = \boldsymbol{A}_{13}$$

Therefore, \eqref{eq:IhG1QMl5an7mHRK4RE5} becomes:

$$\det(\boldsymbol{A})= a_{13}\det(\boldsymbol{A}_{13}) -a_{23}\det(\boldsymbol{A}_{23}) +a_{33}\det(\boldsymbol{A}_{33})$$

This is exactly what we wanted to show \eqref{eq:iFSxbmTqAYneLDEEPzu}. This completes the proof.

Example.

# Performing cofactor expansion along a row and column of 3x3 matrix

Consider the following matrix:

$$\boldsymbol{A}=\begin{pmatrix} 1&1&3\\ 2&4&2\\ 2&1&1\\ \end{pmatrix}$$

Show that the cofactor expansion along the $1$st row and the cofactor expansion along the $3$rd column are equal.

Solution. We know from the Laplace expansion theorem that the cofactor expansion along the $1$st row and the cofactor expansion along the $3$rd column are equal. Let's confirm this.

The cofactor expansion along the $1$st row is:

\begin{align*} \det(\boldsymbol{A})&= 1\begin{vmatrix}4&2\\1&1\end{vmatrix}- 1\begin{vmatrix}2&2\\2&1\end{vmatrix}+ 3\begin{vmatrix}2&4\\2&1\end{vmatrix}\\ &= 1(4-2)- 1(2-4)+ 3(2-8)\\ &=-14 \end{align*}

The cofactor expansion along the $3$rd column is:

\begin{align*} \det(\boldsymbol{A})&= 3\begin{vmatrix}2&4\\2&1\end{vmatrix}- 2\begin{vmatrix}1&1\\2&1\end{vmatrix}+ 1\begin{vmatrix}1&1\\2&4\end{vmatrix}\\ &= 3(2-8)- 2(1-2)+ 1(4-2)\\ &=-14 \end{align*}

Indeed, no matter which row or column we pick for cofactor expansion, we end up with the same value for the determinant!

Example.

# Finding the determinant of 3x3 matrix

Find the determinant of the following matrix:

$$\boldsymbol{A}=\begin{pmatrix} 1&1&2\\ 2&0&5\\ 3&0&1\\ \end{pmatrix}$$

Solution. The Laplace expansion theorem states that the determinant can be computed using cofactor expansion along any row or column. Therefore, we can be smart about the row or column that we pick. We can see that the second column of $\boldsymbol{A}$ is filled mostly with zeros. Therefore, let's perform cofactor expansion along the second column:

\begin{align*} \det(\boldsymbol{A}) &=-1\begin{vmatrix} 2&5\\ 3&1\\ \end{vmatrix}\\ &=-\big((2)(1)-(5)(3)\big)\\ &=13 \end{align*}

Notice how easy it is to find the determinant in this case!

thumb_up
thumb_down
Comment
Citation
Ask a question or leave a feedback...
thumb_up
0
thumb_down
0
chat_bubble_outline
0
settings
Enjoy our search
Hit / to insta-search docs and recipes!