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4.2. Determinants of square matrices

The algorithm for computing determinants of square matrices will be given inductively on the size $n$ of the matrix. That is, we will find determinants of large matrices by computing the determinants of several smaller “sub-matrices”. In the smallest case, $n=1$ , the determinant is “equal” to the matrix, since there is only one coefficient. We also know the determinant of a $2\times 2$ matrix (Definition 4.1.1). Before giving the full definition, let us first state the formula for the determinant of a $3\times 3$ matrix.

Definition 4.2.1 (Determinant of a $3\times 3$ matrix).

Let $A=\begin{pmatrix}a&b&c\\ d&e&f\\ g&h&i\end{pmatrix}\in\operatorname{M}_{{3}}({{\mathbb{R}}})\;$ . Then,

\det A=aei+bfg+cdh-ceg-afh-bdi.

It is a good exercise to check that the following holds, which expresses the $3\times 3$ determinant in terms of three $2\times 2$ determinants:

\det A=a(ei-hf)-b(di-gf)+c(dh-ge)=a\begin{vmatrix}e&f\\ h&i\end{vmatrix}-b\begin{vmatrix}d&f\\ g&i\end{vmatrix}+c\begin{vmatrix}d&e\\ g&h\end{vmatrix}.

This rearranged expression is called the expansion of $\det A$ about the first row of $A$ . An analogous formula will be used to define $4\times 4$ determinants, and higher. Let us point out two things about this expression:

•

The signs alternate, ( $+a,-b,+c$ ) and
•

each element is multiplied by the determinant of the $2\times 2$ matrix obtained by crossing out the row and column containing that element.

For example, the $(1,2)$ coefficient $b$ of $A$ multiplies the determinant

\begin{vmatrix}d&f\\ g&i\end{vmatrix}

which is obtained by excluding the first row and second column of $A$ .

Notation 4.2.2.

To define determinants for larger matrices, we will make use of the following notation. If $A\in\operatorname{M}_{{n}}({{\mathbb{R}}})$ , then $A[i,j]\in\operatorname{M}_{{n-1}}({{\mathbb{R}}})$ is the matrix obtained by deleting the $i$ -th row and $j$ -th column.

Example 4.2.3.

As an example of this notation, let $A=\begin{pmatrix}1&2&3\\ 4&5&6\\ 7&8&9\end{pmatrix}$ . Then we have

$A[1,1]=\begin{pmatrix}5&6\\ 8&9\end{pmatrix},\quad A[3,2]=\begin{pmatrix}1&3\\ 4&6\end{pmatrix}.$

Remark 4.2.4.

One could also check that the following expression holds:

\det A=a\begin{vmatrix}e&f\\ h&i\end{vmatrix}-d\begin{vmatrix}b&c\\ h&i\end{vmatrix}+g\begin{vmatrix}b&c\\ e&f\end{vmatrix}.

This is called the expansion of $\det A$ about the first column, analogous to the situation above. For example, the third matrix is obtained from $A$ by deleting the row and column containing $g$ .

Now we are ready to inductively define the determinant and cofactor for any size matrix. Since we already have a definition of a determinant of a size 2 matrix ( $\det A=ad-bc$ ), we will use that to define the determinant of a size 3 matrix, and then use that to define the determinant of size 4, etc.

Definition 4.2.5 (Determinant of an $n\times n$ matrix).

Let $A=(a_{ij})\in\operatorname{M}_{{n}}({{\mathbb{R}}})$ , and let $1\leq i,j\leq n$ . Assume that the determinant of any $(n-1)\times(n-1)$ matrix is defined.

(i)

The cofactor of the coefficient $a_{ij}$ is the following expression

$A_{ij}=(-1)^{i+j}\det(A[i,j]),$

where $A[i,j]\in\operatorname{M}_{{n-1}}({{\mathbb{R}}})$ is as in Notation 4.2.2.
(ii)

The determinant of $A$ is

$\det A=a_{11}A_{11}+a_{12}A_{12}+\cdots+a_{1n}A_{1n}=\sum_{1\leq i\leq n}a_{1i% }A_{1i}\;.$

This is the expansion of $\det A$ about the first row of $A$ . Equivalently, we can expand about any row or any column. That is, for any given $1\leq k\leq n$ , we have

$\det A=\sum_{1\leq i\leq n}a_{ki}A_{ki}=\sum_{1\leq i\leq n}a_{ik}A_{ik}\;.$
(iii)

This expansion of the determinant in terms of its cofactors $A_{ij}$ is called a cofactor expansion.

Remark 4.2.6.

We will also call $\det(A[i,j])\in{\mathbb{R}}$ (without the sign $(-1)^{i+j}$ ) the minor of the coefficient $a_{ij}$ .

One can check that the computation of $\det A$ for a matrix of size $2$ or $3$ matches that obtained using Definitions 4.1.1 and 4.2.1.

Example 4.2.7.

Let $A=\begin{pmatrix}1&2&3\\ 4&5&6\\ 7&8&9\end{pmatrix}$ , let’s compute the determinant of $A$ , by using the cofactor expansion of the first row. First we compute the cofactors of the coefficients of the first row, $a_{11},a_{12},$ and $a_{13},$ as follows:

$A_{11}=(-1)^{1+1}\det(A[1,1])=\left|\begin{array}[]{cc}5&6\\ 8&9\end{array}\right|=5\cdot 9-6\cdot 8=-3,$

$A_{12}=(-1)^{1+2}\det(A[1,2])=-\left|\begin{array}[]{cc}4&6\\ 7&9\end{array}\right|=-(4\cdot 9-6\cdot 7)=6,$

$A_{13}=(-1)^{1+3}\det(A[1,3])=\left|\begin{array}[]{cc}4&5\\ 7&8\end{array}\right|=4\cdot 8-5\cdot 7=-3.$

The cofactor expansion tells us that determinant of $A$ is

$\det A=a_{11}A_{11}+a_{12}A_{12}+a_{13}A_{13}=1\cdot(-3)+2\cdot 6+3\cdot(-3)=0.$

Remark 4.2.8.

The sign $(-1)^{i+j}$ showing up in the cofactor may be remembered using the “chess-board” pattern:

\begin{array}[]{ccccc}+&-&+&-&\dots\\ -&+&-&+&\dots\\ +&-&+&-&\dots\\ -&+&-&+&\dots\\ \vdots&\vdots&\vdots&\vdots&\ddots\end{array}

In particular, note that the sign of the cofactor of a diagonal coefficient is always positive. Hence, such a cofactor is always equal to its minor.

Example 4.2.9.

It is a good exercise to check that in the $3\times 3$ case, it doesn’t matter which row or column we choose to expand, the resulting expression is always the same as Definition 4.2.1. For example, given the following matrix, expand the second column:

$A=\begin{pmatrix}a&b&c\\ d&e&f\\ g&h&i\end{pmatrix}\quad\hbox{so the cofactor expansion is}\quad bA_{12}+eA_{22% }+hA_{32}.$

Recall that $A_{ij}$ refers to the cofactor, $A_{ij}=(-1)^{i+j}\det(A[i,j])$ . So the above expression becomes

$b(-1)\left|\begin{array}[]{cc}d&f\\ g&i\end{array}\right|+e\left|\begin{array}[]{cc}a&c\\ g&i\end{array}\right|+h(-1)\left|\begin{array}[]{cc}a&c\\ d&f\end{array}\right|=\cdots=\det A.$

Example 4.2.10.

Now let us consider a $4\times 4$ matrix. We will shortly come back to this same example in order to see how the computation of determinants can be simplified further.

$A=\begin{pmatrix}2&0&3&5\\ 0&4&-1&0\\ 1&0&0&1\\ 0&2&1&1\end{pmatrix}.$

Since $\det A$ can be computed by expanding about any row or column of $A$ , let us take the “easiest one”. As the third row has only two non-zero terms, let us expand about it. So, we get

$\det A=a_{31}A_{31}+a_{32}A_{32}+a_{33}A_{33}+a_{34}A_{34}=$

$(-1)^{3+1}(1)\begin{vmatrix}0&3&5\\ 4&-1&0\\ 2&1&1\end{vmatrix}+0+0+(-1)^{3+4}(1)\begin{vmatrix}2&0&3\\ 0&4&-1\\ 0&2&1\end{vmatrix}=\underbrace{\begin{vmatrix}0&3&5\\ 4&-1&0\\ 2&1&1\end{vmatrix}}_{D1}-\underbrace{\begin{vmatrix}2&0&3\\ 0&4&-1\\ 0&2&1\end{vmatrix}}_{D2}.$

So we have reduced the computation to two $3\times 3$ determinants, call them $D_{1}$ and $D_{2}$ . Expanding both of them about their first columns, we get

$\det A=\underbrace{0-4\begin{vmatrix}3&5\\ 1&1\end{vmatrix}+2\begin{vmatrix}3&5\\ -1&0\end{vmatrix}}_{D1}-\underbrace{2\begin{vmatrix}4&-1\\ 2&1\end{vmatrix}+0-0}_{D2}=-4(-2)+2\cdot 5-2\cdot 6=6.$

So the determinant of the $4\times 4$ matrix $A$ is equal to 6.