Title: Default for 7.3 Determinants
Course: HELM Workbooks 2023-2024

Determinant of a \(2\times2\) matrix

The determinant of the matrix \(A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}\) is denoted by \(\begin{vmatrix} a & b \\ c & d \end{vmatrix}\) (note the change from square brackets to vertical lines) and is defined to be the number \(ad-bc\). That is: \[\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad-bc.\]

We can use the notation \(\det(A)\) or \(\lvert A \rvert\) to denote the determinant of \(A\). A common symbol used to represent the determinant of a given matrix is \(\Delta.\)

Exercise

Evaluate the determinants of the following matrices.

1. \(A=[value]\)
2. \(B=[value]\)
3. \(C=[value]\)

1. \(\det{A}=\)
2. \(\det{B}=\)
3. \(\det{C}=\)
Feedback: We use the definition to find the determinant of the following matrices. 1. For \(A=[value],\) \(\det(A) = [value] = [value].\) 2. For \(B=[value],\) \(\det(B) = [value] = [value].\) 3. For \(C=[value],\) \(\det(C) = [value] = [value].\)

Laplace expansion along the top row

This is a technique which can be used to evaluate determinants of any order. In principle, this method can use any row or any column as its starting point. We quote one example: using the top row.

Consider \(\Delta=\begin{vmatrix} 4 & 1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & 2 \end{vmatrix}\).

First we introduce the idea of a minor. Each element in this array of numbers has an associated number called its minor, calculated by removing the column and row in which the element lies and taking the determinant of the remainder. For example consider entry \(a_{23}=3\). To calculate the minor associated with this entry, we strike out the second row and the third column:

[Image: The matrix with rows [4, 1, 1], [1, 2, 3], [3, 1, 2], with a horizontal line crossing out the second row, and a vertical]

which leaves \(\begin{vmatrix} 4 & 1 \\ 3 & 1 \end{vmatrix} = 4 - 3 = 1.\) For the entry \(a_{31}=3\) we strike out the third row and first column:

[Image: The matrix with rows [4, 1, 1], [1, 2, 3], [3, 1, 2], with a horizontal line crossing out row 3, and a vertical line cro]

which leaves \(\begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix} = 3 - 2 = 1.\)

Exercise

Consider the matrix \(A = [value]\). What is the minor of the entry \(a_{22}\)?

The minor of \(a_{22}\) is:
Feedback: The minor of \(a_{22}\) is the \(2\times2\) determinant \(\begin{vmatrix}a_{11} & a_{13} \\ a_{31} & a_{33}\end{vmatrix}=a_{11}a_{33}-a_{31}a_{13}\). For the matrix \(A = [value]\) the minor is: \(\begin{vmatrix}[value] & [value] \\ [value] & [value]\end{vmatrix} = [value] = [value].\)

Cofactors

A cofactor is a minor with a sign attached. The appropriate sign comes from the pattern of signs appropriate to a \(3\times 3\) array: \[\begin{bmatrix}+&-&+\\ -&+&-\\ +&-&+\end{bmatrix}\] (i.e. positive signs on the leading diagonal and the signs "alternate" everywhere else.)

Each entry of a matrix \(A\) has a cofactor associated with it. The cofactor of entry \(a_{11}\) is denoted by \(A_{11}\), that of \(a_{23}\) by \(A_{23}\) and so on.

To obtain the cofactor of an element of a \(3\times 3\) matrix we simply multiply the minor of that element by the corresponding sign from the \(3\times 3\) array of signs.

Hence, for the matrix \(A = \begin{bmatrix} 4 & 1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & 2 \end{bmatrix}\), the cofactor corresponding to \(a_{23}\) is \[A_{23} = - \begin{vmatrix} 4 & 1 \\ 3 & 1 \end{vmatrix} = -1\] and the cofactor corresponding to \(a_{31}\) is \[A_{31} = + \begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix}=1.\]

Exercise

Consider the matrix \(A = [value]\). Compute the cofactors of the following entries:

1. The cofactor of \(a_{22}\) is:
2. The cofactor of \(a_{23}\) is:
Feedback: We are given the matrix \(A = [value].\) 1. The minor of \(a_{22}\) is the \(2\times2\) determinant \(\begin{vmatrix}a_{11} & a_{13} \\ a_{31} & a_{33}\end{vmatrix}=a_{11}a_{33}-a_{31}a_{13}\). In this case, the minor is: \[\begin{vmatrix}[value] & [value] \\ [value] & [value]\end{vmatrix} = [value] = [value]\] The sign in the position of \(a_{22}\) in the array of signs is \(+\), the cofactor of \(a_{22}\) is \(A_{22}=+[value]\). 2. The minor of \(a_{22}\) is the \(2\times2\) determinant \(\begin{vmatrix}a_{11} & a_{12} \\ a_{31} & a_{32}\end{vmatrix}=a_{11}a_{32}-a_{31}a_{12}\). In this case, the minor is: \[\begin{vmatrix}[value] & [value] \\ [value] & [value]\end{vmatrix} = [value] = [value]\] The sign in the position of \(a_{23}\) in the array of signs is \(-\), so the cofactor of \(a_{23}\) is \(A_{23}=[value]\).

Evaluating general determinants

Cofactors are important as it can be shown that the value of the determinant \(\Delta\) of a \(3\times 3\) matrix \(A\) can be found from the formula \(\Delta=a_{11}A_{11}+a_{12}A_{12}+a_{13}A_{13}.\)

In words, "the determinant of a \(3\times 3\) matrix is obtained by multiplying each element of the first row by its corresponding cofactor and then adding the three together". (In fact this rule can be extended to apply to any row or any column and to any order square matrix.)

Key Point

Evaluating General Determinants

If \(A\) is an \(n\times n\) square matrix then: \(\displaystyle{\operatorname{det}(A) = \sum_{j=1}^{n}a_{ij}A_{ij}}\).

In words: The determinant of a square matrix is obtained by multiplying each element of row \(i\) by its corresponding cofactor and then adding these products together.

Example

In the case of \(\Delta=\begin{vmatrix} 4 & 1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & 2 \end{vmatrix}\) we have \(a_{11}=4, a_{12}=1, a_{13}=1,\) and \begin{align*} A_{11} & = + \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} = 4 - 3 = 1, \\ A_{12} & = - \begin{vmatrix} 1 & 3 \\ 3 & 2 \end{vmatrix} = - ( 2 - 9 ) = 7, \\ A_{13} & = + \begin{vmatrix} 1 & 2 \\ 3 & 1 \end{vmatrix} = 1 - 6 = - 5.\end{align*}

Hence, \(\Delta=4\times 1 + 1\times 7 + 1\times -5 = 6\).

Alternatively, choosing to expand along the second row: \begin{align*}\Delta & = a_{21} A_{21} + a_{22} A_{22} + a_{23} A_{23} \\ & = 1 \left( - \begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} \right) + 2 \left( \begin{vmatrix} 4 & 1 \\ 3 & 2 \end{vmatrix} \right) + 3 \left( - \begin{vmatrix} 4 & 1 \\ 3 & 1 \end{vmatrix} \right) \\ & = 6\end{align*} as before.

Exercise

By expanding along the first row, compute the determinant of the matrix \(A = [value]\).

The determinant \(\Delta\) is:
Feedback: For \(A=[value]\), we have: \begin{align*} a_{11} & = [value], & a_{12} & = [value], & a_{13} & = [value], \\ A_{11} & = \det{[value]} = [value], & A_{12} & = \det{[value]} = [value], & A_{13} & = \det{[value]} = [value]. \end{align*} Therefore \[\Delta = a_{11}A_{11}+a_{12}A_{12}+a_{13}A_{13} = [value].\]

Properties of determinants

Often, especially with determinants of large order, we can simplify the evaluation rules. Let's discuss some useful properties of determinants in general.

1. If two rows (or two columns) of a determinant are interchanged then the value of the determinant is multiplied by \(-1\).
For example, \(\begin{vmatrix}4&3\\ 1&2\end{vmatrix}=8-3=5\), but (interchanging columns) \(\begin{vmatrix}3&4\\2&1\end{vmatrix}=3-8=-5\) and (interchanging rows) \(\begin{vmatrix}1&2\\ 4&3\end{vmatrix}=3-8=-5\).
2. The determinant of a matrix \(A\) and the determinant of its transpose \(A^T\) are equal.
\[ \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} = 4 - 6 = -2 = \begin{vmatrix} 1 & 3 \\ 2 & 4 \end{vmatrix} \]
3. If two rows (or two columns) of a matrix \(A\) are equal, then it has zero determinant.
For example, the following determinant has two identical rows:
\[\begin{aligned} \begin{vmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \\ 4 & 5 & 6 \end{vmatrix} &= 1 \times \begin{vmatrix} 2 & 3 \\ 5 & 6 \end{vmatrix} + 2 \times \left(- \begin{vmatrix} 1 & 3 \\ 4 & 6 \end{vmatrix} \right) + 3 \times \begin{vmatrix} 1 & 2 \\ 4 & 5 \end{vmatrix} \\ & = -3 + 2 \times 6 + 3 \times (-3) = 0. \end{aligned}\]
4. If the elements of one row (or one column) of a determinant are multiplied by \(k\), then the resulting determinant is \(k\) times the given determinant.
\[\begin{vmatrix} 1 & 2 & 3 \\ 4 & 8 & 6 \\ 7 & 8 &  9 \end{vmatrix} = 2 \begin{vmatrix} 1 & 2 & 3 \\ 2 & 4 & 3 \\ 7 & 8 & 9 \end{vmatrix}.\]
Note that if one row (or column) of a determinant is a multiple of another row (or column) then the value of the determinant is zero. (This follows from properties 3 and 4.)
For example:
\[\begin{aligned} \begin{vmatrix} 2 & 4 & -1 \\ 4 & 2 & 1 \\ -4 & -8 & 2 \end{vmatrix} & = 2 \times \begin{vmatrix} 2 & 1 \\ -8 & 2 \end{vmatrix} +4 \times \left( -\begin{vmatrix} 4 & 1 \\ -4 & 2 \end{vmatrix} \right) - 1 \times \begin{vmatrix} 4 & 2 \\ -4 & -8 \end{vmatrix} \\ & = 2(12)+4(-12)-(-24)=0. \end{aligned}\]
This is predictable as the 3rd row is \(-2\) times the first row.
5. If we add (or subtract) a multiple of one row (or column) to another, the value of the determinant is unchanged.
Given \(\begin{vmatrix}1&2\\ 4&5\end{vmatrix}\), adding (2 \(\times\) row 1) to (row 2) gives
\[\begin{vmatrix} 1 & 2 \\ 4 + 2 \times 1 & 5 + 2 \times 2 \end{vmatrix} = \begin{vmatrix} 1 & 2 \\ 6 & 9 \end{vmatrix} = 9 - 12 = -3 = \begin{vmatrix} 1 & 2 \\ 4 & 5 \end{vmatrix} \]
6. The determinant of a lower triangular matrix, an upper triangular matrix or a diagonal matrix is the product of the elements on the leading diagonal.
As an example, it is easily confirmed that each of the following determinants has the same value \(1 \times 4 \times 6 = 24\).
\[ \begin{vmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{vmatrix},\qquad \begin{vmatrix} 1 & 0 & 0 \\ 2 & 4 & 0 \\ 3 & 5 & 6 \end{vmatrix},\qquad \begin{vmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 6 \end{vmatrix},\qquad \begin{vmatrix} 4 & 2 & 3 \\ 0 & 1 & 5 \\ 0 & 0 & 6 \end{vmatrix} \]

Example

This exercise is in five parts. Consider the determinant \[\Delta = \det[value]\]

1. Use property 2 to find another matrix whose determinant is equal to \(\Delta\):

2. Now expand along the top row to express \(\Delta\) as the difference of two products, each of a number and a \(3\times 3\) determinant:
\(\Delta=\) \(\times\det\) \(-\) \(\times\det\)
3. Use property 4 to evaluate the first \(3\times3\) determinant:
4. Use property 4 to evaluate the second \(3\times3\) determinant:
5. Deduce the value of \(\Delta\):
Feedback: Property 2 states that the determinant of a matrix is equal to the determinant of its transpose. So the other matrix is just the transpose of the matrix in the question: \[[value]^{T} = [value]\] Expanding along the top row is easiest, because of the two zeros in the row. The formula for the determinant in terms of cofactors tells us: \[\det[value] = ([value])[value]-([value])[value]\] Note the minus sign, which comes from the \(4\times4\) pattern of signs. In the second of the two \(3\times3\) determinants, the second row is double the first row. Hence the determinant is zero, by Property 4. In the first of the two \(3\times3\) determinants, the third column is \([value]\) times the second column. Hence this determinant is also zero. Therefore \(\Delta=0\).

Exercise

In this exercise you will use Laplace expansion along the 1st row to determine the determinant of the following matrix:

\(A=[value]\)

Let the entries in the first row be denoted \(a_{1,1}\), \(a_{1,2}\) and \(a_{1,3}\) in the usual way. First list the cofactors \(A_{i,j}\) of these entries:

\(A_{1,1}=\)

\(A_{1,2}=\)

\(A_{1,3}=\)

Use these values to evaluate the determinant of \(A\):
Feedback: To obtain the cofactor of an element of a \(3 \times 3\) matrix we multiply the minor of that element by the corresponding sign from the \(3 \times 3\) array of signs. For \(a_{1,1}\), the sign is \(+\), and the minor is \(\det[value]\). Hence the cofactor is \([value].\) For \(a_{1,2}\), the sign is \(-\), and the minor is \(\det[value]\). Hence the cofactor is \([value].\) For \(a_{1,3}\), the sign is \(+\) again, and the minor is \(\det[value]\). Hence the cofactor is \([value].\) The determinant of \(A\) is now given by the formula: \[\det(A) = [value] = [value]\]

Exercise

Using any of the properties of determinants to minimise the arithmetic, evaluate the determinants of the following matrices: \[A = [value], \qquad B=[value]\]

1. \(\det(A)=\)
2. \(\det(B)=\)
Feedback: 1. Use Property 4 to take out common factors in the rows and the columns of \(A\): \[\det(A)=720\det[value]\] Note that the first entry in column 3 is a \(1\). This is very useful: use Property 5 to simplify the matrix by adding/subtracting an appropriate multiple of column 3 from columns 1 and 2: \[\det(A)=720\det[value]\] The value of this determinant is most easily found by expanding along the top row. Since two of the entries are \(0\) and the third is \(1\), and the sign associated to the \(1\) is \(+\), the value is just the determinant of \([value]\), which is [value]. Therefore the determinant of \(A\) is \([value]\cdot720=[value].\) 2. The determinant of \(B\) is \(0\) because row 4 is \(2\) times row 1.

Exercise

Find the cofactors of \(x,y,z\) in the following determinant:

\[\begin{vmatrix} [value] & [value] & [value] \\ [value] & [value] & [value] \\ x & y & z \end{vmatrix}\]

Cofactor of \(x\):

Cofactor of \(y\):

Cofactor of \(z\):
Feedback: To obtain the cofactor of an element of a \(3 \times 3\) matrix we multiply the minor of that element by the corresponding sign from the \(3 \times 3\) array of signs. We are given the determinant: \(\begin{vmatrix} [value] & [value] & [value] \\ [value] & [value] & [value] \\ x & y & z \end{vmatrix}.\) For \(x\), the sign is \(+\), and the minor is \(\det [value]\). Hence the cofactor is \([value].\) For \(y\), the sign is \(-\), and the minor is \(\det [value]\). Hence the cofactor is \([value].\) For \(z\), the sign is \(+\) again, and the minor is \(\det [value]\). Hence the cofactor is \([value].\)

Exercise

Using any of the properties of determinants to minimise the arithmetic, evaluate the following determinant:

\[\Delta=\begin{vmatrix} y+z & x+z & x+y \\ x & y & z \\ 1 & 1 & 1 \end{vmatrix}\]

\(\Delta=\)
Feedback: We are given the determinant: \[ \Delta=\begin{vmatrix} y+z & x+z & x+y \\ x & y & z \\ 1 & 1 & 1 \end{vmatrix}.\] The sum of the first and second rows is the row matrix \(\begin{bmatrix}x+y+z & x+y+z & x+y+z\end{bmatrix}\), which is \(x+y+z\) times the third row. Therefore the determinant is zero, by Properties 4 and 5.

Key Point

Evaluating General Determinants

If \(A\) is an \(n\times n\) square matrix then: \(\displaystyle{\operatorname{det}(A) = \sum_{j=1}^{n}a_{ij}A_{ij}}\).

In words: The determinant of a square matrix is obtained by multiplying each element of row \(i\) by its corresponding cofactor and then adding these products together.

In the case of \(\Delta=\begin{vmatrix} 4 & 1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & 2 \end{vmatrix}\) we have \(a_{11}=4, a_{12}=1, a_{13}=1,\) and \begin{align*} A_{11} & = + \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} = 4 - 3 = 1, \\ A_{12} & = - \begin{vmatrix} 1 & 3 \\ 3 & 2 \end{vmatrix} = - ( 2 - 9 ) = 7, \\ A_{13} & = + \begin{vmatrix} 1 & 2 \\ 3 & 1 \end{vmatrix} = 1 - 6 = - 5.\end{align*}

Hence, \(\Delta=4\times 1 + 1\times 7 + 1\times -5 = 6\).

Alternatively, choosing to expand along the second row: \begin{align*}\Delta & = a_{21} A_{21} + a_{22} A_{22} + a_{23} A_{23} \\ & = 1 \left( - \begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} \right) + 2 \left( \begin{vmatrix} 4 & 1 \\ 3 & 2 \end{vmatrix} \right) + 3 \left( - \begin{vmatrix} 4 & 1 \\ 3 & 1 \end{vmatrix} \right) \\ & = 6\end{align*} as before.

Evaluate the determinants of the following matrices.

1. \(A=[value]\)
2. \(B=[value]\)
3. \(C=[value]\)
4. \(D=[value]\)
5. \(E=[value]\)
6. \(F=[value]\)
7. \(G=[value]\)

1. \(\det{A}=\)
2. \(\det{B}=\)
3. \(\det{C}=\)
4. \(\det{D}=\)
5. \(\det{E}=\)
6. \(\det{F}=\)
7. \(\det{G}=\)
Feedback: \[\det(A) = [value] = [value]\] \[\det(B) = [value] = [value]\] \[\det(C) = [value] = [value]\] \[\det(D) = [value] = [value]\] \[\det(E) = [value] = [value]\] \[\det(F) = [value] = [value]\] \[\det(G) = [value] = [value]\]