Section 18.2 Determinants
If we attempted to find the inverse of the general \(2\times 2 \) matrix
\begin{equation*}
A= \begin{pmatrix}
a \amp b \\
c \amp d
\end{pmatrix}
\end{equation*}
we would find that, if \(ad-bc\neq 0 \) the inverse is
\begin{equation*}
A^{-1}=\frac{1}{ad-bc} \begin{pmatrix}
d \amp -b \\
-c \amp a
\end{pmatrix},
\end{equation*}
and if \(ad-bc=0 \) then \(A \) does not have an inverse. Thus for a \(2\times 2\) matrix, \(A \text{,}\) calculating the quantity \(ad-bc \) can act as a test for the invertibility of \(A \text{.}\) This quantity is called the determinant of \(A \) and is denoted by
\begin{equation*}
\det(A) \;\text{or}\; \vert A\vert.
\end{equation*}
Example 18.2.1.
Find the determinant of
\begin{equation*}
A=\begin{pmatrix} 2 \amp -1 \\ 3 \amp 1 \end{pmatrix}.
\end{equation*}
Answer.
Solution.
\begin{equation*}
\begin{vmatrix}
2 \amp -1 \\
3 \amp 1
\end{vmatrix}=2\times 1 -(3\times (-1))=5.
\end{equation*}
Note that since the determinant is not zero this matrix is invertible.
\begin{equation*}
A= \begin{pmatrix}
a \amp b \\
c \amp d
\end{pmatrix}
\end{equation*}
will have an inverse when the vectors \((a,c)^{T} \) and \((b,d)^{T} \) are linearly independent. Now we also know (from Chapter 16) that two vectors in the plane are linearly independent if they define a parallelogram with non-zero area. Finally, from Math1110, we know that the area of the parallelogram defined by the vectors \(\bm{u} = (u_{1},u_{2})^{T} \) and \(\bm{v} = (v_{1},v_{2})^{T} \) is
\begin{equation*}
Area=\vert u_{1}v_{2} - u_{2}v_{1}\vert.
\end{equation*}
Thus matrix \(A \) will have an inverse when \(ad-bc\neq 0 \) , i.e. \(\det(A)\neq 0. \)Let's now apply the same geometric argument to the general \(3 \times 3\) matrix
\begin{equation*}
A=\begin{pmatrix}
u_{1} \amp v_{1} \amp w_{1} \\
u_{2} \amp v_{2} \amp w_{2} \\
u_{3} \amp v_{3} \amp w_{3}
\end{pmatrix}=(\bm{u}, \bm{v}, \bm{w}).
\end{equation*}
Three vectors in space are linearly independent if they define a parallelepiped with non-zero volume. Now, the volume of the parallelepiped formed by the vectors
\begin{equation*}
\bm{u} = (u_{1},u_{2},u_{3})^{T} ,\; \bm{v} = (v_{1},v_{2},v_{3})^{T}, \; \text{and} \; \bm{w} =(w_{1},w_{2},w_{3})^{T}
\end{equation*}
is
\begin{equation*}
Volume=\vert \bm{u} \cdot \bm{v} \times \bm{w} \vert,
\end{equation*}
i.e. the absolute values of the scalar triple product of the vectors, (again see Math1110). Thus the matrix \(A \) will have an inverse when \(\bm{u} \cdot \bm{v} \times \bm{w} \neq 0. \) Hence for a \(3\times 3 \) matrix its determinant is defined as
\begin{equation}
\det(A)=\vert \bm{u} \cdot \bm{v} \times \bm{w} \vert\label{Eq-det_with_three_vectors}\tag{18.2.1}
\end{equation}
Example 18.2.2.
Find the determinant of
\begin{equation*}
A= \begin{pmatrix}
2 \amp 4 \amp 6 \\
3 \amp 2 \amp 1 \\
1 \amp 1 \amp 2
\end{pmatrix}.
\end{equation*}
Answer.
Solution.Let
\(\bm{u}=(2,3,1)^{T},\; \bm{v}=(4,2,1)^{T},\; \bm{w}=(6,1,2)^{T}. \) Then
\begin{equation*}
\bm{v} \times \bm{w}=(4,3,1) ^{T} \times (6,1,2)^{T}=(3,-2,-8)^{T}
\end{equation*}
and hence
\begin{equation*}
\det(A) = (2,3,1) ^{T} \cdot (3,-2,-8)^{T}=(3,-2,-8)^{T}=-8.
\end{equation*}
Theorem 18.2.3.
For the \(2\times 2\) matrix \(\begin{pmatrix}
a \amp b \\
c \amp d
\end{pmatrix}\text{,}\) \(\det \begin{pmatrix} a \amp b \\ c \amp d \end{pmatrix}=ad-bc \text{.}\)
-
For the \(n\times n\) matrix \(A=\begin{pmatrix}
a_{ij}
\end{pmatrix}\) we define the minor, \(M_{ij} \) as the determinant of the \((n-1)\times (n-1) \) matrix obtained by deleting the \(i\)th row and the \(j\)th column of \(A \text{.}\) Then
\begin{equation*}
\det(A)=\sum_{j=1}^{n}(-1)^{i+j}a_{ij}M_{ij}\qquad \text{for any}\qquad i=1,2,3,\ldots,n
\end{equation*}
or
\begin{equation*}
\det(A)=\sum_{i=1}^{n}(-1)^{i+j}a_{ij}M_{ij}\qquad \text{for any}\qquad j=1,2,3, \ldots, n.
\end{equation*}
Example 18.2.4. (Example 18.2.2 revisited).
Find the determinant of
\begin{equation*}
A=\begin{pmatrix}
2 \amp 4 \amp 6 \\
3 \amp 2 \amp 1 \\
1 \amp 1 \amp 2
\end{pmatrix}.
\end{equation*}
Answer.
Solution.Using the first of the formulas given above with
\(i=1: \)
\begin{align*}
\det(A) = \amp (-1)^{1+1} 2 \begin{vmatrix}
2 \amp 1 \\
1 \amp 2
\end{vmatrix} + (-1)^{1+2} 4 \begin{vmatrix}
3 \amp 1 \\
1 \amp 2
\end{vmatrix}+ (-1)^{1+3} 6 \begin{vmatrix}
3 \amp 2 \\
1 \amp 1
\end{vmatrix}\\\\
=\amp 2(4-1)-4(6-1)+6(3-2)\\ \\
=\amp -8.
\end{align*}
Example 18.2.5.
Find the determinant of
\begin{equation*}
A=\begin{pmatrix}
2 \amp 4 \amp 6 \\
0 \amp 2 \amp 1 \\
0 \amp 0 \amp -4
\end{pmatrix}.
\end{equation*}
Answer.
Solution.Using the second of the formulas given above with
\(j=1: \)
\begin{align*}
\det(A) = \amp (-1)^{1+1} 2 \begin{vmatrix}
3 \amp 1 \\
0 \amp -4
\end{vmatrix} + (-1)^{2+1} 0 \begin{vmatrix}
2 \amp 6 \\
0 \amp -4
\end{vmatrix}+ (-1)^{3+1} 0 \begin{vmatrix}
4 \amp 6 \\
3 \amp 1
\end{vmatrix}\\\\
=\amp 2(-12-0)-0+0\\ \\
=\amp -24.
\end{align*}
Notice that for a matrix that is upper triangular the determinant is just the product of the entries on the main diagonal.
Theorem 18.2.6.
To calculate the determinant of the \(n \times n \) matrix \(A \text{:}\)
-
If matrix \(B \) is obtained from matrix \(A \) by interchanging \(2 \) rows then
\begin{equation*}
\det(B)=-\det(A).
\end{equation*}
-
If matrix \(B \) is obtained from matrix \(A \) by multiplying a row of \(A \) by a scalar \(k \) then
\begin{equation*}
\det(B)=k\det(A).
\end{equation*}
-
If matrix \(B \) is obtained from matrix \(A \) by adding a multiple of one row of \(A \) to another then
\begin{equation*}
\det(B)=\det(A).
\end{equation*}
Example 18.2.7. (Example 18.2.2 revisited).
Find the determinant of
\begin{equation*}
A=\begin{pmatrix}
2 \amp 4 \amp 6 \\
3 \amp 2 \amp 1 \\
1 \amp 1 \amp 2
\end{pmatrix}.
\end{equation*}
Answer.
Solution.Using the row reduction method, first reduce
\(A \) to an equivalent upper triangular matrix.
\begin{align*}
\begin{pmatrix}
2 \amp 4 \amp 6 \\
3 \amp 2 \amp 1 \\
1 \amp 1 \amp 2
\end{pmatrix}
\amp \sim
\begin{pmatrix}
2 \amp 4 \amp 6 \\
0 \amp -4 \amp -8 \\
0 \amp -1 \amp -1
\end{pmatrix} \;\;\;
\begin{matrix}
\amp \\
R'_{2} \amp = R_{2}-3\frac{R_{1}}{2} \\
R'_{3} \amp = R_{3}-\frac{R_{1}}{2}
\end{matrix}\\
\amp \sim
\begin{pmatrix}
2 \amp 4 \amp 6 \\
0 \amp -4 \amp -8 \\
0 \amp 0 \amp 1
\end{pmatrix} \;\;\;
\begin{matrix}
\amp \\
\amp \\
R'_{3} \amp = R_{3}-\frac{R_{2}}{4}
\end{matrix}
\end{align*}
Since the only elementary row operation used here was that of adding a multiple of one row to another the determinant of the reduced matrix will be the same as the determinant of
\(A \text{.}\) Thus
\begin{equation*}
\det(A)=2\times (-4)\times 1 = -8.
\end{equation*}
Theorem 18.2.8. Properties of the Determinant.
Let \(A \) and \(B \) be square invertible matrices of order \(n \) and let \(k \) be a real number. Then
\(\displaystyle \det(A^{-1})=\frac{1}{\det(A)}\)
\(\displaystyle \det(A^{T})=\det(A)\)
\(\displaystyle \det(AB)=\det(A) \det(B)\)
\(\displaystyle \det(kA)=k^{n} \det(A)\)
Example 18.2.9.
Calculate the determinant of the following matrices. Which property of determinants does this illustrate?
\begin{equation*}
A=\begin{pmatrix}
-1 \amp 2 \\
3 \amp -4
\end{pmatrix},\;
B=\begin{pmatrix}
-2 \amp 4 \\
6 \amp -8
\end{pmatrix}
\end{equation*}
Solution.Firstly,
\begin{equation*}
\det(A)=(-1)\times (-4)-3\times 2=4-6=-2.
\end{equation*}
Next
\begin{equation*}
\det(B)=(-2)\times (-8)-4\times 6=16-24=-8.
\end{equation*}
Since
\(A \) and
\(B \) are square matrices of order
\(2 \) and
\(B=2A \) the fact that
\(\det(B)=4\det(A) \) illustrates Property
\(4.\) of
Theorem 18.2.8 above.
Theorem 18.2.10.
Consider the non-homogenous system of \(n \) linear equations in \(n \) variables
\begin{equation*}
A \bm{x} = \bm{b},\; \bm{b} \neq 0.
\end{equation*}
The following statements are equivalent:
The system has a unique solution,
\(A \bm{x} =0 \) has only the trivial solution \(\bm{x} =0, \)
The columns of \(A \) are linearly independent,
\(A \) is invertible.
\(\det(A)\neq 0 \text{.}\)
Exercises Example Tasks
1.
Find the determinant of
\begin{equation*}
M=\begin{pmatrix}
1 \amp 1 \amp 2 \\
1 \amp -1 \amp 1 \\
0 \amp 2 \amp 4
\end{pmatrix}
\end{equation*}
Using the minor formula.
Using row reduction to an upper triangular matrix.