Section 7.2 Properties of the Determinant
When we first introduced the determinant we motivated its definition for a \(2\times 2\) matrix by the fact that the value of the determinant is zero if and only if the matrix is singular. We will soon be able to generalize this result to larger matrices, and will eventually establish a formula for the inverse of a nonsingular matrix in terms of determinants.
Recall that we can find the inverse of a matrix or establish that the inverse does not exist by using elementary row operations to carry the given matrix to its reduced row-echelon form. In order to start relating determinants to inverses we need to find out what elementary row operations do to the determinant of a matrix.
Subsection 7.2.1 The Effects of Elementary Row Operations on the Determinant
Recall that there are three elementary row operations:
Switching the order of two rows
Multiplying a row by a non-zero constant
Adding a multiple of one row to another
Elementary row operations are used to carry a matrix to its reduced row-echelon form. In
Exercise 2.1.5.11 we established that elementary row operations are reversible.
In other words, if we know what elementary row operations carried \(A\) to \(\mbox{rref}(A)\text{,}\) we can undo each operation with another elementary row operation to carry \(\mbox{rref}(A)\) back to \(A\text{.}\) This will prove useful for computing the determinant. Computing the determinant of \(\mbox{rref}(A)\) is easy. (Why?) If we know what elementary row operations carry \(\mbox{rref}(A)\) back to \(A\text{,}\) and what effect each of these operations has on the determinant of \(\mbox{rref}(A)\text{,}\) we could find the determinant of \(A\text{.}\)
Exploration 7.2.1.
Let
\begin{equation*}
A=\begin{bmatrix}3\amp -1\amp -1\\3\amp 1\amp -2\\-1\amp 4\amp 2\end{bmatrix}.
\end{equation*}
Problem 7.2.1.
Find \(\det{A}\text{.}\)
Answer.
\begin{equation*}
\det{A}=21.
\end{equation*}
Construct matrix \(B\) by switching the first and the third rows of \(A\text{.}\)
\begin{equation*}
B=\begin{bmatrix}-1\amp 4\amp 2\\3\amp 1\amp -2\\3\amp -1\amp -1\end{bmatrix}.
\end{equation*}
Problem 7.2.2.
Find \(\det{B}\text{.}\)
Answer.
\begin{equation*}
\det{B}=-21
\end{equation*}
Next, try switching consecutive rows. Construct matrix \(B'\) by interchanging the second and third rows of \(A\text{.}\)
\begin{equation*}
B'=\begin{bmatrix}3\amp -1\amp -1\\-1\amp 4\amp 2\\3\amp 1\amp -2\end{bmatrix}.
\end{equation*}
Problem 7.2.3.
Find \(\det{B'}\text{.}\)
Answer.
\begin{equation*}
\det{B'}=-21.
\end{equation*}
It appears that switching any two rows of a matrix produces a determinant that is negative of the determinant of the original matrix. Next, construct matrix \(C\) by multiplying the last row of \(A\) by \(k\text{:}\)
\begin{equation*}
C=\begin{bmatrix}3\amp -1\amp -1\\3\amp 1\amp -2\\-k\amp 4k\amp 2k\end{bmatrix}.
\end{equation*}
Problem 7.2.4.
Find \(\det{C}\text{.}\)
Answer.
\begin{equation*}
\det{C}=21k.
\end{equation*}
It turns out that multiplying the first or the second row of \(A\) by \(k\) yields exactly the same result as this. Finally, construct matrix \(D\) by adding twice row 3 to row 1.
\begin{equation*}
D=\begin{bmatrix}1\amp 7\amp 3\\3\amp 1\amp -2\\-1\amp 4\amp 2\end{bmatrix}.
\end{equation*}
Problem 7.2.5.
Find \(\det{D}\text{.}\)
Answer.
\begin{equation*}
\det{D}=21.
\end{equation*}
This result is particularly surprising. Try a few more variations of this example to convince yourself that adding a multiple of one row to another row does not appear to affect the determinant.
The following theorem generalizes our observations.
Theorem 7.2.6.
Let \(A=\begin{bmatrix}a_{ij}\end{bmatrix}\) be an \(n\times n\) matrix.
If \(B\) is obtained from \(A\) by interchanging two different rows, then
\begin{equation*}
\det{B}=-\det{A}.
\end{equation*}
If \(B\) is obtained from \(A\) by multiplying one of the rows of \(A\) by a non-zero constant \(k\text{.}\) Then
\begin{equation*}
\det{B}=k\det{A}.
\end{equation*}
If \(B\) is obtained from \(A\) by adding a multiple of one row of \(A\) to another row, then
\begin{equation*}
\det{B}=\det{A}.
\end{equation*}
The proof of this theorem is relegated to another time. For a sketch of the proof, you can watch this video:
Lemma 7.2.7.
Let \(A\) be an \(n\times n\) matrix.
If \(A\) has a row of zeros, then \(\det{A}=0\text{.}\)
If two rows of \(A\) are the same, then \(\det{A}=0\text{.}\)
If one row of \(A\) is a scalar multiple of another row, then \(\det{A}=0\text{.}\)
Proof.
[Proof of Part
Item 2]: Suppose rows
\(p\) and
\(q\) of
\(A\) are the same. Let
\(B\) be a matrix obtained from
\(A\) by switching
\(p\) and
\(q\text{.}\) By
Theorem 7.2.6 and
Item 1 we know that
\(\det{B}=-\det{A}\text{.}\) But
\(p\) and
\(q\) are the same, so
\(A=B\text{.}\) But then
\(\det{A}=-\det{A}\text{.}\) We conclude that
\(\det{A}=0\text{.}\)
Because
\(\det{A}=\det{A^T}\text{,}\) we have the following counterpart of
Theorem 7.2.6 for columns.
Theorem 7.2.8. Elementary Column Operations and the Determinant.
Let \(A\) be an \(n\times n\) matrix.
If \(B\) is obtained from \(A\) by interchanging two different columns, then
\begin{equation*}
\det{B}=-\det{A}.
\end{equation*}
If \(B\) is obtained from \(A\) by multiplying one of the columns of \(A\) by a non-zero constant \(k\text{.}\) Then
\begin{equation*}
\det{B}=k\det{A}.
\end{equation*}
If \(B\) is obtained from \(A\) by adding a multiple of one column of \(A\) to another column, then
\begin{equation*}
\det{B}=\det{A}.
\end{equation*}
Subsection 7.2.2 Computing the Determinant Using Elementary Row Operations
What we discovered about the effects of elementary row operations on the determinant will allow us to compute determinants without using the cumbersome process of cofactor expansion.
We discover these tricks throw concrete examples.
Example 7.2.9.
Suppose that a \(6\times 6\) matrix \(A\) is carried to the identity matrix by a sequence of elementary row operations listed below. Find \(\det{A}\text{.}\)
\begin{equation*}
A\xrightarrow{R_2-2R_4}A_1\xrightarrow{R_1\leftrightarrow R_3}A_2\xrightarrow{\frac{1}{2}R_6}A_3\xrightarrow{R_5+R_6}I.
\end{equation*}
Answer.
Let’s take a look at what happens to the determinant of \(A\) one step at a time.
\begin{align*}
\det{A_1} \amp =\det{A}\quad \quad \amp \text{by} \knowl{./knowl/xref/th-elemrowopsanddet.html}{\text{Theorem 7.2.6}}\knowl{./knowl/xref/item-addmultotherrowdet.html}{\text{Item 3}}, \\
\det{A_2} \amp =-\det{A_1}=-\det{A}\quad\quad \amp \text{by} \knowl{./knowl/xref/th-elemrowopsanddet.html}{\text{Theorem 7.2.6}}\knowl{./knowl/xref/item-rowswapanddet.html}{\text{Item 1}}, \\
\det{A_3} \amp =\frac{1}{2}\det{A_2}=-\frac{1}{2}\det{A}\quad\quad \amp \text{by} \knowl{./knowl/xref/th-elemrowopsanddet.html}{\text{Theorem 7.2.6}}\knowl{./knowl/xref/item-rowconstantmultanddet.html}{\text{Item 2}}, \\
\det{I} \amp =\det{A_3}=-\frac{1}{2}\det{A}\quad\quad \amp \text{by} \knowl{./knowl/xref/th-elemrowopsanddet.html}{\text{Theorem 7.2.6}}\knowl{./knowl/xref/item-addmultotherrowdet.html}{\text{Item 3}}.
\end{align*}
\begin{equation*}
-\frac{1}{2}\det{A}=1.
\end{equation*}
Therefore \(\det{A}=-2\text{.}\)
Example 7.2.10.
Let
\begin{equation*}
A=\begin{bmatrix}3\amp 0\amp -9\\10\amp 5\amp 2\\8\amp 4\amp 2 \end{bmatrix}
\end{equation*}
Find \(\det{A}\) by applying elementary row operations to reduce \(A\) to its row-echelon form.
Answer.
\begin{align}
A \amp = \left[\begin{array}{ccc}
3\amp 0\amp -9\\10\amp 5\amp 2\\8\amp 4\amp 2
\end{array}{ccc} \right]
\amp \xrightarrow{(1/3)R_1}
\left[ \begin{array}{ccc}
1\amp 0\amp -3\\10\amp 5\amp 2\\8\amp 4\amp 2
\end{array}{ccc} \right ] \tag{7.2.1}\\
\amp \amp \xrightarrow{(1/2)R_3}
\left[ \begin{array}{ccc}
1\amp 0\amp -3\\10\amp 5\amp 2\\4\amp 2\amp 1
\end{array}{ccc} \right]\tag{7.2.2}\\
\amp \amp \xrightarrow
\left [ \begin{array}{ccc}
1\amp 0\amp -3\\0\amp 5\amp 32\\0\amp 2\amp 13
\end{array}{ccc} \right ]\tag{7.2.3}\\
\amp \amp \xrightarrow{(1/5)R_2}
\left [ \begin{array}{ccc}
1\amp 0\amp -3\\0\amp 1\amp 32/5\\0\amp 2\amp 13
\end{array}{ccc} \right ]\tag{7.2.4}\\
\amp \amp \xrightarrow{R_3-2R_2}
\left [ \begin{array}{ccc}
1\amp 0\amp -3\\0\amp 1\amp 32/5\\0\amp 0\amp 1/5
\end{array}{ccc} \right ].\tag{7.2.5}
\end{align}
We stop when we get to a row-echelon form of
\(A\) because we can see that its determinant is
\(\frac{1}{5}\) (see
Theorem 7.1.13).
The following table summarizes the effect of each elementary row operation on the determinant.
\begin{equation*}
\begin{array}{|c|c|}
\hline \text{Matrix}\amp \text{Determinant}\\ \hline A\amp \det{A}\\ \hline (1) \amp \frac{1}{3}\det{A}\\
\hline (2) \amp \frac{1}{2}\cdot\frac{1}{3}\det{A}\\
\hline (3) \amp \frac{1}{2}\cdot\frac{1}{3}\det{A}\\
\hline (4)\amp \frac{1}{5}\cdot\frac{1}{2}\cdot\frac{1}{3}\det{A}\\
\hline (5) \amp \frac{1}{5}\cdot\frac{1}{2}\cdot\frac{1}{3}\det{A} \\ \hline
\end{array}
\end{equation*}
Since the determinant of the row-echelon form of \(A\) in step 5 is \(\frac{1}{5}\text{,}\) we have
\begin{equation*}
\frac{1}{5}=\frac{1}{5}\cdot\frac{1}{2}\cdot\frac{1}{3}\det{A}.
\end{equation*}
Therefore
\begin{equation*}
\det{A}=6.
\end{equation*}
You should verify this result by direct computation using cofactors.
Subsection 7.2.3 Properties of the Determinant
We begin by summarizing the properties of determinants we introduced in previous sections.
Fact 7.2.11.
The determinant of a triangular matrix is equal to the product of the main diagonal entries. (
Theorem 7.1.13)
The determinant of the transpose is equal to the determinant of the matrix. (
Theorem 7.1.12)
If a matrix contains a row of zeros, then its determinant is equal to 0. (
Lemma 7.2.7)
If two rows of a matrix are the same, then the determinant of the matrix is equal to 0. (
Lemma 7.2.7)
If one row of a matrix is a scalar multiple of another row, then the determinant of the matrix is equal to 0. (
Lemma 7.2.7)
If \(B\) is obtained from \(A\) by interchanging two different rows, then
\begin{equation*}
\det{B}=-\det{A}.
\end{equation*}
If \(B\) is obtained from \(A\) by multiplying one of the rows of \(A\) by a non-zero constant \(k\text{.}\) Then
\begin{equation*}
\det{B}=k\det{A}.
\end{equation*}
If \(B\) is obtained from \(A\) by adding a multiple of one row of \(A\) to another row, then
\begin{equation*}
\det{B}=\det{A}.
\end{equation*}
In this section we will prove the following important results:
A square matrix is singular if and only if its determinant is equal to 0.
The determinant of a product is the product of the determinants.
To get us started, we need the following lemma.
Lemma 7.2.12.
Let \(A\) be a square matrix, and let \(E\) be an elementary matrix, then
\begin{equation*}
\det{EA}=\det{E}\det{A}.
\end{equation*}
Proof.
Recall that if \(E\) is obtained from \(I\) using an elementary row operation, then the same elementary row operation carries \(A\) to \(EA\text{.}\) There are three types of elementary row operations and three types of elementary matrices, so we will have to consider three cases.
Case 1. Suppose \(E\) is obtained from \(I\) by interchanging two rows, then
\begin{equation*}
\det{E}=-1\quad\text{and}\quad \det{EA}=-\det{A},
\end{equation*}
so
\begin{equation*}
\det{EA}=\det{E}\det{A}.
\end{equation*}
Case 2. Suppose \(E\) is obtained from \(I\) by multiplying one of the rows of \(I\) by a non-zero constant \(k\text{,}\) then
\begin{equation*}
\det{E}=k\quad\text{and}\quad \det{EA}=k\det{A}
\end{equation*}
so
\begin{equation*}
\det{EA}=\det{E}\det{A}.
\end{equation*}
Case 3. Suppose \(E\) is obtained from \(I\) by adding a scalar multiple of one row to another row, then
\begin{equation*}
\det{E}=1\quad\text{and}\quad \det{EA}=\det{A},
\end{equation*}
so
\begin{equation*}
\det{EA}=\det{E}\det{A}.
\end{equation*}
Recall that we first introduced determinants in the context of invertibility of \(2\times 2\) matrices. Specifically, we found that
\begin{equation*}
A=\begin{bmatrix}a\amp b\\c\amp d\end{bmatrix}
\end{equation*}
is invertible if and only if \(\det{A}\neq 0\) (a logically equivalent statement is: \(A\) is singular if and only if \(\det{A}=0\)). We are now in the position to prove this result for all square matrices.
Theorem 7.2.13.
A square matrix \(A\) is singular if and only if \(\det{A}=0\text{.}\)
Proof.
Let \(A\) be a square matrix. To determine whether \(A\) is singular we need to find \(\mbox{rref}(A)\text{.}\) In the section on elementary matrices, we found that there exist elementary matrices \(E_1,\ldots ,E_k\) such that
\begin{equation*}
E_k\ldots E_2E_1A=\mbox{rref}(A).
\end{equation*}
Therefore,
\begin{equation*}
\det{(E_k\ldots E_2E_1A)}=\det{\big(\mbox{rref}(A)\big)}.
\end{equation*}
\begin{equation*}
\det{E_k}\ldots \det{E_2}\det{E_1}\det{A}=\det{\big(\mbox{rref}(A)\big)}.
\end{equation*}
Suppose that
\(A\) is singular, then
\(\mbox{rref}(A)\neq I\text{.}\) But then
\(\mbox{rref}(A)\) contains a row of zeros, and
\(\det{\big(\mbox{rref}(A)\big)}=0\) (see
Lemma 7.2.7). Since determinants of elementary matrices are non-zero, we conclude that
\(\det{A}=0\text{.}\)
Conversely, suppose \(\det{A}=0\text{,}\) then
\begin{equation*}
\det{\big(\mbox{rref}(A)\big)}=\det{E_k}\ldots \det{E_2}\det{E_1}\det{A}=0.
\end{equation*}
It follows that \(\mbox{rref}(A)\neq I\text{,}\) so \(A\) is singular.
Let us see this in action in a concrete case.
Example 7.2.14.
Determine whether \(A\) is an invertible matrix without using elementary row operations.
\begin{equation*}
A=\begin{bmatrix}3\amp 4\amp -8\\1\amp 8\amp -10\\1\amp -2\amp 1\end{bmatrix}.
\end{equation*}
Answer.
Compute the determinant of
\(A\text{.}\) You will find that
\(\det{A}=0\text{.}\) By
Theorem 7.2.13 we conclude that
\(A\) is not invertible.
We now draw our attention to products and how they behave for determinants.
Theorem 7.2.15.
Let \(A\) and \(B\) be square matrices, then
\begin{equation*}
\det{AB}=\det{A}\det{B}
\end{equation*}
Proof.
Suppose
\(A\) is invertible, then
\(A\) can be written as a product of elementary matrices, see also
Theorem 4.5.8.
\begin{equation*}
A=E_1E_2\ldots E_k.
\end{equation*}
\begin{align*}
\det{AB}\amp =\det{(E_1E_2\ldots E_kB)} \\
\amp =\det{E_1}\det{E_2}\ldots \det{E_k}\det{B} \\
\amp =\det{(E_1E_2\ldots E_k)}\det{B} \\
\amp =\det{A}\det{B}.
\end{align*}
Now suppose that \(A\) is not invertible. Then \(AB\) is also not invertible. %Needs proof So, \(\det{A}=0\) and \(\det{AB}=0\text{.}\) Thus \(\det{AB}=0=\det{A}\det{B}\text{.}\)
Theorem 7.2.16.
Let \(A\) be a nonsingular matrix, then
\begin{equation*}
\det{A^{-1}}=\frac{1}{\det{A}}.
\end{equation*}
Exercises 7.2.4 Exercises
1.
2.
Let \(A\) be an \(n\times n\) matrix. Show that
\begin{equation*}
\det{kA}=k^n\det{A}.
\end{equation*}
3.
4.
Prove that if one row of a matrix is a linear combination of two other rows of the matrix, then the determinant of the matrix is 0.
5.
Find \(\det{A}\) and \(\det{B}\) using elementary row operations, where
\begin{equation*}
A=\begin{bmatrix}3\amp 1\amp -3\\12\amp 5\amp -5\\4\amp 2\amp 1\end{bmatrix} \quad \text{and} \quad B=\begin{bmatrix}3\amp 2\amp 2\\2\amp 3\amp 3\\1\amp 1\amp 1\end{bmatrix}.
\end{equation*}
Answer.
\(\det{A}=1 \) and \(\det{B}=0 \) .
Exercise Group.
Each of the following matrices is an elementary matrix.
What elementary row operation does this matrix perform?
Compute the determinant of the matrix in two different ways:
By cofactor expansion.
By thinking about how the given matrix was obtained from the identity matrix.
6.
\begin{equation*}
E_1=\begin{bmatrix}1\amp 0\amp 0\\0\amp 2\amp 0\\0\amp 0\amp 1\end{bmatrix}.
\end{equation*}
Hint.
Answer.
\begin{equation*}
\det{E_1}=2.
\end{equation*}
7.
\begin{equation*}
E_2=\begin{bmatrix}1\amp 0\amp 4\\0\amp 1\amp 0\\0\amp 0\amp 1\end{bmatrix}.
\end{equation*}
Answer.
\begin{equation*}
\det{E_2}=1.
\end{equation*}
8.
\begin{equation*}
E_3=\begin{bmatrix}0\amp 0\amp 1\\0\amp 1\amp 0\\1\amp 0\amp 0\end{bmatrix}.
\end{equation*}
Answer.
\begin{equation*}
\det{E_3}=-1.
\end{equation*}
Exercise Group.
Without doing written computations, determine whether the given matrix is singular.
11.
Show that all matrices of the form
\begin{equation*}
\begin{bmatrix}x\amp x+1\amp x+2\\x+3\amp x+4\amp x+5\\x+6\amp x+7\amp x+8\end{bmatrix}
\end{equation*}
are singular.
12.
Find values of \(x\) for which the given matrix is singular.
\begin{equation*}
\begin{bmatrix}2-x\amp 1\\5\amp 6-x\end{bmatrix}
\end{equation*}
List values of \(x\) in an increasing order.
Answer.
\begin{equation*}
x=1\quad x=7
\end{equation*}
13.
Suppose \(A\) and \(B\) are \(2\times 2\) matrices such that \(\det{A}=2\) and \(\det{B}=\frac{1}{3}\text{.}\) Find each of the following.
\(\displaystyle \det{AB^{-1}}\)
\(\displaystyle \det{(AB)^{-1}}\)
\(\displaystyle \det{2AB}\)
Answer.
\(\displaystyle \det{AB^{-1}}=6\)
\(\displaystyle \det{(AB)^{-1}}=3/2\)
\(\displaystyle \det{2AB}=8/3\)
14.
Prove or give a counterexample.
\begin{equation*}
\det{(A+B)}=\det{(A^T+B^T)}
\end{equation*}
15.
16.
Suppose \(A\) is an invertible matrix such that
\begin{equation*}
A^{-1}=A^T
\end{equation*}
Find \(\det{A}\) if we know that \(\det{A}\lt 0\text{.}\)
