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Coordinated Linear Algebra

Anna Davis, Paul Zachlin, Allan Donsig, Levi Heath, Mikkel Munkholm

Section 5.2 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 5.2.1 The Effects of Elementary Row Operations on the Determinant

Recall that there are three elementary row operations:
  1. Switching the order of two rows
  2. Multiplying a row by a non-zero constant
  3. 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 1.1.6.12 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 5.2.1.

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 5.2.1.
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 5.2.2.
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 5.2.3.
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 5.2.4.
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 5.2.5.
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.
The proof of this theorem is relegated to another time. For a sketch of the proof, you can watch this video:
The following lemma is a useful consequence of Item 1 and Item 2 of Theorem 5.2.6.

Proof.

We will prove Item 2. Parts Item 1 and Item 3 are left as exercises.
[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 5.2.6 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 5.2.6 for columns.

Subsection 5.2.2 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 through concrete examples.

Example 5.2.9.

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 5.2.6}}\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 5.2.6}}\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 5.2.6}}\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 5.2.6}}\knowl{./knowl/xref/item-addmultotherrowdet.html}{\text{Item 3}}. \end{align*}
Recall that \(\det{I}=1\) (Corollary 5.1.14). This gives us
\begin{equation*} -\frac{1}{2}\det{A}=1. \end{equation*}
Therefore \(\det{A}=-2\text{.}\)

Example 5.2.10.

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 = \begin{bmatrix} 3\amp 0\amp -9\\10\amp 5\amp 2\\8\amp 4\amp 2 \end{bmatrix} \amp \xrightarrow{(1/3)R_1} \begin{bmatrix} 1\amp 0\amp -3\\10\amp 5\amp 2\\8\amp 4\amp 2 \end{bmatrix} \tag{5.2.1}\\ \amp \amp \xrightarrow{(1/2)R_3} \begin{bmatrix} 1\amp 0\amp -3\\10\amp 5\amp 2\\4\amp 2\amp 1 \end{bmatrix}\tag{5.2.2}\\ \amp \amp \xrightarrow{R_2-10R_1} \begin{bmatrix} 1\amp 0\amp -3\\0\amp 5\amp 32\\0\amp 2\amp 13 \end{bmatrix}\tag{5.2.3}\\ \amp \amp \xrightarrow{(1/5)R_2} \begin{bmatrix} 1\amp 0\amp -3\\0\amp 1\amp 32/5\\0\amp 2\amp 13 \end{bmatrix}\tag{5.2.4}\\ \amp \amp \xrightarrow{R_3-2R_2} \begin{bmatrix} 1\amp 0\amp -3\\0\amp 1\amp 32/5\\0\amp 0\amp 1/5 \end{bmatrix}.\tag{5.2.5} \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 5.1.13).
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 5.2.3 Properties of the Determinant

We begin by summarizing the properties of determinants we introduced in previous sections.
In this section we will prove the following important results:
  1. A square matrix is singular if and only if its determinant is equal to 0.
  2. The determinant of a product is the product of the determinants.
To get us started, we need the following lemma.

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.

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*}
By repeated application of Lemma 5.2.12, we find that
\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 5.2.7). 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 5.2.14.

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 5.2.13 we conclude that \(A\) is not invertible.
We now draw our attention to products and how they behave for determinants.

Proof.

Suppose \(A\) is invertible, then \(A\) can be written as a product of elementary matrices, see also Theorem 3.3.8.
\begin{equation*} A=E_1E_2\ldots E_k. \end{equation*}
Then, by repeated application of Lemma 5.2.12, we get
\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 \(\det{AB}=0=\det{A}\det{B}\text{.}\)
The following theorem is a nice consequence of Theorem 5.2.15. We leave the proof to the reader, see Exercise 5.2.4.15.

Exercises 5.2.4 Exercises

2.

Let \(A\) be an \(n\times n\) matrix. Show that
\begin{equation*} \det{kA}=k^n\det{A}. \end{equation*}

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.
  1. What elementary row operation does this matrix perform?
  2. Compute the determinant of the matrix in two different ways:
  3. By cofactor expansion.
  4. 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.
9.
\begin{equation*} A=\begin{bmatrix}0\amp 0\amp -3\\2\amp 3\amp -1\\0\amp 0\amp 10\end{bmatrix} \end{equation*}
  • \(A\) is singular
  • \(A\) is nonsingular
10.
\begin{equation*} A=\begin{bmatrix}2\amp -3\amp -1\\0\amp 4\amp 5\\0\amp 0\amp -3\\\end{bmatrix} \end{equation*}
  • \(A\) is singular
  • \(A\) is nonsingular

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.
  1. \(\displaystyle \det{AB^{-1}}\)
  2. \(\displaystyle \det{(AB)^{-1}}\)
  3. \(\displaystyle \det{2AB}\)
Answer.
  1. \(\displaystyle \det{AB^{-1}}=6\)
  2. \(\displaystyle \det{(AB)^{-1}}=3/2\)
  3. \(\displaystyle \det{2AB}=8/3\)

14.

Prove or give a counterexample.
\begin{equation*} \det{(A+B)}=\det{(A^T+B^T)} \end{equation*}

16.

Suppose \(A\) is an invertible matrix such that
\begin{equation*} A^{-1}=A^T \end{equation*}
Find \(\det{A}\) if we know that \(\det{A}\lt 0\text{.}\)
Answer.
\(\det{A}=-1\)
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