Defining the Determinant

Section 4.1 Defining the Determinant

In the previous chapter, we learned an algorithm for finding the inverse of an invertible \(n\times n\) matrix \(A\text{.}\) This algorithm also told us when a matrix is not invertible—that is, when it is singular. The centerpiece of the algorithm was the row reduction of an \(n\times 2n\) matrix.

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In this section, we will learn about the useful and powerful function called the determinant. The determinant by itself will not give us the inverse of a matrix, but it will reveal whether or not a matrix is invertible.

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Subsection 4.1.1 The Definition of the Determinant

In Exercise 1.3.11, we saw that when a matrix \(A = \begin{bmatrix} a \amp b \\ c \amp d \end{bmatrix}\) has the property that \(ad-bc \neq 0\text{,}\) \(A\) is row equivalent to \(I_2\text{.}\) (In that context we were working over \(\rr\text{,}\) but this is true over any field.) The converse is also true, though we haven’t yet established it.

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Connecting this fact with Theorem 3.3.13, when \({ad-bc \neq 0}\text{,}\) we see that the \(2\times 2\) matrix \(A\) is invertible. (We will see that the converse of this statement is also true.)

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This quantity \(ad-bc\) for \(A\) is special—it is called the determinant of \(A\text{,}\) and this relationship between the determinant and invertibility also holds for larger matrices. The goal of this section is to define a number for any \(n\times n\) matrix which functions the same way that \(ad-bc\) does for a \(2\times 2\) matrix.

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Definition 4.1.1.

For an \(m\times n\) matrix \(A\text{,}\) we define \(A_{ij}\) to be the submatrix of \(A\) that results from deleting row \(i\) and column \(j\) from \(A\text{.}\)

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Example 4.1.2.

If \(A\) is the following matrix,

\begin{equation*} A = \begin{bmatrix} 3 \amp -7 \amp -4 \amp -7 \\ 2 \amp 0 \amp -8 \amp -2 \\ 8 \amp 6 \amp 2 \amp 2 \end{bmatrix}\text{,} \end{equation*}

then the submatrices \(A_{13}\) and \(A_{22}\) are

\begin{equation*} A_{13} = \begin{bmatrix} 2 \amp 0 \amp -2 \\ 8 \amp 6 \amp 2 \end{bmatrix}, \hspace{6pt} A_{22} = \begin{bmatrix} 3 \amp -4 \amp -7 \\ 8 \amp 2 \amp 2 \end{bmatrix}\text{.} \end{equation*}

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Perhaps surprisingly, this bit of notation is all we need to define the determinant. We note that this is a recursive definition, which means that the calculation happens in stages by reducing the size of the matrix.

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Definition 4.1.3.

Suppose that \(A \in M_n(\ff)\) with \(n \ge 1\text{,}\) and let the entries of \(A\) be denoted \(a_{ij}\text{.}\) Then the determinant of \(A\text{,}\) denoted \(\det(A)\) or \(|A|\text{,}\) is defined as follows. If \(n=1\text{,}\) then \(\det(A) = a_{11}\text{.}\) If \(n \ge 2\text{,}\) then

\begin{align*} \det(A) \amp = a_{11}\det(A_{11}) - a_{12}\det(A_{12}) + \cdots + (-1)^{1+n}a_{1n}\det(A_{1n})\\ \amp = \sum_{j=1}^n (-1)^{1+j}a_{1j}\det(A_{1j})\text{.} \end{align*}

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To calculate the determinant of an \(n\times n\) matrix, we need to know the determinant of a lot of \((n-1)\times (n-1)\) submatrices—in fact, we need to know the determinant of \(n\) submatrices. The reader will appreciate that when we begin with a small matrix, this is a manageable calculation; when the matrix is larger, carrying out this calculation by hand would be a decidedly less pleasant task.

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Note 4.1.4.

When \(n=2\text{,}\) the formula for the determinant of \(A\) reduces to the familiar expression with which we opened this section. Note that, in this case, \(A_{11} = [a_{22}]\) and \(A_{12} = [a_{21}]\) are both \(1\times 1\) matrices. We have

\begin{align*} \det(A) \amp = a_{11}\det(A_{11}) - a_{12}\det(A_{12}) \\ \amp = a_{11}a_{22} - a_{12}a_{21}\text{.} \end{align*}

This last expression is precisely \(ad-bc\) with different symbols.

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We will put our new definition to use in the following example.

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Example 4.1.5.

Let \(A\) be the following matrix over \(\rr\text{:}\)

\begin{equation*} A = \begin{bmatrix} 2 \amp -4 \amp -1 \\ 6 \amp -6 \amp 0 \\ -5 \amp -3 \amp -5 \end{bmatrix}\text{.} \end{equation*}

We will find \(\det(A)\) using the formula:

\begin{align*} \det(A) \amp = 2 \begin{vmatrix} -6 \amp 0 \\ -3 \amp -5 \end{vmatrix} - (-4) \begin{vmatrix} 6 \amp 0 \\ -5 \amp -5 \end{vmatrix} + (-1) \begin{vmatrix} 6 \amp -6 \\ -5 \amp -3 \end{vmatrix}\\ \amp = 2(30-0) + 4(-30-0) - (-18-30)\\ \amp = 60 - 120 + 48 = -12\text{.} \end{align*}

We see that \(\det(A) = -12\text{.}\)

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Note 4.1.6.

For the purposes of what follows, we will introduce another bit of notation. If \(A = [a_{ij}]\text{,}\) then the \((i,j)\)-cofactor of \(A\text{,}\) denoted by \(C_{ij}\text{,}\) is

\begin{equation*} C_{ij} = (-1)^{i+j}\det(A_{ij})\text{.} \end{equation*}

Our first determinant formula could then be written

\begin{equation*} \det(A) = \sum_{j=1}^n a_{1j}C_{1j}\text{.} \end{equation*}

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Subsection 4.1.2 Additional Ways to Calculate the Determinant

The definition of the determinant uses the first row of the matrix as its “home base” for calculations. For this reason, the original formula is sometimes referred to as cofactor expansion along the first row of \(A\text{.}\) A perhaps surprising result, and one we will not prove here because of its complicated nature, asserts that the determinant can be calculated using expansion along any row, not just the first.

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Theorem 4.1.7.

Suppose that \(A = [a_{ij}]\) and that \(A \in M_n(\ff)\) with \(n \ge 2\text{.}\) Then \(\det(A)\) can be calculated using cofactor expansion across any row of \(A\text{.}\) In other words, for any \(1 \le i \le n\text{,}\) we have

\begin{equation*} \det(A) = \sum_{j=1}^n a_{ij}C_{ij}\text{.} \end{equation*}

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Because we can target any row of a matrix in our calculation of the determinant, rows which contain several zero entries are particularly attractive.

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Example 4.1.8.

We consider the following matrix \(A\text{:}\)

\begin{equation*} A = \begin{bmatrix} 2 \amp 1 \amp -3 \\ 0 \amp 4 \amp 0 \\ -1 \amp -5 \amp -2 \end{bmatrix} \text{.} \end{equation*}

Since we can expand along any row to calculate the determinant, and since the second row contains two zeros, we will expand along the second row. The zeros make it unnecessary to calculate the cofactors \(C_{21}\) and \(C_{23}\text{:}\)

\begin{align*} \det(A) \amp = 0\cdot C_{21} + 4\cdot C_{22} + 0\cdot C_{23}\\ \amp = (-1)^{2+2}(4) \begin{vmatrix} 2 \amp -3 \\ -1 \amp -2 \end{vmatrix}\\ \amp = 4 (-4-3) = -28\text{.} \end{align*}

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In addition to expanding along any row of a matrix to calculate the determinant, we can also use any column. Instead of proving this directly using the definition of the determinant, we will show that a matrix and its transpose have the same determinant. The result about using columns to calculate the determinant will follow.

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Theorem 4.1.9.

Let \(A \in M_n(\ff)\text{.}\) Then \(\det(A) = \det(A^T)\text{.}\)

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Note 4.1.10.

While we have the machinery to prove this result here, the proof we could offer is clever but not exactly intuitive for some students. (See Beezer

⁶

linear.ups.edu

for a good example of such a proof.) Instead, we will take advantage of results in the next section and prove this result in Subsection 4.2.3. The reader should be aware that some results between the statement and proof of Theorem 4.1.9 depend on the truth of this theorem. We have resolved the possible inconsistency for ourselves knowing that we could prove this result now but are choosing to offer a clearer proof later.

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Corollary 4.1.11.

Suppose that \(A = [a_{ij}]\) and that \(A \in M_n(\ff)\) with \(n \ge 2\text{.}\) Then \(\det(A)\) can be calculated using cofactor expansion along any column of \(A\text{.}\) In other words, for any \(1 \le i \le n\text{,}\) we have

\begin{equation*} \det(A) = \sum_{j=1}^n a_{ji}C_{ji}\text{.} \end{equation*}

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Proof.

We note that column \(i\) of \(A\) is the same as row \(i\) of \(A^T\text{.}\) So, calculating \(\det(A)\) using expansion along column \(i\) is the same as calculating \(\det(A^T)\) using expansion along row \(i\text{.}\) But since \(\det(A^T) = \det(A)\text{,}\) this calculation will result in \(\det(A)\text{.}\)

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The following example shows how we can use this flexibility in calculating the determinant.

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Example 4.1.12.

We will calculate the determinant of the following matrix:

\begin{equation*} A = \begin{bmatrix} 3 \amp 1 \amp 0 \amp -1 \\ -2 \amp -1 \amp 2 \amp 4 \\ 1 \amp 3 \amp 0 \amp -3 \\ 2 \amp -2 \amp 0 \amp 0 \end{bmatrix} \text{.} \end{equation*}

Since the third column of \(A\) has three zeros, we will expand along that column:

\begin{equation*} \det(A) = \sum_{i=1}^4 a_{i3}C_{i3} = 2 \cdot C_{23}\text{.} \end{equation*}

Just for this example, we will rename \(A_{23} = B\text{,}\) so \(\det(A) = -2\det(B)\text{.}\) Here is \(B\text{:}\)

\begin{equation*} B = \begin{bmatrix} 3 \amp 1 \amp -1 \\ 1 \amp 3 \amp -3 \\ 2 \amp -2 \amp 0 \end{bmatrix}\text{.} \end{equation*}

We will take advantage of the zero in the \((3,3)\)-entry of \(B\) by expanding along the third row to calculate \(\det(B)\text{:}\)

\begin{align*} \det(B) \amp = (-1)^{3+1} (2) \begin{vmatrix} 1 \amp -1 \\ 3 \amp -3 \end{vmatrix} + (-1)^{3+2}(-2) \begin{vmatrix} 3 \amp -1 \\ 1 \amp -3 \end{vmatrix}\\ \amp = 2 (0) + 2(-8) = -16\text{.} \end{align*}

Since we determined that \(\det(A) = -2\det(B)\text{,}\) this means that \(\det(A) = 32\text{.}\)

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We will close this section with one final result, the proof of which will be Exercise 4.1.4.10. The truth of this proposition also relies on rows/columns with many zeros. By a “triangular” matrix we mean either an upper triangular or a lower triangular matrix.

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Proposition 4.1.13.

Let \(A \in M_n(\ff)\) be a triangular matrix. Then \(\det(A)\) is the product of the entries along the main diagonal of \(A\text{.}\)

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Reading Questions 4.1.3 Reading Questions

1.

Compute the determinant of the matrix \(A\) by cofactor expansion across the first row. Write out all of your calculations.

\begin{equation*} A = \begin{bmatrix} 2 \amp 1 \amp -1 \\ 0 \amp 3 \amp 4 \\ 1 \amp -1 \amp 5 \end{bmatrix}\text{.} \end{equation*}

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2.

Consider the following matrix \(B\text{:}\)

\begin{equation*} B = \begin{bmatrix} 2 \amp 3 \amp 1 \\ 6 \amp -2 \amp -1 \\ 0 \amp 0 \amp 4 \end{bmatrix}\text{.} \end{equation*}

Compute the determinant of \(B\) by cofactor expansion down the third column. Write out all of your calculations.
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There is a better row or column to use for this calculation. Which one is it? Choose that row or column and compute the determinant of \(B\) by cofactor expansion along that row or column. Write out all of your calculations. Why is this easier?
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Exercises 4.1.4 Exercises

1.

For each of the following matrices over \(\rr\text{,}\) find the determinant by hand using cofactor expansion along a row or column of your choice.

\(\displaystyle A = \begin{bmatrix} 3 \amp 3 \amp -0.5 \\ 2 \amp -3 \amp 0 \\ -1 \amp 2 \amp 2 \end{bmatrix}\)
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\(\displaystyle A = \begin{bmatrix} 2 \amp -4 \amp 4 \amp -1 \\ -1 \amp 0 \amp -4 \amp 3 \\ 2 \amp 2 \amp -4 \amp 0 \\ 4 \amp 0 \amp -2 \amp 1 \end{bmatrix}\)
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2.

For each of the following matrices over \(\rr\text{,}\) find the determinant by hand using cofactor expansion along a row or column of your choice.

\(\displaystyle A = \begin{bmatrix} -3 \amp 3 \amp 0 \\ 0 \amp 4 \amp -2 \\ -1 \amp -4 \amp 2 \end{bmatrix}\)
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\(\displaystyle A = \begin{bmatrix} -2 \amp 4 \amp 0 \amp 0 \\ 0.5 \amp -3.5 \amp -2 \amp -4 \\ 0.5 \amp 0 \amp 0 \amp -2 \\ 1 \amp -2 \amp 3 \amp -3 \end{bmatrix}\)
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Answer.

\(\displaystyle 6\)
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\(\displaystyle 66\)
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3.

Consider the following matrix \(A\) with entries in \(\ff_5\text{:}\)

\begin{equation*} A = \begin{bmatrix} 3 \amp 1 \amp 4 \\ 0 \amp 4 \amp 1 \\ 4 \amp 2 \amp 4 \end{bmatrix}\text{.} \end{equation*}

Find \(\det(A)\) by hand using cofactor expansion along a row or column of your choice. (The answer should be a number in \(\ff_5\text{.}\))

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4.

For each of the following, write down the elementary matrix that performs the given elementary row operation on a \(3\times 3\) matrix. Then, calculate the determinant of that elementary matrix.

Multiply row 2 by \(-3\)
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Switch rows 2 and 3
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Replace the third row with the sum of the third row and seven times the first row
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Answer.

The elementary matrix is

\begin{equation*} E = \begin{bmatrix} 1 \amp 0 \amp 0 \\ 0 \amp -3 \amp 0 \\ 0 \amp 0 \amp 1 \end{bmatrix}\text{,} \end{equation*}

and \(\det(E) = -3\text{.}\)

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The elementary matrix is

\begin{equation*} E = \begin{bmatrix} 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \\ 0 \amp 1 \amp 0 \end{bmatrix}\text{,} \end{equation*}

and \(\det(E) = -1\text{.}\)

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The elementary matrix is

\begin{equation*} E = \begin{bmatrix} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 7 \amp 0 \amp 1 \end{bmatrix}\text{,} \end{equation*}

and \(\det(E) = 1\text{.}\)

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5.

Let \(A = \begin{bmatrix} 3 \amp 1 \\ 2 \amp -2 \end{bmatrix}\text{.}\) Write down the matrix \(5A\text{.}\) How are \(\det(A)\) and \(\det(5A)\) related?

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6.

Let \(A = \begin{bmatrix} a \amp b \\ c \amp d \end{bmatrix}\) be a matrix over \(\rr\) and let \(k \in \rr\text{.}\) Find a formula that relates \(\det(kA)\) to \(k\) and \(\det(A)\text{.}\)

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Solution.

We note that \(kA = \begin{bmatrix} ka \amp kb \\ kc \amp kd \end{bmatrix}\text{,}\) and

\begin{equation*} \det(kA) = kakd - kbkc = k^2ad - k^2bc = k^2(ad-bc) = k^2\det(A)\text{.} \end{equation*}

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7.

Let \(A = \begin{bmatrix} 1 \amp -2 \\ 2.5 \amp c \end{bmatrix}\) be a matrix over \(\rr\text{.}\) Find a value of \(c\) so that \({\det(A) = 0}\) or explain why this is not possible.

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8.

Let \(A\) be the following matrix over \(\rr\text{:}\)

\begin{equation*} A = \begin{bmatrix} 2 \amp -2 \amp 1 \\ 3 \amp -2 \amp 0 \\ 2 \amp 3 \amp c \end{bmatrix} \text{.} \end{equation*}

Find a value of \(c\) so that \(\det(A) = 0\) or explain why this is not possible.

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Answer.

We must have \(c = -\frac{13}{2}\) in order to have \(\det(A) = 0\text{.}\)

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9.

Let \(A\) be the following matrix over \(\rr\text{:}\)

\begin{equation*} A = \begin{bmatrix} x-5 \amp 1 \\ 4 \amp x-2 \end{bmatrix}\text{.} \end{equation*}

Find all values of \(x\) such that \(\det(A) = 0\text{.}\)

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Answer.

The values are \(x=1\) and \(x=6\text{.}\)

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Writing Exercises

10.

Prove that the determinant of a triangular matrix is the product of the entries along the main diagonal.

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11.

Prove that if \(A\) is a \(2\times 2\) matrix and if \(B\) is the result of switching the rows in \(A\text{,}\) then \(\det(B) = -\det(A)\text{.}\)

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12.

Suppose that \(A\) is a \(2\times 2\) matrix and that \(B\) is the result of applying the replace row operation to \(A\text{.}\) Prove that \(\det(B) = \det(A)\text{.}\)

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Answer.

We let \(A = \begin{bmatrix} a \amp b \\ c \amp d \end{bmatrix}\text{,}\) and we consider the case of adding \(k\) times row 1 to row 2. We will call the resulting matrix \(B\text{,}\) and it has this appearance:

\begin{equation*} B = \begin{bmatrix} a \amp b \\ ka + c \amp kb + d \end{bmatrix}\text{.} \end{equation*}

We know that \(\det(A) = ad-bc\text{,}\) and we will now calculate \(\det(B)\text{:}\)

\begin{align*} \det(B) \amp = a(kb+d) - b(ka+c)\\ \amp = kab + ad - kab - bc\\ \amp = ad - bc = \det(A)\text{.} \end{align*}

Technically we have only proved one of the two necessary cases to complete this proof.

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