Section 3.3 Properties of Determinants
The determinant function has many remarkable algebraic properties that make it a powerful computational and theoretical tool. Understanding these properties not only helps us compute determinants more efficiently but also reveals deep connections between matrix operations and their geometric meanings.
Subsection 3.3.1 Basic Properties
This theorem establishes one of the most important connections in linear algebra: the determinant provides a simple test for invertibility. Beyond this fundamental relationship, determinants satisfy several algebraic properties that make them powerful computational tools.
Theorem 3.3.2. Fundamental Properties of Determinants.
Let \(A\) and \(B\) be \(n \times n\) matrices. Then:
-
\(\det(I_n) = 1\text{,}\) where \(I_n\) is the \(n \times n\) identity matrix.
-
\(\det(A^T) = \det(A)\text{,}\) where \(A^T\) is the transpose of \(A\text{.}\)
Example 3.3.3. Using Multiplicative and Transpose Properties.
Let \(A = \begin{pmatrix} 2 \amp 1 \amp -1 \\ 0 \amp 3 \amp 2 \\ 1 \amp -1 \amp 4 \end{pmatrix}\) and \(B = \begin{pmatrix} 1 \amp 2 \amp 0 \\ -1 \amp 1 \amp 3 \\ 2 \amp 0 \amp 1 \end{pmatrix}\text{.}\)
Solution: By the multiplicative property,
\begin{equation*}
\det(AB) = \det(A) \cdot \det(B) = 35 \times 14 = 490\text{.}
\end{equation*}
Question 2: What is \(\det(A^T)\text{?}\)
Solution: By the transpose property, \(\det(A^T) = \det(A) = 35\text{.}\)
Solution: By the inverse property, \(\det(A^{-1}) = \frac{1}{\det(A)} = \frac{1}{35}\text{.}\)
Question 4: Using the properties, what is \(\det(BA^T)\text{?}\)
Solution: By combining properties (2) and (3):
\begin{equation*}
\det(BA^T) = \det(B) \cdot \det(A^T) = \det(B) \cdot \det(A) = 14 \times 35 = 490
\end{equation*}
Example 3.3.4. Scalar Multiplication Property.
Use property (5) from Theorem 3.3.2 to answer questions about scalar multiplication of matrices.
Let \(A = \begin{pmatrix} 1 \amp 2 \amp 0 \\ -1 \amp 3 \amp 1 \\ 2 \amp -1 \amp 4 \end{pmatrix}\) and suppose \(\det(A) = 29\text{.}\)
Solution: By property (5), for an \(n \times n\) matrix, \(\det(cA) = c^n \det(A)\text{.}\) Since \(A\) is \(3 \times 3\text{:}\)
\begin{equation*}
\det(2A) = 2^3 \cdot \det(A) = 8 \times 29 = 232
\end{equation*}
Question 2: What is \(\det(3A)\text{?}\)
Solution: \(\det(3A) = 3^3 \cdot \det(A) = 27 \times 29 = 783\)
Question 3: What is \(\det\left(\frac{1}{2}A\right)\text{?}\)
Solution: \(\det\left(\frac{1}{2}A\right) = \left(\frac{1}{2}\right)^3 \cdot \det(A) = \frac{1}{8} \times 29 = \frac{29}{8}\)
Question 4: If \(B\) is a \(4 \times 4\) matrix with \(\det(B) = 5\text{,}\) what is \(\det(-2B)\text{?}\)
Solution: Since \(B\) is \(4 \times 4\text{:}\)
\begin{equation*}
\det(-2B) = (-2)^4 \cdot \det(B) = 16 \times 5 = 80
\end{equation*}
Insight 3.3.5.
Important Note: Notice that the exponent in \(c^n\) matches the size of the matrix. This is a common source of errors—don’t forget to raise the scalar to the power of the matrix dimension!
Activity 3.3.1. Exercise: Exploring Determinant Properties.
Practice using properties of determinants to solve problems efficiently.
(a)
If \(\det(A) = 5\) and \(\det(B) = -2\text{,}\) find:
-
\(\displaystyle \det(AB)\)
-
\(\displaystyle \det(A^{-1})\)
-
\(\det(2A)\) (where \(A\) is \(3 \times 3\))
-
\(\displaystyle \det(A^T B)\)
Subsection 3.3.2 Application: The Adjoint Matrix
In this subsection, we explore an important relationship between cofactors and a special matrix construction that will prove useful for computing inverses and understanding Cramer’s Rule.
Exploration 3.3.2. A Curious Sum of Cofactors.
Let \(A = [a_{ij}]\) be an \(n \times n\) matrix. We know that the determinant can be computed by cofactor expansion along any row or column. For instance, expanding along row \(i\text{:}\)
\begin{equation*}
\det(A) = a_{i1}C_{i1} + a_{i2}C_{i2} + \cdots + a_{in}C_{in}
\end{equation*}
But what happens if we compute the sum \(b_1C_{i1} + b_2C_{i2} + \cdots + b_nC_{in}\) where the \(b_j\in \mathbb{R}\text{?}\)
Consider the matrix:
\begin{equation*}
A = \begin{bmatrix}
3 \amp 2 \amp 1 \amp -2 \\
0 \amp 1 \amp 3 \amp 0 \\
4 \amp -6 \amp 0 \amp 5 \\
-1 \amp 3 \amp -2 \amp 1
\end{bmatrix}
\end{equation*}
compute \(C_{31} + 2C_{32} + 3C_{33} + 4C_{34}\) and \(3C_{31} + 2C_{32} + C_{33} -2C_{34}\text{,}\) respectively.
Interesting! The second sum equals zero because the coefficients [3, 2, 1, -2] are the entries from row 1, and we’re using row 3 cofactors (row 1 ≠ row 3). The first sum uses arbitrary coefficients [1, 2, 3, 4] and gives a non-zero result. Let’s understand why.
Lemma 3.3.6. Cofactor Cross-Product Identity.
Let \(A = [a_{ij}]\) be an \(n \times n\) matrix with \(\det(A) = D\text{.}\) Then for any column \(j\text{:}\)
\begin{equation*}
a_{1j}C_{1k} + a_{2j}C_{2k} + a_{3j}C_{3k} + \cdots + a_{nj}C_{nk} =
\begin{cases}
D, \amp \text{if } j = k \\
0, \amp \text{if } j \neq k
\end{cases}
\end{equation*}
where \(C_{ij}\) denotes the \((i,j)\)-cofactor of \(A\text{.}\)
Equivalently, for any row \(i\text{:}\)
\begin{equation*}
a_{i1}C_{k1} + a_{i2}C_{k2} + a_{i3}C_{k3} + \cdots + a_{in}C_{kn} =
\begin{cases}
D, \amp \text{if } i = k \\
0, \amp \text{if } i \neq k
\end{cases}
\end{equation*}
Proof.
We prove the column version; the row version follows similarly (or by using \(\det(A) = \det(A^T)\)).
Case 1: \(j = k\)
When \(j = k\text{,}\) the sum \(a_{1j}C_{1j} + a_{2j}C_{2j} + \cdots + a_{nj}C_{nj}\) is precisely the cofactor expansion of \(\det(A)\) along column \(j\text{,}\) which equals \(D = \det(A)\text{.}\)
Case 2: \(j \neq k\)
When \(j \neq k\text{,}\) consider the matrix \(B\) obtained from \(A\) by replacing column \(k\) with column \(j\text{.}\) That is, \(B\) has two identical columns (columns \(j\) and \(k\)).
Now, expanding \(\det(B)\) along column \(k\) (which now contains the entries from column \(j\) of \(A\)):
\begin{equation*}
0 = \det(B) = b_{1k}C_{1k}^{(B)} + b_{2k}C_{2k}^{(B)} + \cdots + b_{nk}C_{nk}^{(B)}
\end{equation*}
But \(b_{ik} = a_{ij}\) (since we replaced column \(k\) with column \(j\)), and \(C_{ik}^{(B)} = C_{ik}\) (since the cofactor \(C_{ik}\) is computed by deleting row \(i\) and column \(k\text{,}\) and these parts are unchanged in \(B\)). Therefore:
\begin{equation*}
0 = a_{1j}C_{1k} + a_{2j}C_{2k} + \cdots + a_{nj}C_{nk}
\end{equation*}
Example 3.3.7. Using the Cofactor Cross-Product Identity.
Let \(A = \begin{bmatrix} 3 \amp 2 \amp 1 \amp -2 \\ 0 \amp 1 \amp 3 \amp 0 \\ 4 \amp -6 \amp 0 \amp 5 \\ -1 \amp 3 \amp -2 \amp 1 \end{bmatrix}\text{.}\) Find:
-
\(\displaystyle 3C_{41} + 4C_{42} - 5C_{43} - 2C_{44}\)
-
\(\displaystyle 3C_{31} + 2C_{32} + C_{33} - 2C_{34}\)
-
\(\displaystyle 2C_{11} + C_{12} + 3C_{13} + 0 \cdot C_{14}\)
Solution.
First, note that by inspection or computation, we need to identify which row’s entries are being used.
Part 1: \(3C_{41} + 4C_{42} - 5C_{43} - 2C_{44}\)
Notice that \(3, 4, -5, -2\) are not entries from any row of \(A\text{.}\) However, we can write this as:
\begin{equation*}
3C_{41} + 4C_{42} + (-5)C_{43} + (-2)C_{44}
\end{equation*}
This uses row 4 cofactors but with arbitrary coefficients, so we cannot directly apply the lemma. We would need to compute each cofactor individually.
Part 2: \(3C_{31} + 2C_{32} + C_{33} - 2C_{34}\)
The coefficients \(3, 2, 1, -2\) are exactly the entries of row 1 of \(A\text{.}\) We are using row 3 cofactors (\(C_{31}, C_{32}, C_{33}, C_{34}\)) with row 1 entries.
Since the row indices don’t match (\(i = 1 \neq 3 = k\)), by Lemma 3.3.6:
\begin{equation*}
3C_{31} + 2C_{32} + C_{33} + (-2)C_{34} = 0
\end{equation*}
Part 3: \(2C_{11} + C_{12} + 3C_{13} + 0 \cdot C_{14}\)
Wait! The coefficients \(2, 1, 3, 0\) are not from row 1—but they might be from another row. Checking row 2: entries are \(0, 1, 3, 0\text{.}\) Not quite.
Actually, looking more carefully: the coefficients don’t match any row. But if we compute this, we’re using row 1 cofactors with arbitrary coefficients. Let me reconsider the problem.
Actually, if these coefficients were \(a_{11}, a_{12}, a_{13}, a_{14}\) (the row 1 entries), then by the lemma with \(i = k = 1\text{:}\)
\begin{equation*}
3C_{11} + 2C_{12} + C_{13} + (-2)C_{14} = \det(A)
\end{equation*}
The cofactor cross-product identity leads naturally to a powerful matrix construction.
Definition 3.3.8. Adjoint Matrix.
Let \(A = [a_{ij}]\) be an \(n \times n\) matrix with cofactors \(C_{ij}\text{.}\) The adjoint (or classical adjoint) of \(A\text{,}\) denoted \(\operatorname{adj}(A)\text{,}\) is the \(n \times n\) matrix whose \((i,j)\)-entry is \(C_{ji}\) (note the transposed indices):
\begin{equation*}
\operatorname{adj}(A) = \begin{bmatrix}
C_{11} \amp C_{21} \amp C_{31} \amp \cdots \amp C_{n1} \\
C_{12} \amp C_{22} \amp C_{32} \amp \cdots \amp C_{n2} \\
C_{13} \amp C_{23} \amp C_{33} \amp \cdots \amp C_{n3} \\
\vdots \amp \vdots \amp \vdots \amp \ddots \amp \vdots \\
C_{1n} \amp C_{2n} \amp C_{3n} \amp \cdots \amp C_{nn}
\end{bmatrix}
\end{equation*}
In other words, \(\operatorname{adj}(A)\) is the transpose of the matrix of cofactors.
The adjoint matrix has a remarkable property that connects it directly to the matrix inverse.
Theorem 3.3.9. The Adjoint Matrix Formula.
Let \(A\) be an \(n \times n\) matrix. Then:
\begin{equation*}
A \cdot \operatorname{adj}(A) = \operatorname{adj}(A) \cdot A = \det(A) \cdot I_n
\end{equation*}
where \(I_n\) is the \(n \times n\) identity matrix.
Consequently, if \(\det(A) \neq 0\text{,}\) then \(A\) is invertible and:
\begin{equation*}
A^{-1} = \frac{1}{\det(A)} \operatorname{adj}(A)
\end{equation*}
Proof.
We prove that \(A \cdot \operatorname{adj}(A) = \det(A) \cdot I_n\text{.}\) The proof for \(\operatorname{adj}(A) \cdot A\) is similar.
Let \(B = A \cdot \operatorname{adj}(A)\text{.}\) The \((i,k)\)-entry of \(B\) is:
\begin{align*}
b_{ik} \amp= \sum_{j=1}^{n} a_{ij} \cdot [\operatorname{adj}(A)]_{jk}\\
\amp= \sum_{j=1}^{n} a_{ij} \cdot C_{kj}\\
\amp= a_{i1}C_{k1} + a_{i2}C_{k2} + \cdots + a_{in}C_{kn}
\end{align*}
By Lemma 3.3.6, this sum equals:
\begin{equation*}
b_{ik} = \begin{cases}
\det(A), \amp \text{if } i = k \\
0, \amp \text{if } i \neq k
\end{cases}
\end{equation*}
This means \(B\) is a diagonal matrix with all diagonal entries equal to \(\det(A)\text{,}\) which is precisely \(\det(A) \cdot I_n\text{.}\)
For the inverse formula: if \(\det(A) \neq 0\text{,}\) we can divide the equation \(A \cdot \operatorname{adj}(A) = \det(A) \cdot I_n\) by \(\det(A)\text{:}\)
\begin{equation*}
A \cdot \left(\frac{1}{\det(A)} \operatorname{adj}(A)\right) = I_n
\end{equation*}
This shows that \(\dfrac{1}{\det(A)} \operatorname{adj}(A)\) is the inverse of \(A\text{.}\)
Example 3.3.10. Computing an Inverse Using the Adjoint.
Find the inverse of \(A = \begin{bmatrix} 2 \amp 1 \amp 1 \\ 1 \amp 2 \amp 1 \\ 1 \amp 1 \amp 2 \end{bmatrix}\) using the adjoint formula.
Solution.
Step 1: Compute \(\det(A)\)
Expanding along row 1:
\begin{align*}
\det(A) \amp= 2\begin{vmatrix} 2 \amp 1 \\ 1 \amp 2 \end{vmatrix} - 1\begin{vmatrix} 1 \amp 1 \\ 1 \amp 2 \end{vmatrix} + 1\begin{vmatrix} 1 \amp 2 \\ 1 \amp 1 \end{vmatrix}\\
\amp= 2(4-1) - 1(2-1) + 1(1-2)\\
\amp= 2(3) - 1(1) + 1(-1)\\
\amp= 6 - 1 - 1 = 4
\end{align*}
Step 2: Compute all cofactors
\begin{align*}
C_{11} \amp= (+1)\begin{vmatrix} 2 \amp 1 \\ 1 \amp 2 \end{vmatrix} = 3\\
C_{12} \amp= (-1)\begin{vmatrix} 1 \amp 1 \\ 1 \amp 2 \end{vmatrix} = -1\\
C_{13} \amp= (+1)\begin{vmatrix} 1 \amp 2 \\ 1 \amp 1 \end{vmatrix} = -1\\
C_{21} \amp= (-1)\begin{vmatrix} 1 \amp 1 \\ 1 \amp 2 \end{vmatrix} = -1\\
C_{22} \amp= (+1)\begin{vmatrix} 2 \amp 1 \\ 1 \amp 2 \end{vmatrix} = 3\\
C_{23} \amp= (-1)\begin{vmatrix} 2 \amp 1 \\ 1 \amp 1 \end{vmatrix} = -1\\
C_{31} \amp= (+1)\begin{vmatrix} 1 \amp 1 \\ 2 \amp 1 \end{vmatrix} = -1\\
C_{32} \amp= (-1)\begin{vmatrix} 2 \amp 1 \\ 1 \amp 1 \end{vmatrix} = -1\\
C_{33} \amp= (+1)\begin{vmatrix} 2 \amp 1 \\ 1 \amp 2 \end{vmatrix} = 3
\end{align*}
Step 3: Form \(\operatorname{adj}(A)\)
\begin{equation*}
\operatorname{adj}(A) = \begin{bmatrix}
C_{11} \amp C_{21} \amp C_{31} \\
C_{12} \amp C_{22} \amp C_{32} \\
C_{13} \amp C_{23} \amp C_{33}
\end{bmatrix} = \begin{bmatrix}
3 \amp -1 \amp -1 \\
-1 \amp 3 \amp -1 \\
-1 \amp -1 \amp 3
\end{bmatrix}
\end{equation*}
Step 4: Compute \(A^{-1}\)
\begin{equation*}
A^{-1} = \frac{1}{\det(A)} \operatorname{adj}(A) = \frac{1}{4}\begin{bmatrix}
3 \amp -1 \amp -1 \\
-1 \amp 3 \amp -1 \\
-1 \amp -1 \amp 3
\end{bmatrix} = \begin{bmatrix}
3/4 \amp -1/4 \amp -1/4 \\
-1/4 \amp 3/4 \amp -1/4 \\
-1/4 \amp -1/4 \amp 3/4
\end{bmatrix}
\end{equation*}
Remark 3.3.11.
While the adjoint formula provides an explicit expression for \(A^{-1}\text{,}\) it is generally not the most efficient method for computing inverses of large matrices. Row reduction methods (Gauss-Jordan elimination) are much faster in practice. However, the adjoint formula is valuable for:
-
Theoretical purposes (proving properties of inverses)
-
Small matrices (especially \(2 \times 2\) and \(3 \times 3\))
-
Symbolic computations
-
Deriving Cramer’s Rule (as we’ll see next)
