- 3.1.1 Determinant
- 3.1.2 Domains and Codomains
- 3.1.3 Subsets, Regions, and Images
- 3.1.4 A Note About Computing Determinants
- 3.1.5 The Determinant of the Identity
- 3.1.6 Alternating
- 3.1.7 Multilinear
- Property 3 - Multilinear
- Example 4 - Scaling Volume
- Figure 5 - Scaled Image of Unit Square
- Example 5 - Factors
- Exercise 4 - Compute the Determinant
- Figure 6 - Exercise 4 Image of Unit Square
- Example 6 - Terms
- Figure 7 - Example 6 Image of Sum
- Exercise 5 - Compute the Determinant
- Figure 8 - Exercise 5 Image of Unit Square
- 3.1.8 2x2 Determinant
- 3.1.9 Row Properties
- Example 7 - Duplicate Rows
- Theorem 2 - Property 4
- Proof - Property 4
- Example 8 - Type III Row Operation
- Theorem 3 - Property 5
- Proof - Property 5
- Exercise 6 - Compute the Determinant
- Example 9 - Row of Zeros
- Theorem 4 - Property 6
- Proof - Property 6
- Corollary 5 - Linearly Dependent Row
- Proof - Corollary 5
- 3.1.10 Direct Properties
- 3.1.11 Function Properties
- 3.1.12 Properties of Determinant (the List)
- 3.1.13 Exercises
- Exercise 11 - Scaling and Orientation
- Exercise 12 - Determinant of a Scalar Multiple
- Exercise 13 - Effect of Row Operations
- Exercise 14 - Determinant Algebra
- Exercise 15 - Recognizing a Singular Matrix
- Exercise 16 - Solve for a Singular Matrix
- Exercise 17 - A Common Misconception
- Exercise 18 - A 6×6 Determinant by Gaussian Elimination
- Exercise 19 - Lemma: Odd-Order Skew-Symmetric Matrices are Singular
- Exercise 20 - Lemma: Determinant of a Power
- copyleft
Video Lecture 1: The Determinant.¶
The Determinant
Definition¶
Determinant
The determinant of an $n\times n$ matrix is the scaling factor of the change in volume(area) in $n$-dimensional space taking orientation into account such that:
- $\det(I_n) = 1$, where $I_n$ is the $n\times n$ identity matrix.
- The determinant is alternating.
The exchange of two rows multiplies the determinant by $-1$. This refers to orientation. - The determinant is multilinear.
The determinant is best understood in terms of this geometrical notion and its properties.
Video Lecture 2: Properties of Determinant.¶
Properties of DeterminantWe denote determinant of $n\times n$ matrix $A$ by
$$ \det(A) = |A| $$
Notice that we use the absolute value bars even though the determinant of a matrix may have a negative value indicating that the orientation of space has been changed by the linear transformation encoded by the matrix.
We apologize for the length of this section. Fully explaining all ten properties of the determinant, together with their corollaries, simply takes a lot of text. The good news is that the two video lectures above cover everything that follows.
Everyone should read A Note About Computing Determinants. Sacrificing up to 9/10 points and wasting time using the Laplace Expansion on a test will ruin your day.
If you have watched both lectures and feel comfortable with the material, you may skip straight to the final section, 10 Properties of Determinant, which gathers every property in one place for quick reference. If, on the other hand, a particular idea in the videos left you with questions, work through the corresponding part of this page, where each property is developed, illustrated, and proved in detail.
Example 1 - Geometry of a Determinant¶
Let $A$ be the matrix,
$$\begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix}.$$
| Determinant | Graph |
|---|---|
| $$ \det\left(A\right) = \left|\,A\,\right| = \det\left(\begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix}\right) = \begin{vmatrix} 3 & 0 \\ 0 & 2 \end{vmatrix} = 6. $$ |
 |
Under a linear transformation, the images of parallel lines remain parallel and evenly spaced. Hence one need only plot the images of the four corners of the unit square. The line segments connecting the corners of the unit square must be mapped to the line segments connecting the images of each corner. This yields the image of the unit square, $A(D)$.
$$ A(D) = \left\{\, \mathbf{y}\in\mathbb{R}^2\,:\,\mathbf{y}=A\mathbf{x},\ \text{for some } x\in D\,\right\} $$
This indicates that the area of any region $S\subset\mathbb{R}^2$ will be stretched to a region six times the original area of $S$ and orientation will be preserved.
Let us take a minute to discuss some vocabulary for matrices and the linear transformations they encode. Whenever one writes the definition of a function, one is expected to include
Definition¶
Function
A function is a relation between two sets.
- A set of valid inputs for the function called its domain
- A set of valid outputs for the function called its codomain
- A rule or set of rules that defines exactly one output in the codomain for every input in the domain.
Functions are usually described by an equation that represents the rule that maps each element of the domain to a unique element of the codomain.
$$ y = f(x) $$
The output $y$ for an input $x$ is called the image of $x$ under $f$. Often the domain and codomain are not explicitly defined. When presented with an equation, one infers the domain to the largest set of elements $x$ for which the rule $f(x)$ makes sense.
Since not all elements of the codomain need to have an element of the domain mapped to them, there is latitude in the definition of the codomain. Using the equation, one usually infers the codomain to be the set of all possible objects of the type defined by the equation.
Rule three declares that every element of the domain must be mapped to some element of the codomain, and each element of the domain must be mapped to only one element of the codomain. When graphing functions with one real input and one real output on the Cartesian plane, this is often referred to as the vertical line test.
We are not always interested in the entire domain of a function; we are often interested in the what happens to some subset of the domain to which the function is applied. Using the verb applied here connotes the idea of motion in space. However it can also imply that a process has occurred that somehow results in the appropriate output for each input.
By the time one studies differential or integral calculus much of these details are simply implied by an equation or expression. For example when one reads about the real function
$$
y = \sqrt{9 - x^2},
$$
the reader is expected to determine the definition of the function just from the equation.
The domain of the implied function is the interval $[-3,3]$. Square brackets indicate that the endpoints $\pm3$ are included in the interval; that is the interval is a closed interval.
The codomain of the implied function is the real line $\mathbb{R}$, or the interval of the entire real line $(-\infty,\infty)$.
The rule for a function usually requires a variable that represents each element of the equation defining the rule.
The independent variable $x$ represents an input from the domain $[-3,3]$ so $-3\le x\le 3$
The dependent variable $y$ represents an output in the codomain and the value of the output can be computed using the equation.
$$ y = f(x) = \sqrt{9 - x^2} $$
When the author and reader agree, a function doesn't really require an explicit name like $f$. In many cases the output variable can be used as a substitute for the name of the function.
$$
y' = \dfrac{-x}{\sqrt{9-x^2}}
$$
Likewise, one typically writes a matrix expecting the reader to infer properties about the linear transformation represented by the matrix. For example
$$
A = \begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix}
$$
This $2\times 2$ matrix represents a linear transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$.
- The domain of the linear transformation is all of the vector space $\mathbb{R}^2$.
- The codomain of the linear transformation is all of the vector space $\mathbb{R}^2$.
- The rule for computing an output vector $y$ from an input vector $x$ is given by
$$ \mathbf{y} = A\mathbf{x} = \begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 3x_1 \\ 2x_2 \end{bmatrix} $$
- In the case of a square $n\times n$ matrix, the domain and codomain are the same vector space $\mathbb{R}^n$. In this case one writes that matrix $A$ represents a linear transformation on $\mathbb{R}^n$.
Exercise 1 - Determine Domain and Codomain¶
Determine the domain and codomain of the linear transformations represented by each matrix:
(a) $\begin{bmatrix}\ \ 3\ &\ \ 0\ &\ \ 2\ & -5\ &\ \ 7\ & -7\ \\ -6\ &\ \ 9\ & -5\ &\ \ 0\ &\ \ 9\ & -7\ \\ -7\ & -3\ &\ \ 5\ &\ \ 5\ &\ \ 1\ & -5\ \end{bmatrix}$
(b) $\begin{bmatrix}\ \ 6\ & -1\ &\ \ 5\ \\ -5\ & -3\ &\ \ 5\ \\ \ \ 8\ &\ \ 6\ & -2\ \\ -3\ &\ \ 2\ &\ \ 1\ \\ -6\ &\ \ 1\ & -8\ \\ -5\ &\ \ 8\ & -8\ \\ \ \ 2\ & -4\ &\ \ 1\ \end{bmatrix}$
(c) $\begin{bmatrix} -7\ &\ \ 6\ &\ \ 3\ &\ \ 8\ & -1\ &\ \ 7\ &\ \ 8\ \\ \ \ 1\ & -4\ &\ \ 4\ & -7\ & -7\ & -8\ & -6\ \\ -1\ &\ \ 1\ &\ \ 5\ &\ \ 6\ &\ \ 9\ & -2\ & -4\ \\ -9\ & -6\ & -1\ &\ \ 1\ & -9\ & -5\ & -7\ \\ -3\ &\ \ 2\ & -8\ &\ \ 9\ &\ \ 5\ &\ \ 6\ & -7\ \\ -6\ & -5\ & -5\ & -8\ &\ \ 6\ & -1\ &\ \ 7\ \end{bmatrix}$
View Solution
(a) Domain: $\mathbb{R}^7$, Codomain: $\mathbb{R}^3$
(b) Domain: $\mathbb{R}^3$, Codomain: $\mathbb{R}^7$
(c) Domain: $\mathbb{R}^7$, Codomain: $\mathbb{R}^6$
Definition¶
Set
A set is a collection of things. In this class our sets are usually sets of vectors or scalars.
Definition¶
Subset
A subset of a set $U$ is a collection of any of the elements of $U$. If all of the elements of set $A$ are also in set $U$, then $A$ is a subset of $U$, and this relationship is denoted
$$ A\subset U $$
Some subsets are trivial
$\varnothing$ denotes the empty set and has no elements in it. The empty set $\ \varnothing = \left\{\,\right\}$. For any set $U$, $\ \varnothing\subset U$ no matter what is set $U$ (even if $\ U = \varnothing$).
$U\subset U$; that is every set is a subset of itself because every element of the set $\ U$ is also an element of $\ U$.
Some writers distinguish between strict subset $\subset$ and conditional subset $\subseteq$.
In linear algebra every vector space contains at least one element, the zero vector $\mathbf{0}$. A vector space with only this one element $\left\{\mathbf{0}\right\}$ is called a trivial vector space.
Some subsets are intended to restrict our attention to a specific region of the domain.
$[0,3]\subset[-3,3]$, perhaps we are only interested in the nonnegative inputs.
A region is a set (or subset) with desirable properties. In linear algebra the term region is often used to refer to a subset of a vector space whose graph is an area or volume.
For example the unit square in $\mathbb{R}^2$ is the subset of the plane $\mathbb{R}^2$ that consists of a square with vertices $(0,0)$, $(1,0)$, $(1,1)$, and $(0,1)$.
The unit square is often called a region of the plane or a domain. In this case we want to restrict the domain of our linear transformation to only those vectors that lie in the unit square. A vector lies in the unit square when, in standard position, the tip sits in the unit square. The vector $\mathbf{v}$ has standard coordinates so that
$$ \mathbf{v} = \begin{bmatrix} v_1 \\ v_2 \end{bmatrix},\ 0\le v_1\le 1,\ 0\le v_2\le 1 $$
A set can be described by listing the elements of the set such as:
$$ \left\{\, 0,\ 3,\ 6,\ 9,\ \dots \,\right\} $$
Usually sets are too large to list all of the elements so we describe the set using set builder notation.
Definition¶
Set Builder Notation
Set builder notation is a standardized method of describing sets using mathematical logic. The set description lies between two curly brackets (braces) and consists of three parts:
- a variable declaration, and optionally a formula or domain that specifies the numbers we are considering,
- a vertical bar or colon separates the variable declaration from the predicate, and
- a predicate the elements of the domain must satisfy to be members of the set.
If the variable declaration includes a formula, then the elements of the set are the results of the formula applied the values for which the predicate is true.
If the variable declaration includes membership to a set or domain, then the set consists of those values for which both the membership statement and the predicate are true. If
$$ S = \left\{\,x\in\mathbb{Z}\,:\,x > 0\,\right\},$$
then $Q(x)=x\in\mathbb{Z}$ and $P(x)=(x > 0)$ must be true for $x$ to be a member of set $S$. An equivalent definition of set $S$ takes the form
$$ S = \left\{\,x\,:\,(x\in\mathbb{Z})\ \land\ (x > 0) \,\right\}.$$
It is common to read the vertical bar or colon, "such that", "where", "with the property that" or "for which".
The following are equivalent:
$$
\begin{align*}
U &= \left\{\,\mathbf{v}\in\mathbb{R}^2\,|\,0\le v_1\le 1,\ 0\le v_2\le 1\,\right\} \\
\\
&= \left\{\,\mathbf{v}\in\mathbb{R}^2\,:\,0\le v_1\le 1,\ 0\le v_2\le 1\,\right\} \\
\\
&= \left\{\, \mathbf{v}\text{ such that }\mathbf{v}\in\mathbb{R}^2\text{ and }0\le v_1\le1\text{ and }0\le v_2\le1 \,\right\}
\end{align*}
$$
When we apply the linear transformation $A$ to the unit square we obtain a region in the codomain $\mathbb{R}^2$.
Definition¶
Image
The image of a subset $S\subset D$ of the domain of a function is the set of all images of each element of $S$,
$$f(S) = \left\{\, f(x)\,:\,x\in S \,\right\}$$
If $A$ is an $m\times n$ matrix and $S\subset\mathbb{R}^n$, the image of $S$ under $A$ is,
$$A(S) = \left\{\, A\mathbf{x} \,:\, \mathbf{x}\in S \,\right\}$$
Example 2 - Images of Vectors¶
If $A$ is an $m\times n$ matrix, then
- The image of the zero vector is $A\mathbf{0}=\mathbf{0}$, the zero vector.
- The image of the empty set $\ \varnothing$ is the empty set, $A(\varnothing) = \varnothing$.
- The image of the line segment from the origin $\mathbf{0}$ to vector $\mathbf{p}\in\mathbb{R}^n$ in the domain is the line segment from the origin to vector $A\mathbf{p}\in\mathbb{R}^m$ in the codomain.
The images of parallel lines stay parallel and evenly spaced.
Exercise 2 - Images of Line Segments¶
Show algebraically that for $m\times n$ matrix $A$ the image of a straight line in $\mathbb{R}^n$ by the linear transformation $\mathbf{y} = A\mathbf{x}$ is a straight line in $\mathbb{R}^m$ or a point.
View Solution
Suppose that $A$ is an $m\times n$ matrix and $\ell\,:\,[0,1]\rightarrow\mathbb{R}^n$ is the equation of a line segment $S$ from $\mathbf{p}_0$ to $\mathbf{p}_1$. Then $\ell$ has the parameterization $$\ell(t) = \mathbf{p}_0(1-t) + \mathbf{p}_1t$$ The image of $S$ under $A$, $A(S) = \left\{\, \mathbf{y}\in\mathbb{R}^m \,:\, \mathbf{y}=A\mathbf{x},\ \mathbf{x}\in S \,\right\}$. Thus for each $\mathbf{x}\in S$, for some $t\in[0,1]$, $$\mathbf{x} = \mathbf{p}_0(1-t) + \mathbf{p}_1t$$ Hence, $$ \mathbf{y} = A\mathbf{x} = A\left(\mathbf{p}_0(1-t) + \mathbf{p}_1t\right) = A\mathbf{p}_0(1-t) + A\mathbf{p}_1t$$ If $A\mathbf{p}_0 = A\mathbf{p}_1$, then $A(S)$ is a point. Otherwise $\mathbf{y}$ is on the line segment from $A\mathbf{p}_0$ to $A\mathbf{p}_1$ in $\mathbb{R}^m$. In this case, each image vector $\mathbf{y}\in A(S)$ is on the line segment parameterized by $\ell_A\,:\,[0,1]\rightarrow\mathbb{R}^m$, where $$ \ell_A(t) = A\mathbf{p}_0(1-t) + A\mathbf{p}_1t $$There are several techniques for people to compute the determinant of an $n\times n$ matrix.
- Gaussian elimination and the properties of determinants
- The Laplace Expansion
- Leibniz Formula
In your homework, quizzes or exams always use the first method of computing the determinant of an $n\times n$ matrix. There are only a few instances when use of the Laplace Expansion will be useful. The Leibniz formula is not very useful at all. Our textbook introduces the Laplace Expansion in 3.1 and the properties of determinant in 3.2. The online notes and the video lectures introduce properties first and the Laplace expansion second. That is due to the fact that understanding determinants will aid us in solving problems with them.
In this situation both computers and people use Gaussian elimination and the properties of determinants to compute them. To get lots of points on your homework assignments and tests, students will want to use this method to demonstrate mastery of these properties.
The Laplace expansion is useful for computing determinants of $2\times 2$ matrices and cross products of two 3-dimensional vectors. It can be used strategically with the other properties of determinants to simplify computations. However one does not compute a $6\times 6$ determinant by computing $\frac{6!}{2} = 360$ separate $2\times 2$ determinants, or $120$ distinct $3\times 3$ determinants. The room for computational errors is huge, and it will consume so much time on an exam that finishing will not be possible.
Do not confuse the Laplace Expansion with the definition of the determinant given above.
The identity matrix represents the identity linear transformation that does not change volumes at all. So the scaling factor for the identity matrix is $1$ because the volume of a parallelepiped divided by itself is $1$.
Property 1¶
The determinant of an $n\times n $ identity matrix is one. This is part of the definition of determinant.
$$\det(I_n) = 1$$
Notice the use of the word volume for any dimension. One can use the term volume as a generic term.
- The one-dimensional term for content is length.
- The two-dimensional term is area.
- The three-dimensional term is volume.
There are specific formulas for $n$ dimensional spheres, and the $n$ dimensional hypercubes. Generically we refer to $n$-dimensional content in $n$-dimensional space as a volume. In mathematics, even a length or area is a specific dimension of volume.
The determinant of any identity matrix is $1$. This is part of the definition of determinant. We will see that
$$\det(O) = 0.$$
This is a consequence of the definition of determinant.
Example 3 - Exchanging Rows¶
Let us consider matrix $A$ defined as before
$$A = \begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix}.$$
This is the matrix used in our video and the determinant is $6$ because $\ihat$ is stretched by a factor of $3$ to $3\ihat$, and $\jhat$ is stretched by a factor of $2$ to $2\jhat$. This leaves us with a rectangle with base $3$ and height $2$. This new area represents the image of the unit square under the transformation $A$. The new area is $3\times 2 = 6$ so the determinant is $\frac{6}{1} = 6$.
Now let us flip the rows so that we have a new matrix
$$B = \begin{bmatrix} 0 & 2 \\ 3 & 0 \end{bmatrix}$$
This matrix sends $\ihat$ to $3\jhat$ and $\jhat$ to $2\ihat$. This linear transformation represented by matrix B changes the orientation of the basis vector $\ihat$ and $\jhat$ so that $B\,\jhat$ is to the right of $B\,\ihat$ instead of on the left.
The determinant of matrix $B$ is
$$ \text{det}(B) = |B| = \begin{vmatrix} 0 & 2 \\ 3 & 0 \end{vmatrix} = -6 $$
We can now compute the determinant of any elementary permutation matrix, that is any type I elementary matrix. Since an elementary permutation matrix E exchanges exactly one pair of rows of the identity matrix, it changes the orientation of the domain, thus
$$ \text{det}(E)=|E|= -1 $$
We can also compute the determinant of any permutation matrix P by counting the number of row exchanges.
Property 2¶
If permutation matrix $P$ exchanges $n$ pairs of rows of the identity matrix, then
$$ \text{det}(P) = |P| = \left\{\begin{array}{rcl}\ \ 1 & \ & \text{if $n$ is even} \\ -1 & \ & \text{if $n$ is odd} \end{array} \right. = (-1)^n $$
Like property 1, this behavior is part of the definition of determinant.
Exercise 3 - Determinant of a Permutation Matrix¶
Consider the $4\times 4$ matrix that exchanges row 1 and 4, and rows 2 and 3 is given by
$$ P = \begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix} $$
View Solution
This constitutes two row exchanges so the determinant of $P$ results in$$ \text{det}(P) = |P| = (-1)^2 = 1 $$
The Multilinearity Property connects two ideas in this course. A linear transformation applies to the entire vector, or matrix. Gauss-Jordan elimination consists of three types of row operations. Determinant is multilinear because like Gauss-Jordan elimination axiom $3$ applies only to one row at a time. This one-row-at-a-time property is not quite linear like a linear transformation. Each row operation on a matrix has a different effect on the value of the determinant.
Performing a type I row operation, exchanging two rows of a matrix, changes the sign of the determinant.
Property 3¶
Determinant is Multilinear
As with properties 1 and 2, this is still part of the definition of determinant.
(a) Multiplying a row by a scalar, multiplies the determinant by that scalar.
(b) Adding corresponding rows of two otherwise identical matrices yields the sum of their determinants.
Notice that all of the other rows must be identical for both terms of the sum and to factor a scalar.
Example 4 - Scaling Volume¶
Let us return to our previous example $A = \begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix}$ and compare it to $B = \begin{bmatrix} 3 & 0 \\ 0 & 4 \end{bmatrix}$
Notice that by multiplying only one side of the image rectangle (row of the matrix) by two multiplies the area by two as well.
$$ \text{det}(B) = \begin{vmatrix} 3 & 0 \\ 0 & 4 \end{vmatrix} = \begin{vmatrix} 3 & 0 \\ 2\cdot 0 & 2\cdot 4 \end{vmatrix} = 2\begin{vmatrix} 3 & 0 \\ 0 & 2 \end{vmatrix} = 2\cdot 6 = 12 $$
Example 5 - Factors¶
We can now compute the determinant of a type II elementary matrix.
$$ \begin{align*} \begin{vmatrix} 3 & 0 \\ 0 & 1 \end{vmatrix} &= 3\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 3\cdot 1 = 3 \\ \\ \begin{vmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{vmatrix} &= 4\begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} = 4\cdot 1 = 4 \end{align*} $$
Given an $n\times n$ type II elementary matrix and non-zero scalar $\alpha$
$$ \begin{vmatrix} 1 & 0 & \cdots & 0 & \cdots & 0 & 0 \\ 0 & 1 & \cdots & 0 & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \cdots & \vdots & \vdots \\ 0 & 0 & \cdots & \alpha & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \cdots & \vdots & \vdots \\ 0 & 0 & \cdots & 0 & \cdots & 1 & 0 \\ 0 & 0 & \cdots & 0 & \cdots & 0 & 1 \end{vmatrix} = \alpha|I| = \alpha\cdot 1 = \alpha $$
Computing a determinant requires thinking of Property 3(a) as factoring.
Exercise 4 - Compute the Determinant¶
Use properties 1, 2, and 3(a) to compute the determinant of matrix $A$:
$$ A = \begin{bmatrix} 0 & 3 \\ -4 & 0 \end{bmatrix} $$
View Solution
| $$ \begin{align*} \left|A\right| &= \begin{vmatrix} 0 & 3 \\ -4 & 0 \end{vmatrix} \\ \\ &= \begin{vmatrix} 3\cdot 0 & 3\cdot 1 \\ -4 & 0 \end{vmatrix} \\ \\ &= 3\begin{vmatrix} 0 & 1 \\ -4\cdot 1 & -4\cdot 0 \end{vmatrix} \\ \\ &= 3(-4)\begin{vmatrix} 0 & 1 \\ 1 & 0 \end{vmatrix} \\ \\ &= 3(-4)(-1) = 12 \end{align*} $$ |
 |
Example 6 - Terms¶
Suppose $A = \begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix}$ and now $B = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$
In figure 7, we are increasing the area (volume) of the rectangle by adding to the length but not the height. This is important to property 3(b). We only add one of the rows of our two matrices. All of the other rows of each matrix are identical. In this special case do we get the sum of two areas or $\det(A) + \det(B)$. In this example,
$$\det\left( \begin{bmatrix} {\color{royalblue}3}+{\color{crimson}2} & {\color{royalblue}0}+{\color{crimson}0} \\ 0 & 2 \end{bmatrix} \right) = \det\left( \begin{bmatrix} {\color{royalblue}3} & {\color{royalblue}0} \\ 0 & 2 \end{bmatrix} \right) + \det\left( \begin{bmatrix} {\color{crimson}2} & {\color{crimson}0} \\ 0 & 2 \end{bmatrix} \right) = 6 + 4 = 10 $$
Exercise 5 - Compute the Determinant¶
Use properties 1-3 of the determinant to compute $|A|$ where
$$ A = \begin{bmatrix} 5 & 2 \\ 0 & 2 \end{bmatrix} $$
View Solution
| $$ \begin{align*} \left|A\right| = \begin{vmatrix} 5 & 2 \\ 0 & 2 \end{vmatrix} &= \begin{vmatrix} {\color{royalblue}5}+{\color{crimson}0} & {\color{royalblue}0}+{\color{crimson}2} \\ 0 & 2 \end{vmatrix} \\ \\ &= \begin{vmatrix} {\color{royalblue}5} & {\color{royalblue}0} \\ 0 & 2 \end{vmatrix} + \begin{vmatrix} {\color{crimson}0} & {\color{crimson}2} \\ 0 & 2 \end{vmatrix} \\ \\ &= (5)(2)\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} + 0 \\ \\ &= 10(1) = 10 \end{align*} $$ |
 |
From the three properties above we can derive all of the other properties and formulas for determinants. In particular, we can now derive the equation for the determinant of any $2\times 2$ matrix.
Theorem 1¶
The Laplace Expansion of a $2\times 2$ matrix
The determinant of a $2\times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is given by the formula
$$ \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc $$
We will find in section 3.2 that this is the Laplace Expansion of a $2\times 2$ matrix.
Proof of Theorem 1¶
Let matrix $A$ be defined by
$$ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} $$
Computing the determinant using the three properties of the definition
$$ \begin{align*} \text{det}(A) = |A| &= \begin{vmatrix} a & b \\ c & d \end{vmatrix} \\ \\ &= \begin{vmatrix} a + {\color{darkblue}0} & 0 + {\color{darkblue}b} \\ c & d \end{vmatrix} \\ \\ &= \begin{vmatrix} a & 0 \\ c & d \end{vmatrix} + \begin{vmatrix} 0 & b \\ c & d \end{vmatrix} \qquad &\text{Property 3(b)} \\ \\ &= \begin{vmatrix} a & 0 \\ c & d \end{vmatrix} + (-1) \begin{vmatrix} {\color{darkblue} c} & {\color{darkblue} d} \\ {\color{darkblue} 0} & {\color{darkblue} b} \end{vmatrix} &\text{Property 2} \\ \\ &= \begin{vmatrix} a & 0 \\ 0 + {\color{darkblue}c} & d + {\color{darkblue}0} \end{vmatrix} - \begin{vmatrix} c + {\color{darkblue}0} & 0 + {\color{darkblue}d} \\ 0 & b \end{vmatrix} \\ \\ &= \begin{vmatrix} a & 0 \\ 0 & d \end{vmatrix} + \begin{vmatrix} a & 0 \\ {\color{darkblue}c} & {\color{darkblue}0} \end{vmatrix} - \begin{vmatrix} c & 0 \\ 0 & b \end{vmatrix} - \begin{vmatrix} {\color{darkblue}0} & {\color{darkblue}d} \\ 0 & b \end{vmatrix} \qquad &\text{Property 3(b)} \\ \\ &= {\color{darkblue}a}\begin{vmatrix} 1 & 0 \\ 0 & d \end{vmatrix} + 0 - {\color{darkblue}c}\begin{vmatrix} 1 & 0 \\ 0 & b \end{vmatrix} - 0 \qquad &\text{Property 3(a)} \\ \\ &= a{\color{darkblue}d}\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} - c{\color{darkblue}b}\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} \qquad &\text{Property 3(a)} \\ \\ &= ad - bc \qquad\qquad\qquad &\text{Property 1}\ \blacksquare \end{align*} $$
The first three properties of determinants are axioms from the definition of determinant. The next three properties are consequences of these axioms. Each is a theorem we must prove using the first three properties.
Property 4¶
Example 7 - Duplicate Rows¶
Consider matrix $B$ given by
$$ B = \begin{bmatrix} -1 & \ \ 4\ &\ \ 8\ \\ \ \ 1\ &\ \ 8\ & -1\ \\ -1\ &\ \ 4\ &\ \ 8\ \end{bmatrix} $$
We can use property 2 to exchange rows 1 and 3 and obtain
$$ |B| = \begin{vmatrix} -1 & \ \ 4\ &\ \ 8\ \\ \ \ 1\ &\ \ 8\ & -1\ \\ -1\ &\ \ 4\ &\ \ 8\ \end{vmatrix} = -\begin{vmatrix} -1 & \ \ 4\ &\ \ 8\ \\ \ \ 1\ &\ \ 8\ & -1\ \\ -1\ &\ \ 4\ &\ \ 8\ \end{vmatrix} = -|B| $$
In other words, exchanging the first and third rows using property 2 gives us the negative of the original matrix; however, the first and third rows are identical so the matrix is unchanged. This algebra results in the equation
$$ |B| = -|B| $$
There is only one real (or complex) scalar equal to its own negative, zero. Hence $|B| = 0$. Notice that we use property 2 to derive property 4.
Theorem 2¶
Property 4
If an $n\times n$ matrix $A$ has two equal rows, then the determinant $|A| = 0$.
Proof of Property 4¶
Let matrix $A$ be an $n\times n$ matrix with two identical rows, $\mathbf{a}^j = \mathbf{a}^k$, $1\le j < k\le n$. Then exchanging these two rows does not change the matrix. However property 2 of determinants requires that exchanging rows changes the sign of the determinant. Hence
$$ |A| = \begin{vmatrix} \mathbf{a}^1 \\ \vdots \\ \mathbf{a}^j \\ \vdots \\ \mathbf{a}^k \\ \vdots \\ \mathbf{a}^n \end{vmatrix} = -\begin{vmatrix} \mathbf{a}^1 \\ \vdots \\ \mathbf{a}^k \\ \vdots \\ \mathbf{a}^j \\ \vdots \\ \mathbf{a}^n \end{vmatrix} = -|A| $$
For real or complex numbers, $|A|=-|A|$ implies that $|A|=0$. $\blacksquare$
Property 5¶
Example 8 - Type III Row Operation¶
Property 2 tells us how a type I row operation affects the value of a determinant and property 3(a) informs us of the effect of a type II row operation. Let us consider a type III row operation. If we add a nonzero multiple t of row 1 to row 2 of a $2\times 2$ matrix
$$ B = \begin{bmatrix} a & b \\ c & d \end{bmatrix} $$
we obtain a new matrix
$$ C = \begin{bmatrix} a & b \\ c + t a & d + t b \end{bmatrix} $$
Computing the determinant of matrix $C$ one obtains
$$ \begin{align*} |C| &= \begin{vmatrix} a & b \\ c + t a & d + t b \end{vmatrix} \\ \\ &= \begin{vmatrix} a & b \\ c & d \end{vmatrix} + \begin{vmatrix} a & b \\ t a & t b \end{vmatrix} \qquad &\text{Property 3(b)} \\ \\ &= \begin{vmatrix} a & b \\ c & d \end{vmatrix} + t\begin{vmatrix} a & b \\ a & b \end{vmatrix} \qquad &\text{Property 3(a)} \\ \\ &= \begin{vmatrix} a & b \\ c & d \end{vmatrix} + t\cdot 0 \qquad &\text{Property 4} \\ \\ &= |B| \end{align*} $$
Theorem 3¶
Property 5
Performing a type III row operation on a matrix does not change the value of the determinant!
Proof of Property 5¶
Suppose that $A$ is an $n\times n$ matrix and $t\in\mathbb{R}$ is a scalar. Let $1\le j < k\le n$ and perform a type III row operation on row $k$ by subtracting $t\mathbf{a}^j$ from row $\mathbf{a}^k$. Using property 3 and 4,
$$ \begin{vmatrix} \mathbf{a}^1 \\ \vdots \\ \mathbf{a}^j \\ \vdots \\ \mathbf{a}^k - t\mathbf{a}^j \\ \vdots \\ \mathbf{a}^n \end{vmatrix} = \begin{vmatrix} \mathbf{a}^1 \\ \vdots \\ \mathbf{a}^j \\ \vdots \\ \mathbf{a}^k \\ \vdots \\ \mathbf{a}^n \end{vmatrix} + \begin{vmatrix} \mathbf{a}^1 \\ \vdots \\ \mathbf{a}^j \\ \vdots \\ -t\mathbf{a}^j \\ \vdots \\ \mathbf{a}^n \end{vmatrix} = |A| -t\begin{vmatrix} \mathbf{a}^1 \\ \vdots \\ \mathbf{a}^j \\ \vdots \\ \mathbf{a}^j \\ \vdots \\ \mathbf{a}^n \end{vmatrix} = |A| - t\cdot0 = |A| $$
Hence a type III row operation has no effect on the value of the determinant of a matrix. $\blacksquare$
Exercise 6 - Compute the Determinant¶
Use properties 1-5 of the determinant to compute $|A|$ where
$$ A = \begin{bmatrix}\ \ 2\ &\ \ 6\ &\ \ 2\ \\ -2\ &\ \ 6\ &\ \ 3\ \\ \ \ 8 &\ 36\ &\ \ 0\ \end{bmatrix} $$
View Solution
$$ \begin{align*} \begin{vmatrix}\ \ 2 & 6 & 2 \\ -2 & 6 & 3 \\ \ \ 8 & 36 & 0\ \end{vmatrix} &= \begin{vmatrix}\ \ 2 & 6 & 2 \\ \ \ 0 & 12 & 5 \\ \ \ 8 & 36 & 0\ \end{vmatrix} \qquad &\begin{array}{l} \text{Property 5} \\ R_2 + R_1 \end{array} \\ \\ &= \begin{vmatrix}\ \ 2 & 6 & 2 \\ \ \ 0 & 12 & 5 \\ \ \ 0 & 12 & -8\ \end{vmatrix} \qquad &\begin{array}{l} \text{Property 5} \\ R_3 - 4R_1 \end{array} \\ \\ &= \begin{vmatrix}\ \ 2 & 6 & 2 \\ \ \ 0 & 12 & 5 \\ \ \ 0 & 0 & -13\ \end{vmatrix} \qquad &\begin{array}{l} \text{Property 5} \\ R_3 - R_2 \end{array} \\ \\ &= \begin{vmatrix}\ \ 2 & 6 & 2 \\ \ \ 0 & 12 & 0 \\ \ \ 0 & 0 & -13\ \end{vmatrix} \qquad &\begin{array}{l} \text{Property 5} \\ R_2 + \frac{5}{13}R_3 \end{array} \\ \\ &= \begin{vmatrix}\ \ 2 & 6 & 0 \\ \ \ 0 & 12 & 0 \\ \ \ 0 & 0 & -13\ \end{vmatrix} \qquad &\begin{array}{l} \text{Property 5} \\ R_1 + \frac{2}{13}R_3 \end{array} \\ \\ &= \begin{vmatrix}\ \ 2 & 0 & 0 \\ \ \ 0 & 12 & 0 \\ \ \ 0 & 0 & -13\ \end{vmatrix} \qquad &\begin{array}{l} \text{Property 5} \\ R_1 - \frac{1}{2}R_2 \end{array} \\ \\ &= 2(12)(-13)\begin{vmatrix}\ \ 1 & 0 & 0 \\ \ \ 0 & 1 & 0 \\ \ \ 0 & 0 & 1\ \end{vmatrix} \qquad &\begin{array}{l} \text{Property 3(a)} \\ \end{array} \\ \\ &= 2(12)(-13)(1) = 24(-13) = -240-72 = -312 \end{align*} $$Property 6¶
Example 9 - Row of Zeros¶
Consider matrix $B$ defined by
$$ B = \begin{bmatrix} 0 & 0 \\ 3 & 2 \end{bmatrix} $$
If we use property 3(a)
$$|B| = \begin{vmatrix} 0 & 0 \\ 3 & 2 \end{vmatrix} = \begin{vmatrix} 5280\cdot 0 & 5280\cdot 0 \\ 3 & 2 \end{vmatrix} = 5280\begin{vmatrix} 0 & 0 \\ 3 & 2 \end{vmatrix} = 5280\cdot|B|$$
Subtracting $|B|$ from both sides yields
$$ 0 = 5279\cdot|B| $$
Hence $|B| = 0$.
Proof of Property 6¶
Let $A$ be an $n\times n$ matrix with a row of zeros, and $t=0$. Let $1\le k\le n$ and row $k$ be row of zeros, $\mathbf{a}^k=\mathbf{0}^\mathsf{T}$. Multiplying row $k$ by $t$ also multiplies the determinant by $t$ using property 3(a). However $t\mathbf{a}^k = \mathbf{0}^\mathsf{T}$. Hence $|A|=t|A|=0|A|=0$. $\blacksquare$
Corollary 5¶
Linearly Dependent Row
If $n\times n$ matrix $A$ has a linearly dependent row, then $\det(A)=0$.
Proof of Corollary 5¶
Let $A$ be an $n\times n$ matrix with at least one row that is a linear combination of the others. Let $1\le k\le n$ and $\mathbf{a}^k$ be the row that is a linear combination of the other rows of $A$. Performing type III row operations on row $k$, one can reduce row $k$ to a row of zeros. Hence $|A|=0$. $\blacksquare$
Example 10 - Diagonal Matrix¶
Consider a diagonal matrix
$$ A = \begin{bmatrix} a_{11} & 0 & \cdots & 0 & 0 \\ 0 & a_{22} & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & a_{n-1,n-1} & 0 \\ 0 & 0 & \cdots & 0 & a_{nn} \end{bmatrix} $$
Using property 3(a), $n$ times we obtain
$$ \begin{align*} |A| &= \begin{vmatrix} a_{11} & 0 & \cdots & 0 & 0 \\ 0 & a_{22} & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & a_{n-1,n-1} & 0 \\ 0 & 0 & \cdots & 0 & a_{nn} \end{vmatrix} = a_{11}a_{22}\cdots a_{n-1,n-1}a_{nn} \begin{vmatrix} 1 & 0 & \cdots & 0 & 0 \\ 0 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 1 & 0 \\ 0 & 0 & \cdots & 0 & 1 \end{vmatrix} \\ \\ &= a_{11}a_{22}\cdots a_{n-1,n-1}a_{nn}\left|I_n\right| = \displaystyle\prod_{k=1}^n a_{kk} \end{align*} $$
Let us consider an upper triangular matrix
$$ B = \begin{bmatrix} b_{11} & * & \cdots & * & * \\ 0 & b_{22} & \cdots & * & * \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & b_{n-1,n-1} & * \\ 0 & 0 & \cdots & 0 & b_{nn} \end{bmatrix} $$
Here the splat $*$ indicates the placement of a real number. We are running out of letters so in linear algebra we use the splat to represent an element of a matrix whose value is unimportant to the property of the matrix in question. The values of each element of the lower triangle are all zero. This defines matrix $B$ as an upper triangular matrix. The elements in the upper triangle can be any real number without affecting our result.
Two results may occur:
- If all of the diagonal elements are nonzero then we will reduce matrix $B$ to a diagonal matrix without changing the value of the determinant so
$$\det(B) = \prod_{k=1}^n b_{kk}$$
- If there are diagonal elements that are equal to zero, then there is a zero diagonal element with largest index $k$, $b_{kk}$. Our determinant becomes
$$ |B| = \begin{vmatrix} b_{11} & * & \cdots & * & * \\ 0 & b_{22} & \cdots & * & * \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 0 & * \\ 0 & 0 & \cdots & 0 & b_{nn} \end{vmatrix} = \begin{vmatrix} b_{11} & 0 & \cdots & * & 0 \\ 0 & b_{22} & \cdots & * & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 0 & 0 \\ 0 & 0 & \cdots & 0 & b_{nn} \end{vmatrix} = 0 $$
The last zero diagonal element will find itself in a row of zeros, hence by property 6, the value of the determinant is zero. The result is that the determinant of this upper triangular matrix is still the product of the diagonal elements.
Theorem 6¶
Property 7
The determinant of an $n\times n$ upper triangular, lower triangular, or diagonal matrix is equal to the product of its diagonal elements.
Exercise 7 - Compute the Determinant¶
Compute the determinant of the following matrix.
$$ A = \begin{bmatrix}\ \ 2\ &\ \ 3\ &\ 12\ & -2\ \\ 0 &\ \ 3\ & -1\ &\ \ 7\ \\ \ \ 0\ &\ \ 0\ & \ \ 8 &\ 36\ \\ \ \ 0\ &\ \ 0\ &\ \ 0\ & -3\end{bmatrix} $$
View Solution
$$ |A| = \begin{vmatrix}\ \ 2\ &\ \ 3\ &\ 12\ & -2\ \\ 0 &\ \ 3\ & -1\ &\ \ 7\ \\ \ \ 0\ &\ \ 0\ & \ \ 8 &\ 36\ \\ \ \ 0\ &\ \ 0\ &\ \ 0\ & -3\end{vmatrix} = 2(3)(8)(-3) = 6(-24) = -144 $$Theorem 7¶
Property 8
The determinant of an $n\times n$ matrix is equal to zero if and only if the matrix is singular.
Exercise 8 - Determine if a Matrix is Invertible¶
Determine if the following matrix is invertible.
$$ A = \begin{bmatrix}\ \ 2\ &\ \ 6\ &\ \ 2\ \\ -2\ &\ \ 6\ &\ \ 3\ \\ \ \ 8 &\ 36\ &\ \ 0\ \end{bmatrix} $$
View Solution
Matrix $A$ is from Exercise 6.$$ |A| = \begin{vmatrix}\ \ 2\ &\ \ 6\ &\ \ 2\ \\ -2\ &\ \ 6\ &\ \ 3\ \\ \ \ 8 &\ 36\ &\ \ 0\ \end{vmatrix} = -312 \neq 0 $$
Since the determinant of matrix $A$ is nonzero, matrix $A$ is nonsingular and thus invertible.
Property 9¶
Theorem 8¶
Property 9
The determinant of a product of $n\times n$ matrices is the product of the determinants of the matrices.
Proof of Property 9¶
If you watched the video by Grant Sanderson you should realize that the product of two $n\times n$ matrices, $A$ and $B$, is the composition of two linear transformations they represent. For every vector $\mathbf{x}\in\mathbb{R}^n$:
$$ AB\mathbf{x} = A\left(B\mathbf{x}\right) $$
When we calculate the determinant of the composition $AB$ of two non-singular matrices, we compute the scaling factor of the product. However we can also think of this as the scaling factor for matrix $A$ applied to the scaling factor for matrix $B$. Recall that the determinant for matrix $B$ applied to the unit square $D$ is given by
$$\det(B) = \frac{\text{area of }B(D)}{\text{area of }D},$$
where areas are signed (orientation aware) areas. The determinant of matrix $A$ is also the ratio,
$$\det(A) = \frac{\text{area of }A(\,B(D)\,)}{\text{area of }B(D)}$$
Hence
$$ \text{det}(AB) = \frac{\text{area of }(AB)(D)}{\text{area of }D} = \frac{\text{area of }(AB)(D)}{\text{area of }A(\,B(D)\,)}\cdot\frac{\text{area of }A(\,B(D)\,)}{\text{area of }D} = \det(A)\det(B) $$
This is the one line that Grant Sanderson alluded to at the end of the video.
If matrix $A$ is singular, then each column of matrix $AB$ is a linear combination of the columns of matrix $A$. The number of linearly independent columns in matrix $A$ is $r = \textrm{rank}(A) < n$. The product matrix $AB$ has $n$ columns, however, they are linear combinations of only $r$ linearly independent vectors, at most, $n-r>0$ of the columns are dependent. Hence matrix $AB$ is singular, so $\det(AB)=0 = 0\det(B) = \det(A)\det(B)$.
If matrix $B$ is singular, then each row of matrix $AB$ is a linear combination of the rows of matrix $B$. Again, the number of linearly independent rows in matrix $B$ is $r =\textrm{rank}(B) < n$. Thus $AB$ can have at most $r$ linearly independent rows. Hence matrix $AB$ is singular, so $\det(AB)=0= \det(A)0 = \det(A)\det(B)$.
In all three cases $\det(AB) = \det(A)\det(B)$. $\blacksquare$
Corollary 9¶
The Determinant of the Inverse
If matrix A is a nonsingular matrix then $|A|\neq 0$, and
$$\det(A^{-1}) = \frac{1}{\det(A)}$$
Proof of Corollary 9¶
If $A$ is an $n\times n$ non-singular matrix, then $\det(A)\neq 0$ and
$$ 1 = \text{det}(I_n) = \text{det}\left(A^{-1}A\right) = \text{det}\left(A^{-1}\right)\,\text{det}(A). $$
Dividing both sides by $\det(A)$ yields the result,
$$ \dfrac{1}{\text{det}(A)} = \text{det}\left(A^{-1}\right).\ \blacksquare $$
Exercise 9 - Compute the Determinant¶
What is the determinant of the inverse of the following matrix?
$$ A = \begin{bmatrix}\ \ 2\ &\ \ 6\ &\ \ 2\ \\ -2\ &\ \ 6\ &\ \ 3\ \\ \ \ 8 &\ 36\ &\ \ 0\ \end{bmatrix} $$
View Solution
$$|A| = \begin{vmatrix}\ \ 2\ &\ \ 6\ &\ \ 2\ \\ -2\ &\ \ 6\ &\ \ 3\ \\ \ \ 8 &\ 36\ &\ \ 0\ \end{vmatrix} = -312$$Thus
$$\left|A^{-1}\right| = \frac{1}{-312} = -\frac{1}{312}$$
Property 10¶
Proof of Property 10¶
Let $A$ be an $n\times n$ matrix. We can factor $A$ into a product of a lower triangular and upper triangular matrix. In some matrices row interchanges are required before factoring. This is the $LU$-Decomposition
$$ PA = LU $$
where $L$ is a lower triangular matrix with 1 in each of the diagonal entries, $U$ is an upper triangular matrix, and $P$ is a permutation matrix. Recall $\det(P) = (-1)^m$, where $m$ is the necessary number of row interchanges. Hence
$$ |PA| = (-1)^m|A| = |LU| = |L|\cdot|U| = 1\cdot\displaystyle\prod_{k=1}^n u_{kk} = \displaystyle\prod_{k=1}^n u_{kk} $$
Now let's use the properties of determinants to compute the determinant of $A^T$. Using property 7,
$$ \left|A^T\right| = (-1)^m\left|\left(LU\right)^T\right| = (-1)^m\left|U^TL^T\right| = (-1)^m\left|U^T\right|\cdot\left|L^T\right| = (-1)^m|U|\,|L| = |A|.\ \blacksquare $$
Exercise 10 - Compute the Determinant¶
Compute the determinant of $A^T$ where
$$ A = \begin{bmatrix}\ \ 2\ &\ \ 6\ &\ \ 2\ \\ -2\ &\ \ 6\ &\ \ 3\ \\ \ \ 8 &\ 36\ &\ \ 0\ \end{bmatrix} $$
View Solution
$$ \left|A^T\right| = |A| = -312 $$The 10 Properties of Determinant¶
Property 10 tells us that every row property is also a column property. In our list we recognize those properties that have dual column properties as well.
Axiomatic Properties¶
These properties simply recognize the axioms in the definition of determinants
- $\det(I_n) = 1$
- Alternating Property: exchanging a pair of rows (or columns) changes the sign of the determinant
- Multilinear Property:
(a) scaling a row (or column) by a factor scales the determinant by the same factor
(b) adding a row (or column) vector to exactly one row (column) produces the sum of two determinants
Direct Properties¶
Direct properties are consequences of the three axioms (properties) of determinants. They are theorems about determinants that are proved directly from the axioms.
- If an $n\times n$ matrix has two identical rows (columns), then its determinant is zero.
- Adding a multiple of one row (column) to another row (column) has no effect on the value of the determinant.
- If an $n\times n$ matrix has a row (column) of zeros, then its determinant is zero.
- The determinant of an $n\times n$ upper triangular, lower triangular, or diagonal matrix is equal to the product of its diagonal elements.
- The determinant of an $n\times n$ matrix is equal to zero if and only if the matrix is singular.
Function Properties¶
Function properties relate the determinant to the linear transformations that $n\times n$ matrices represent.
- The determinant of a product of $n\times n$ matrices is the product of the determinants of the matrices.
- The determinant of the transpose of a matrix is equal to the determinant of the matrix.
Exercise 11 - Scaling and Orientation¶
A $2\times 2$ matrix $A$ represents a linear transformation with $\det(A) = -5$.
(a) By what factor does $A$ scale the area of a region in the plane?
(b) Does $A$ preserve or reverse orientation?
View Solution
(a) Area is scaled by the absolute value of the determinant, $|\det(A)| = 5$. Every region has its area multiplied by $5$.(b) Because $\det(A) = -5 < 0$, the transformation reverses orientation (it flips the plane, as in Example 3).
Exercise 12 - Determinant of a Scalar Multiple¶
Let
$$ A = \begin{bmatrix} 1 & 2 & 0 \\ 3 & -1 & 2 \\ 0 & 4 & 1 \end{bmatrix}. $$
(a) Compute $\det(A)$.
(b) Without multiplying out the matrix $-3A$, use Property 3(a) to compute $\det(-3A)$.
View Solution
(a) Reducing with row operations (or expanding the first column), $$\det(A) = \begin{vmatrix} 1 & 2 & 0 \\ 3 & -1 & 2 \\ 0 & 4 & 1 \end{vmatrix} = -15.$$(b) Each of the $n = 3$ rows of $A$ is multiplied by $-3$. By Property 3(a) we factor $-3$ out of each row, once per row: $$\det(-3A) = (-3)^3\det(A) = -27\,(-15) = 405.$$
In general, for an $n\times n$ matrix and scalar $c$, $$\det(cA) = c^n\det(A).$$ A common error is to write $\det(cA) = c\det(A)$; that scales only one row, not all $n$ of them.
Exercise 13 - Effect of Row Operations¶
Suppose $A$ is a $4\times 4$ matrix with $\det(A) = 7$. Find the determinant of the matrix obtained from $A$ by each of the following, stating the property you used.
(a) exchanging two rows
(b) multiplying one row by $5$
(c) adding $3$ times row 1 to row 2
(d) multiplying the entire matrix by $2$
(e) transposing the matrix
(f) the matrix $A^2$
View Solution
(a) $-7$ (Property 2, exchanging rows changes the sign).(b) $5\cdot 7 = 35$ (Property 3(a)).
(c) $7$ (Property 5, a type III operation does not change the determinant).
(d) $2^4\cdot 7 = 112$ ($\det(cA) = c^n\det(A)$ with $n = 4$).
(e) $7$ (Property 10, $\det(A^\mathsf{T}) = \det(A)$).
(f) $7^2 = 49$ (Property 9, $\det(A^2) = \det(A)\det(A)$).
Exercise 14 - Determinant Algebra¶
Let $A$ and $B$ be $3\times 3$ matrices with $\det(A) = 4$ and $\det(B) = -3$. Compute each determinant.
(a) $\det(AB)$
(b) $\det\!\left(A^{-1}\right)$
(c) $\det\!\left(A^3\right)$
(d) $\det\!\left(A^\mathsf{T}B^2\right)$
(e) $\det\!\left(A^{-1}B^\mathsf{T}\right)$
View Solution
(a) $\det(AB) = \det(A)\det(B) = (4)(-3) = -12$.(b) $\det\!\left(A^{-1}\right) = \dfrac{1}{\det(A)} = \dfrac{1}{4}$.
(c) $\det\!\left(A^3\right) = \det(A)^3 = 4^3 = 64$.
(d) $\det\!\left(A^\mathsf{T}B^2\right) = \det(A)\det(B)^2 = (4)(9) = 36$.
(e) $\det\!\left(A^{-1}B^\mathsf{T}\right) = \dfrac{1}{\det(A)}\det(B) = \dfrac{-3}{4} = -\dfrac{3}{4}$.
Exercise 15 - Recognizing a Singular Matrix¶
Without doing a full computation, explain why the following matrix is singular, then confirm that its determinant is zero.
$$ A = \begin{bmatrix} 1 & 2 & 3 & 1 \\ 0 & 1 & -1 & 2 \\ 2 & 3 & 7 & 0 \\ 1 & 0 & 5 & -3 \end{bmatrix} $$
View Solution
Row 3 is a linear combination of the other rows: $\mathbf{a}^3 = 2\mathbf{a}^1 - \mathbf{a}^2$, since $$2(1,2,3,1) - (0,1,-1,2) = (2,3,7,0).$$ By Corollary 5, a matrix with a row that is a linear combination of the others has determinant zero, so $\det(A) = 0$ and $A$ is singular (Property 8). Performing the type III operation $\mathbf{a}^3 \leftarrow \mathbf{a}^3 - 2\mathbf{a}^1 + \mathbf{a}^2$ produces a row of zeros, confirming $\det(A) = 0$.Exercise 16 - Solve for a Singular Matrix¶
For which value(s) of $x$ is the matrix
$$ A = \begin{bmatrix} x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x \end{bmatrix} $$
singular?
View Solution
A matrix is singular exactly when its determinant is zero (Property 8). Expanding, $$\det(A) = x^3 - 3x + 2 = (x-1)^2(x+2).$$ Setting $\det(A) = 0$ gives $x = 1$ (a repeated root) or $x = -2$. For these values the rows become linearly dependent: at $x = 1$ all three rows are equal, and at $x = -2$ the rows sum to the zero vector.Exercise 17 - A Common Misconception¶
A student claims that $\det(A + B) = \det(A) + \det(B)$ for all square matrices. Give a counterexample using two $2\times 2$ matrices, and explain when adding matrices does split the determinant into a sum.
View Solution
Let $$A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \qquad B = \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}.$$ Then $\det(A) = 1$ and $\det(B) = 6$, so $\det(A) + \det(B) = 7$. However, $$A + B = \begin{bmatrix} 3 & 0 \\ 0 & 4 \end{bmatrix}, \qquad \det(A+B) = 12 \neq 7.$$ The determinant is not additive over matrix addition. The additivity in Property 3(b) applies only when the two matrices are identical except in a single row (or column); then that one row splits and the determinant becomes a sum of two determinants.Exercise 18 - A $6\times 6$ Determinant by Gaussian Elimination¶
Use Gaussian elimination and the properties of determinants to compute the determinant of the nonsingular matrix
$$ A = \begin{bmatrix} \ \ 2 & \ \ 1 & -1 & \ \ 0 & \ \ 3 & \ \ 1 \\ \ \ 4 & \ \ 5 & \ \ 0 & -1 & \ \ 6 & \ \ 4 \\ -2 & \ \ 2 & \ \ 2 & \ \ 3 & -2 & \ \ 0 \\ \ \ 6 & \ \ 0 & -7 & 13 & \ \ 9 & \ \ 2 \\ \ \ 2 & \ \ 7 & \ \ 4 & -2 & \ \ 1 & 11 \\ \ \ 0 & \ \ 3 & \ \ 1 & -5 & \ \ 7 & -3 \end{bmatrix}. $$
View Solution
We reduce $A$ to upper triangular form using only type III row operations, which by Property 5 leave the determinant unchanged. No row exchanges are needed, so the sign never changes. $$ \begin{align*} |A| &= \begin{vmatrix} \ \ 2 & \ \ 1 & -1 & \ \ 0 & \ \ 3 & \ \ 1 \\ \ \ 4 & \ \ 5 & \ \ 0 & -1 & \ \ 6 & \ \ 4 \\ -2 & \ \ 2 & \ \ 2 & \ \ 3 & -2 & \ \ 0 \\ \ \ 6 & \ \ 0 & -7 & 13 & \ \ 9 & \ \ 2 \\ \ \ 2 & \ \ 7 & \ \ 4 & -2 & \ \ 1 & 11 \\ \ \ 0 & \ \ 3 & \ \ 1 & -5 & \ \ 7 & -3 \end{vmatrix}\begin{array}{l} R_2-2R_1 \\ R_3+R_1 \\ R_4-3R_1 \\ R_5-R_1 \\ \end{array} = \begin{vmatrix} \ \ 2 &\ \ 1 & -1 &\ \ 0 &\ \ 3 &\ \ 1 \\ \ \ 0 &\ \ 3 &\ \ 2 & -1 &\ \ 0 &\ \ 2 \\ \ \ 0 &\ \ 3 &\ \ 1 &\ \ 3 &\ \ 1 &\ \ 1 \\ \ \ 0 & -3 & -4 &\ 13 &\ \ 0 & -1 \\ \ \ 0 &\ \ 6 &\ \ 5 & -2 & -2 &\ 10 \\ \ \ 0 &\ \ 3 &\ \ 1 & -5 &\ \ 7 & -3 \end{vmatrix}\begin{array}{l} \\ \\ R_3-R_2 \\ R_4+R_2 \\ R_5-2R_2 \\ R_6-R_2 \end{array} \\ \\ &= \begin{vmatrix} \ \ 2 &\ \ 1 & -1 &\ \ 0 &\ \ 3 &\ \ 1 \\ \ \ 0 &\ \ 3 &\ \ 2 & -1 &\ \ 0 &\ \ 2 \\ \ \ 0 &\ \ 0 & -1 &\ \ 4 &\ \ 1 & -1 \\ \ \ 0 &\ \ 0 & -2 &\ 12 &\ \ 0 &\ \ 1 \\ \ \ 0 &\ \ 0 &\ \ 1 &\ \ 0 & -2 &\ \ 6 \\ \ \ 0 &\ \ 0 & -1 & -4 &\ \ 7 & -5 \end{vmatrix}\begin{array}{l} \\ \\ \\ R_4-2R_3 \\ R_5+R_3 \\ R_6-R_3 \end{array} = \begin{vmatrix} \ \ 2 &\ \ 1 & -1 &\ \ 0 &\ \ 3 &\ \ 1 \\ \ \ 0 &\ \ 3 &\ \ 2 & -1 &\ \ 0 &\ \ 2 \\ \ \ 0 &\ \ 0 & -1 &\ \ 4 &\ \ 1 & -1 \\ \ \ 0 &\ \ 0 &\ \ 0 &\ \ 4 & -2 &\ \ 3 \\ \ \ 0 &\ \ 0 &\ \ 0 &\ \ 4 & -1 &\ \ 5 \\ \ \ 0 &\ \ 0 &\ \ 0 & -8 &\ \ 6 & -4 \end{vmatrix}\begin{array}{l} \\ \\ \\ \\ R_5-R_4 \\ R_6+2R_4 \end{array} \\ \\ &= \begin{vmatrix} \ \ 2 &\ \ 1 & -1 &\ \ 0 &\ \ 3 &\ \ 1 \\ \ \ 0 &\ \ 3 &\ \ 2 & -1 &\ \ 0 &\ \ 2 \\ \ \ 0 &\ \ 0 & -1 &\ \ 4 &\ \ 1 & -1 \\ \ \ 0 &\ \ 0 &\ \ 0 &\ \ 4 & -2 &\ \ 3 \\ \ \ 0 &\ \ 0 &\ \ 0 &\ \ 0 &\ \ 1 &\ \ 2 \\ \ \ 0 &\ \ 0 &\ \ 0 &\ \ 0 &\ \ 2 &\ \ 2 \end{vmatrix}\begin{array}{l} \\ \\ \\ \\ \\ R_6-2R_5 \end{array} = \begin{vmatrix} \ \ 2 &\ \ 1 & -1 &\ \ 0 &\ \ 3 &\ \ 1 \\ \ \ 0 &\ \ 3 &\ \ 2 & -1 &\ \ 0 &\ \ 2 \\ \ \ 0 &\ \ 0 & -1 &\ \ 4 &\ \ 1 & -1 \\ \ \ 0 &\ \ 0 &\ \ 0 &\ \ 4 & -2 &\ \ 3 \\ \ \ 0 &\ \ 0 &\ \ 0 &\ \ 0 &\ \ 1 &\ \ 2 \\ \ \ 0 &\ \ 0 &\ \ 0 &\ \ 0 &\ \ 0 & -2 \end{vmatrix} \end{align*} $$ The result is upper triangular, so by Property 7 the determinant is the product of the diagonal entries. Since no operation changed the value, $$\det(A) = (2)(3)(-1)(4)(1)(-2) = 48.$$Compare this with the Laplace expansion, which would require evaluating $\tfrac{6!}{2} = 360$ separate $2\times 2$ determinants. A handful of integer row operations is far faster and far less error prone — the method you will want on the midterm.
Exercise 19 - Lemma: Odd-Order Skew-Symmetric Matrices are Singular¶
A matrix is skew-symmetric if $A^\mathsf{T} = -A$. Prove that if $A$ is an $n\times n$ skew-symmetric matrix and $n$ is odd, then $\det(A) = 0$ (so $A$ is singular).
View Solution
By Property 10, $\det(A) = \det\!\left(A^\mathsf{T}\right)$. Since $A$ is skew-symmetric, $A^\mathsf{T} = -A$, so $$\det(A) = \det\!\left(A^\mathsf{T}\right) = \det(-A).$$ Using $\det(cA) = c^n\det(A)$ with $c = -1$, $$\det(-A) = (-1)^n\det(A).$$ Because $n$ is odd, $(-1)^n = -1$, and therefore $$\det(A) = -\det(A) \quad\Longrightarrow\quad 2\det(A) = 0 \quad\Longrightarrow\quad \det(A) = 0.$$ Hence $A$ is singular (Property 8). $\blacksquare$Note that the hypothesis that $n$ is odd is essential: the skew-symmetric matrix $\begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$ has determinant $4 \neq 0$.
Exercise 20 - Lemma: Determinant of a Power¶
Prove that for any $n\times n$ matrix $A$ and any positive integer $k$, $$\det\!\left(A^k\right) = \det(A)^k.$$
View Solution
We argue by induction on $k$.Base case ($k = 1$): $\det\!\left(A^1\right) = \det(A) = \det(A)^1$.
Inductive step: Suppose $\det\!\left(A^k\right) = \det(A)^k$ for some positive integer $k$. Since $A^{k+1} = A^k A$, Property 9 (the determinant of a product is the product of the determinants) gives $$\det\!\left(A^{k+1}\right) = \det\!\left(A^k A\right) = \det\!\left(A^k\right)\det(A) = \det(A)^k\,\det(A) = \det(A)^{k+1}.$$ By the principle of mathematical induction, $\det\!\left(A^k\right) = \det(A)^k$ for every positive integer $k$. $\blacksquare$
If $A$ is invertible, the result extends to $k = 0$ (since $A^0 = I_n$ and $\det(A)^0 = 1$) and to negative integers using Corollary 9.
