- 3.2.1 Minor Matrices and Cofactors
- 3.2.2 Laplace Expansion
- 3.2.3 Adjugate
- 3.2.4 Properties of Adjugate
- 3.2.5 Inverse and Adjugate
- Theorem 2 - Adjugate and Inverse
- Exercise 7 - Adjugate and Inverse 2-1
- Exercise 8 - Adjugate and Inverse 2-2
- Exercise 9 - Adjugate and Inverse 2-3
- Exercise 10 - Adjugate and Inverse 2-4
- Theorem 3 - Adjugate and Product
- Proof - Theorem 3-1
- Exercise 11 - Adjugate and Product 2-2
- Exercise 12 - Adjugate and Product 2-3
- 3.2.6 Exercises
- Exercise 13 - Compute the Determinant
- Exercise 14 - Compute the Determinant
- Exercise 15 - Compute the Determinant
- Exercise 16 - Compute the Determinant
- Exercise 17 - Compute the Determinant
- Exercise 18 - Compute the Determinant
- Exercise 19 - Compute the Determinant
- Exercise 20 - Compute the Determinant
- Exercise 21 - Compute the Determinant
- Exercise 22 - Compute the Determinant
- 3.2.7 Complex-Valued Matrices and Determinants
- 3.2.8 More Exercises
- copyleft
Video Lecture 1: Determinant Formulas and Cofactors.¶
Determinant Formulas and CofactorsLeibniz Formula for Determinants (The Big Formula)¶
The first formula for calculating determinant was developed by Leibniz. Leibniz Formula is very tedious and bulky.
$$\det(A) = \sum_{\tau\in S_n} \textrm{sgn}(\tau)\prod_{i=1}^n a_{i\tau(i)} = \sum_{\sigma\in S_n} \textrm{sgn}(\sigma)\prod_{i=1}^n a_{\sigma(i)i}$$
where $\textrm{sgn}$ is the sign function (signature, signum, parity) of
permutations in the
permutation group $S_n$. The permutation group $S_n$ is the set of permutations of the first $n$ positive integers.
Evaluating a determinant using Leibniz formula requires $n!\cdot n$ multiplications and additions. A $3\times 3$ matrix has $3!=6$ terms and each term is a product of $3$ factors. A $4\times 4$ matrix requires $4!\cdot4=96$ multiplications and additions, a $5\times 5$ requires $5!\cdot 5=600$, and this is getting wildly impractical.
Pierre-Simon Laplace noticed a pattern in Leibniz' formula called the
Laplace expansion, or cofactor expansion.
Example 1 - A 3x3 Determinant¶
Consider the $3\times 3$ matrix $A$ defined by
$$ A = \begin{bmatrix}\ 4\ &\ \ 2\ & -3\ \\ \ 2\ & -1\ &\ \ 2\ \\ \ 2\ &\ \ 1\ & -4\ \end{bmatrix} = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} $$
There are $3!$ permutations of $\left\{\,1,2,3\,\right\}$,
$$ \begin{align*} \pi_1 &= (1, 2, 3 ) \\ \pi_2 &= (1, 3, 2 ) \\ \pi_3 &= (2, 1, 3 ) \\ \pi_4 &= (2, 3, 1 ) \\ \pi_5 &= (3, 1, 2 ) \\ \pi_6 &= (3, 2, 1 ) \\ \end{align*} $$
Here $\pi$ is for permutation and $S_3 = \left\{\,\pi_1,\ \pi_2,\ \pi_3,\ \pi_4,\ \pi_5,\ \pi_6 \,\right\}$. Using Leibniz' formula,
$$ \begin{align*} \det(A) &= \sum_{\tau\in S_n} \textrm{sgn}(\tau)\prod_{i=1}^n a_{i\tau(i)} \\ \\ &= \textrm{sgn}(\pi_1)a_{1\pi_1(1)}a_{2\pi_1(2)}a_{3\pi_1(3)} + \textrm{sgn}(\pi_2)a_{1\pi_2(1)}a_{2\pi_2(2)}a_{3\pi_2(3)} + \textrm{sgn}(\pi_3)a_{1\pi_3(1)}a_{2\pi_3(2)}a_{3\pi_3(3)} \\ \\ &\qquad+\ \textrm{sgn}(\pi_4)a_{1\pi_4(1)}a_{2\pi_4(2)}a_{3\pi_4(3)} + \textrm{sgn}(\pi_5)a_{1\pi_5(1)}a_{2\pi_5(2)}a_{3\pi_5(3)} + \textrm{sgn}(\pi_6)a_{1\pi_6(1)}a_{2\pi_6(2)}a_{3\pi_6(3)} \\ \\ &= a_{11}a_{22}a_{33} - a_{11}a_{23}a_{32} - a_{12}a_{21}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32} - a_{13}a_{22}a_{31} \\ \\ &= (4)(-1)(-4) - (4)(2)(1) - (2)(2)(-4) + (2)(2)(2) + (-3)(2)(1) - (-3)(-1)(2) \\ \\ &= 16 - 8 + 16 + 8 - 6 - 6 = 8 + 24 - 12 = 20 \end{align*} $$
If we pick a row (or column), there must be a factor from the selected row (or column) in each permutation $a_{i\tau(i)}$ ( or $a_{\sigma(i)i}$ ). If we choose the first row, then $a_{11}$ appears in two terms, $a_{12}$ appears in two terms, and $a_{13}$ appears in two terms. Grouping the terms together and factoring reveals
$$ \begin{align*} \det(A) &=\quad\, a_{11}\left( a_{22}a_{33} - a_{23}a_{32} \right) \\ &\quad\ + a_{12}\left( -a_{21}a_{33} + a_{23}a_{31} \right) \\ &\quad\ + a_{13}\left( a_{21}a_{32} - a_{22}a_{31} \right) \\ \end{align*} $$
We call the elements of our chosen row (or column) the factors, and the sum in the parentheses cofactors. This sum can be visualized
$$
\begin{vmatrix} a_{11} & \Large{\circ} & \Large{\circ} \\ \Large{\circ} & a_{22} & a_{23} \\ \Large{\circ} & a_{32} & a_{33} \end{vmatrix}
+ \begin{vmatrix} \Large{\circ} & a_{12} & \Large{\circ} \\ a_{21} & \Large{\circ} & a_{23} \\ a_{31} & \Large{\circ} & a_{33} \end{vmatrix} + \begin{vmatrix} \Large{\circ} & \Large{\circ} & a_{13} \\ a_{21} & a_{22} & \Large{\circ} \\ a_{31} & a_{32} & \Large{\circ} \end{vmatrix}
$$
Definition¶
Minor
Let $A = \left[ a_{ij} \right]$ be an $n\times n$ matrix. For every element $a_{ij}$ of matrix $A$, the minor (minor matrix) $A_{ij}$ is the submatrix defined by removing both the $i^{\text{th}}$ row and the $j^{\text{th}}$ column from matrix $A$.
We can identify nine $2\times 2$ submatrices of $3\times 3$ matrix $A$. For example consider $a_{\color{#CC0099}{1}\color{darkblue}{1}} = 4$. If one removes row one and column one, from matrix $A$,
The result is a $2\times 2$ minor $A_{11}$ is given by
$$ A_{11} = \begin{bmatrix} -1\ &\ \ 2\ \\ \ \ 1\ & -4 \end{bmatrix} $$
Notice that because the minor $A_{11}$ is a matrix, we use a capital letter. However there is a minor associated with every element of the original $n\times n$ matrix. As a result, we denote the minor submatrix using the same subscripts as the element of the matrix. In this way an $n\times n$ matrix will have $n^2$ minor submatrices, one for each element of the matrix.
Definition¶
Sign Function
The sign, signum, signature, or parity is the sign of the permutation from Leibniz' formula. In the Laplace expansion, there is a shortcut for computing the parity.
$$ \textrm{sgn}(a_{ij}) = (-1)^{i+j} $$
Hence the parity of the term in the Laplace expansion depends only on the indices of the factor.
This creates a pattern for the sign function in an $n\times n$ matrix
$$ \begin{bmatrix} \ +\ &\ -\ &\ +\ &\ -\ &\ \ &\ \cdots\ \\ \ -\ &\ +\ &\ -\ &\ +\ &\ \ &\ \cdots\ \\ \ +\ &\ -\ &\ +\ &\ -\ &\ \ &\ \cdots\ \\ \ -\ &\ +\ &\ -\ &\ +\ &\ \ &\ \cdots \ \\ \ \vdots\ &\ \vdots\ &\ \vdots\ &\ \vdots\ &\ \ddots\ &\ \vdots\ \\ \ \cdots\ &\ \cdots\ &\ \cdots\ &\ \cdots\ &\ \ &\ +\ \end{bmatrix} $$
Definition¶
Cofactor
For an $n\times n$ matrix $A = \left[ a_{ij} \right]$, the cofactor of an element of matrix $A$ is defined by the formula
$$ C_{ij} = \textrm{Cof}(a_{ij}) = \textrm{sgn}(a_{ij})\det(A_{ij}) = (-1)^{i+j}|A_{ij}| $$
In our example,
$$ \begin{align*} \det(A) &= a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13} \\ \\ &= (4)\begin{vmatrix} -1\ &\ \ 2\ \\ \ \ 1\ & -4\ \end{vmatrix} - (2)\begin{vmatrix}\ \ 2\ &\ \ 2\ \\ \ \ 2\ & -4\ \end{vmatrix} + (-3)\begin{vmatrix}\ \ 2\ & -1\ \\ \ \ 2\ &\ \ 1\ \end{vmatrix} \\ \\ &= (4)(4-2) - (2)(-8-4) - (3)(2+2) \\ \\ &= (4)(2) + (2)(12) - (3)(4) = 8 + 24 - 12 = 20 \end{align*} $$
Definition¶
Laplace Expansion
Let $A = \left[ a_{ij} \right]$ be an $n\times n$ matrix. The Laplace expansion or Laplace formula for the determinant of matrix $A$ can be expressed as one of $2n$ weighted sums.
The Laplace expansion along row $i$ is given by
$$\det(A) = \sum_{j=1}^n a_{ij}\,\textrm{Cof}(a_{ij}) = \sum_{j=1}^n a_{ij}\,\text{sgn}(a_{ij})\det(A_{ij}) = \sum_{j=1}^n a_{ij}(-1)^{i+j}\left|A_{ij}\right|$$
The Laplace expansion along column $k$ is given by
$$\det(A) = \sum_{j=1}^n a_{jk}\,\textrm{Cof}(a_{jk}) = \sum_{j=1}^n a_{jk}\,\textrm{sgn}(a_{jk})\det(A_{jk}) = \sum_{j=1}^n a_{jk}(-1)^{j+k}\left|A_{jk}\right|$$
It is important to understand that the Laplace Expansion of the determinant is not the definition of determinant; it is a property of the determinant. Leibniz formula, and the Laplace or Cofactor expansion are introduced first in most textbooks, but they are only computing strategies that result from the geometric definition.
How Does One Use the Laplace Expansion¶
Video Lecture 2: Computing Determinants with Cofactors.¶
Computing Determinants with CofactorsCofactor expansion used with the properties of determinants greatly simplify calculations.
Example 2 - Compute The Determinant¶
Compute the determinant of the $3\times 3$ matrix from example 1 using all of the properties of determinants including the Laplace expansion.
$$ |A| = \begin{vmatrix}\ 4\ &\ \ 2\ & -3\ \\ \ 2\ & -1\ &\ \ 2\ \\ \ 2\ &\ \ 1\ & -4\ \end{vmatrix} $$
Take the time to study this matrix. Look for ways to use the properties of determinants to simplify your work. You have already seen two different ways to compute the determinant in this example.
- Row two and row three look very similar. Perhaps subtract row two from row three and 2 times row two from row one. These type III row operations have no effect on the value of the determinant.
$$ \begin{align*} |A| &= \begin{vmatrix}\ 4\ &\ \ 2\ & -3\ \\ \ 2\ & -1\ &\ \ 2\ \\ \ 2\ &\ \ 1\ & -4\ \end{vmatrix}\begin{array}{l} R_1 - 2R_2 \\ \\ R_3-R_2 \end{array} = \begin{vmatrix}\ 0\ &\ 4\ & -7\ \\ \ 2\ & -1\ &\ 2\ \\ \ 0\ &\ 2\ & -6\ \end{vmatrix} \end{align*} $$
$$ \begin{align*} |A| &= 0\begin{vmatrix} -1\ &\ \ 2\ \\ \ \ 2\ & -6\ \end{vmatrix} - 2\begin{vmatrix}\ 4\ & -7\ \\ \ 2\ & -6\ \end{vmatrix}\begin{array}{l} R_1-R_2 \\ \\ \end{array} + 0\begin{vmatrix}\ 4\ & -7\ \\ -1\ &\ 2\ \end{vmatrix} \\ \\ &= 0 -2\begin{vmatrix}\ 2\ & -1\ \\ \ 2\ & -6\ \end{vmatrix}\begin{array}{l} \\ R_2-R_1 \end{array} + 0 = -2\begin{vmatrix}\ 2\ & -1\ \\ \ 0\ & -5\ \end{vmatrix} \\ \\ &= (-2)(2)(-5) = 20 \end{align*} $$
- Since row 1 and row 3 are so similar, subtract 2 times row 3 from row 1
$$ \begin{align*} |A| &= \begin{vmatrix}\ 4\ &\ \ 2\ & -3\ \\ \ 2\ & -1\ &\ \ 2\ \\ \ 2\ &\ \ 1\ & -4\ \end{vmatrix}\begin{array}{l} R_1 - 2R_3 \\ \\ \\ \end{array} = \begin{vmatrix}\ 0\ &\ \ 0\ &\ 5\ \\ \ 2\ & -1\ &\ \ 2\ \\ \ 2\ &\ \ 1\ & -4\ \end{vmatrix} \\ \\ \end{align*} $$
$$ \begin{align*} |A| &= 0|A_{11}| - 0|A_{12}| + 5\begin{vmatrix}\ 2\ & -1\ \\ \ 2\ &\ 1\ \end{vmatrix}\begin{array}{l} \\ R_2-R_1 \end{array} \\ \\ &= 0 - 0 + 5\begin{vmatrix}\ 2\ & -1\ \\ \ 0\ &\ 2\ \end{vmatrix} = 5(2)(2) = 20 \end{align*} $$
- Notice that column 1 and column 2 are similar. Instead subtract 2 times column 2 from column 1:
$$ \begin{align*} |A| &= \begin{vmatrix}\ 4\ &\ \ 2\ & -3\ \\ \ 2\ & -1\ &\ \ 2\ \\ \ 2\ &\ \ 1\ & -4\ \end{vmatrix}\begin{array}{l} C_1 - 2C_2 \\ \ \\ \ \end{array} = \begin{vmatrix}\ \ 0\ &\ \ 2\ & -3\ \\ \ 4\ & -1\ &\ \ 2\ \\ \ 0\ &\ \ 1\ & -4\ \end{vmatrix} \end{align*} $$
$$ \begin{align*} |A| &= 0 - 4\begin{vmatrix}\ \ 2\ & -3\ \\ \ \ 1\ & -4 \ \end{vmatrix} + 0 = -4(-8 + 3) = 20 \end{align*} $$
We do not compute the determinants of the minor submatrices $A_{11}$ and $A_{31}$ because their associated factors are zero.
Definition¶
Cofactor Matrix
The cofactor matrix associated with an $n\times n$ matrix $A$ is the matrix of cofactors. Recall that each cofactor
$$ C_{ij} = \textrm{sgn}(a_{ij})\det(A_{ij}) = (-1)^{i+j}|A_{ij}| $$
$$ \text{Cof}(A) := \begin{bmatrix} C_{ij} \end{bmatrix} = \begin{bmatrix} \ C_{11}\ &\ C_{12}\ &\ \cdots\ &\ C_{1n}\ \\ \ C_{21}\ &\ C_{22}\ &\ \cdots\ &\ C_{2n}\ \\ \ \vdots\ &\ \vdots\ &\ \ddots\ &\ \vdots\ \\ \ C_{n1}\ &\ C_{n2}\ &\ \cdots\ &\ C_{nn} \end{bmatrix} $$
Example 3 - Compute The Cofactor Matrix¶
Compute the cofactor matrix of matrix $A$ in Example 1:
$$
\begin{align*}
\textrm{Cof}(A) &= \begin{bmatrix} {\color{royalblue}\mathbf{+}}\begin{vmatrix} -1\ &\ \ 2\ \\ \ \ 1\ & -4\ \end{vmatrix} & {\color{crimson}\mathbf{-}}\begin{vmatrix}\ \ 2\ &\ \ 2\ \\ \ \ 2\ & -4\ \end{vmatrix} & {\color{royalblue}\mathbf{+}}\begin{vmatrix}\ \ 2\ & -1\ \\ \ \ 2\ &\ \ 1\ \end{vmatrix} \\ {\color{crimson}\mathbf{-}}\begin{vmatrix}\ \ 2\ & -3\ \\ \ \ 1\ & -4\ \end{vmatrix} & {\color{royalblue}\mathbf{+}}\begin{vmatrix}\ \ 4\ & -3\ \\ \ \ 2\ & -4\ \end{vmatrix} & {\color{crimson}\mathbf{-}}\begin{vmatrix}\ \ 4\ &\ \ 2\ \\ \ \ 2\ &\ \ 1\ \end{vmatrix} \\ {\color{royalblue}\mathbf{+}}\begin{vmatrix}\ \ 2\ & -3\ \\ -1\ &\ \ 2\ \end{vmatrix} & {\color{crimson}\mathbf{-}}\begin{vmatrix}\ \ 4\ & -3\ \\ \ \ 2\ &\ \ 2\ \end{vmatrix} & {\color{royalblue}\mathbf{+}}\begin{vmatrix}\ \ 4\ &\ \ 2\ \\ \ \ 2\ & -1\ \end{vmatrix} \end{bmatrix} \\
\\
&= \begin{bmatrix}\ \ 2\ &\ \ 12\ &\ \ 4\ \\ \ \ 5\ & -10\ &\ \ 0\ \\ \ \ 1\ & -14\ & -8\ \end{bmatrix}
\end{align*}
$$
Definition¶
Adjugate
The adjugate matrix associated with an $n\times n$ matrix $A$, is the transpose of the cofactor matrix.
$$ \text{adj}(A) := \text{Cof}(A)^\mathsf{T} = \begin{bmatrix} C_{ji} \end{bmatrix} = \begin{bmatrix} \ C_{11}\ &\ C_{21}\ &\ \cdots\ &\ C_{n1}\ \\ \ C_{12}\ &\ C_{22}\ &\ \cdots\ &\ C_{n2}\ \\ \ \vdots\ &\ \vdots\ &\ \ddots\ &\ \vdots\ \\ \ C_{1n}\ &\ C_{2n}\ &\ \cdots\ &\ C_{nn} \end{bmatrix} $$
The adjugate matrix or adjunct was historically called the adjoint or matrix adjoint. There are still old publications that refer to the adjugate as the classical adjoint. If you read adjoint of a matrix in an older publication insert classical in your notes. The term adjoint is unfortunately used in a wide variety of situations including three in linear algebra. We will refrain from using the term adjoint in this course.
These terms, adjugate, adjunct, and adjoint are used both as nouns and as adjectives in the literature. (sigh). I will use these terms in both parts of speech in order to acclimatize everyone to their uses.
Example 4 - Compute the Adjugate¶
Compute the adjugate of matrix $A$ in Example 1:
$$ \textrm{adj}(A) = \textrm{Cof}(A)^\mathsf{T} = \begin{bmatrix}\ \ 2\ &\ \ 5\ &\ \ 1\ \\ \ \ 12\ & -10\ & -14\ \\ \ \ 4\ &\ \ 0\ & -8\ \end{bmatrix} $$
Exercise 1 - Compute the Inverse¶
Compute the inverse of matrix $A = \begin{bmatrix}\ \ 7\ & -3\ \\ \ \ 2\ &\ \ 0\ \end{bmatrix}$
View Solution
A really slick application of the adjugate matrix is computing the inverse of a $2\times2$ matrix on a test. The adjugate of a $2\times 2$ is very straight-forward because the minor of each element is a scalar: $$ \textrm{adj}(A) = \begin{bmatrix}\ \ 0\ & -2\ \\ \ \ 3\ &\ \ 7\ \end{bmatrix}^\mathsf{T} = \begin{bmatrix}\ \ 0\ &\ \ 3\ \\ -2\ &\ \ 7\ \end{bmatrix} $$ So the diagonal elements switch places, and the anti-diagonal elements get multiplied by $-1$. Now $$ A^{-1} = \frac{1}{\det(A)}\textrm{adj}(A) = \frac{1}{6}\begin{bmatrix}\ \ 0\ &\ \ 3\ \\ -2\ &\ \ 7\ \end{bmatrix} $$ Done! (You should check by multiplying $A$ by $A^{-1}$)Theorem 1¶
Properties of Adjugate
Let $A$ be an $n\times n$ matrix and $c\in\mathbb{R}$ be a scalar, then
$A\,\textrm{adj}(A) = \det(A)I_n$.
$\textrm{adj}(I) = I$.
$\textrm{adj}(O) = O$.
$\textrm{adj}(A^\mathsf{T}) = \textrm{adj}(A)^\mathsf{T}$.
$\textrm{adj}(cA) = c^{n-1}\textrm{adj}(A)$.
If $n > 1$, then $A$ is non-singular if and only if $|\textrm{adj}(A)|\neq0$.
Proof of Theorem 1-1¶
Suppose $A$ is an $n\times n$ matrix. Recall the Laplace expansion of the determinant of $A$ using the $i^{\text{th}}$ row of matrix $A$:
$$ \begin{align*} |A| &= \sum_{j=1}^n a_{ij}\textrm{Cof}(a_{ij}) = \sum_{j=1}^n a_{ij}(-1)^{i+j}|A_{ij}| = \sum_{j=1}^n a_{ij}C_{ij} \\ \\ &= a_{i1}C_{i1} + a_{i2}C_{i2} + \cdots + a_{in}C_{in} \\ \\ &= \begin{bmatrix} a_{i1}\ & a_{i2}\ & \cdots\ & a_{in}\ \end{bmatrix} \begin{bmatrix} C_{i1}\ \\ C_{i2}\ \\ \vdots\ \\ C_{in}\ \end{bmatrix} \\ \\ &= \mathbf{a}^i\left(\text{adj}\mathbf{(A)}\right)_i \end{align*} $$
The determinant of matrix $A$ is the product of the $i^{\text{th}}$ row of matrix $A$ and the $i^{\text{th}}$ column of its adjugate, $\textrm{adj}(A)$.
What happens if we multiply the $i^{\text{th}}$ row of matrix $A$ times another column of $\textrm{adj}(A)$? When $k\neq i$, what is the value of
$$\mathbf{a}^i\left(\textrm{adj}(A)\right)_k\ \text{?}$$
The key here is to remember that when computing the Laplace expansion of $\det(A)$ along the $k^{\text{th}}$ row, the $k^{\text{th}}$ row is never used for any minor. Every minor $A_{kj}$, $1\le j\le n$ is obtained by removing the $k^{\text{th}}$ row and the $j^{\text{th}}$ column. So let us construct a new matrix $B$ such that the rows of $B$ are all identical to the rows of $A$, except the $k^{th}$ row. We'll define the $k^{\text{th}}$ row of matrix $B$ to be $\mathbf{a}^i$. Hence for $1\le j\le n$,
$$ \mathbf{b}^j = \left\{ \begin{array}{ccc} \mathbf{a}^j & \text{ if } & j\neq k \\ \mathbf{a}^i & \text{ if } & j=k \end{array} \right.\ . $$
Notice that due to this construction, the cofactors along the $k^{\text{th}}$ row of both matrices are the same. For $1\le j\le n$,
$$ \textrm{Cof}(b_{kj}) = \textrm{Cof}(a_{kj}) = C_{kj}. $$
Thus
$$ \mathbf{a}^i\left(\text{adj}(A)\right)_k = \sum_{j=1}^n a_{ij}C_{kj} = \sum_{j=1}^n b_{ij}C_{kj} = \mathbf{b}^i\left(\text{adj}(B)\right)_k. $$
Furthermore, since rows $i$ and $k$ are the same in matrix $B$ we have
$$ \mathbf{a}^i\left(\text{adj}(A)\right)_k = \mathbf{b}^i\left(\text{adj}(B)\right)_k = \mathbf{b}^k\left(\text{adj}(B)\right)_k = |B| = 0. $$
Since this is true for any row $k$ of matrix $A$ where $k\neq i$, we have
$$ \mathbf{a}^i\left(\text{adj}(A)\right)_j = \left\{ \begin{array}{ccc} \det(A) & \text{ if } & i=j \\ 0 & \text{ if } & i\neq j \end{array} \right.\ . $$
Therefore the matrix product
$$ A\,\textrm{adj}(A) = \left[ \mathbf{a}^i\left( \textrm{adj}(A) \right)_j \right] = \left[ \det(A)\delta_{ij} \right] = \det(A)I_n.\ \blacksquare $$
Exercise 2 - Theorem 1-2¶
Prove Theorem 1-2.
View Solution
Suppose that $A=I_n$ for some positive integer $n$. Using Theorem 1, part 1. $$ \textrm{adj}(I_n) = I_n\,\textrm{adj}(I_n) = |I_n|I_n = I_n. \blacksquare $$Exercise 3 - Theorem 1-3¶
Prove Theorm 1-3.
View Solution
This is left as an exerise for students.Exercise 4 - Theorem 1-4¶
Prove Theorem 1-4.
View Solution
Suppose that $A$ is an $n\times n$ matrix. If $C_{ij} = (-1)^{i+j}|A_{ij}|$, then $C_{ji} = (-1)^{j+i}|A_{ji}| = (-1)^{i+j}|(A^\mathsf{T})_{ij}|$. Therefore, $$\textrm{adj}(A^\mathsf{T}) = \left[ C_{ji} \right]^\mathsf{T} = \left( \left[ C_{ij} \right]^\mathsf{T} \right)^\mathsf{T} = \left(\textrm{adj}(A)\right)^\mathsf{T}.\ \blacksquare$$Exercise 5 - Theorem 1-5¶
Prove Theorem 1-5.
View Solution
This is left as an exerise for students.Exercise 6 - Theorem 1-6¶
Prove Theorem 1-6.
View Solution
$\Longrightarrow$ Suppose that integer $n>1$, and $A$ is an $n\times n$ singular matrix. Then $|A| = 0$, and at least one row of matrix $A$ is a linear combination of the other rows. Let row $i$, $1\le i\le n$, be a linear combination of the other rows of matrix$A$. Choose $1\le k\le n$ so that $i\neq k$ ($n>1$!). For every $1\le j\le n$, row $i$ of the minor matrix $B_{kj}$ is a linear combination of the other rows of $B_{kj}$. Hence $|B_{kj}|=0$. The means that the cofactor matrix $\textrm{Cof}(A)$ has a row of zeros. Thus $\textrm{adj}(A)$ has a column of zeros. Therefore $|\textrm{adj}(A)|=0$.
$\Longleftarrow$ Suppose that integer $n>1$, and $A$ is an $n\times n$ matrix such that $|\textrm{adj}(A)|=0$, then $|A|^m = |A||\textrm{adj}(A)| = |A|(0) = 0$. Hence $|A|=0$ and $A$ is a singular matrix.
We proved that for integer $n>1$, $n\times n$ matrix $A$ is singular if and only if $|\textrm{adj}(A)|=0$. Therefore for integer $n>1$, $n\times n$ matrix $A$ is non-singular if and only if $|\textrm{adj}(A)|\neq 0$. $\blacksquare$
$$a\,\textrm{adj}([a]) = [a]\textrm{adj}([a]) = |[a]|I_1 = a[1]$$ It is clear that if $a\neq 0$, then $\textrm{adj}([a])=[1]$. It is less clear that $\textrm{adj}([0])=[1]$. We may get a chance to re-visit this later.
Theorem 2¶
Adjugate and Inverse
If $A$ is a non-singular matrix, then
$A^{-1} = \frac{1}{\det(A)}\textrm{adj}(A)$.
$\det(\textrm{adj}(A)) = |A|^{n-1}$.
$\textrm{adj}(A)$ is non-singular and $\textrm{adj}(A)^{-1} = |A|^{-1}A$.
$\textrm{adj}(A)^{-1} = \textrm{adj}(A^{-1})$.
Exercise 7¶
Prove Theorem 2-1.
View Solution
Suppose that $A$ is a non-singular matrix. Then by Theorem 1, $A\,\textrm{adj}(A) = \det(A)I_n$. Since $A$ is non-singular, it is invertible. Multiplying both sides of this equation by $A^{-1}$ yields, $$\textrm{adj}(A) = \det(A)A^{-1}.$$ Using Theorem 3.1.7, $A$ is non-singular if and only if $\det(A)\neq 0$. Dividing both sides by $\det(A)$ give us the result. $$A^{-1} = \frac{1}{\det(A)}\textrm{adj}(A).\ \blacksquare$$Exercise 8¶
Prove Theorem 2-2.
View Solution
This is left as an exerise for students. There are two cases, 1. $A$ is singular, and 2. $A$ is non-singular.Exercise 9¶
Prove Theorem 2-3.
View Solution
Suppose that $A$ is an $n\times n$ non-singular matrix. Then $|A|\neq 0$. By Theorem 2-2, $|\textrm{adj}(A)| = |A|^{n-1} \neq 0$. Hence $\textrm{adj}(A)$ is non-singular. Furthmore $$\textrm{adj}(A) = |A|A^{-1}.$$ This implies that $$\left(\textrm{adj}(A)\right)^{-1} = \left(|A|A^{-1}\right)^{-1} = |A|^{-1}A.\ \blacksquare$$Exercise 10¶
Prove Theorem 2-4.
View Solution
This is left as an exerise for students.Theorem 3¶
Adjugate and Inverse
If $A$ and $B$ are $n\times n$ matrices, then
$\textrm{adj}(AB) = \textrm{adj}(B)\textrm{adj}(A)$.
If $k$ is a non-negative integer, then $\textrm{adj}(A^k) = \textrm{adj}(A)^k$.
If $A$ is non-singular, then for any integer $k\in\mathbb{Z}$, $\textrm{adj}(A^k) = \textrm{adj}(A)^k$.
Proof of Theorem 3-1¶
Let $A$ and $B$ be $n\times n$ matrices. Then if $A$ is singular, by Exercise 3.1.21, $AB$ is also singular. Hence
$$ |A||B| = (0)|B| = 0 = |AB|. $$
Likewise if $B$ is singular, then $|AB| = 0 = |A||B|$.
If both $A$ and $B$ are non-singular, then
$$ \textrm{adj}(B)\textrm{adj}(A) = |B|\,B^{-1}\,|A|\,A^{-1} = |A||B|B^{-1}A^{-1} = |AB|(AB)^{-1} = \textrm{adj}(AB).\ \blacksquare$$
Exercise 11¶
Prove Theorem 3-2.
View Solution
This is left as an exerise for students.Exercise 12¶
Prove Theorem 3-3.
View Solution
This is left as an exerise for students.Compute each determinant by inspection. This means use the properties of the determinant to compute the value of the determinant using mental arithmetic.
Exercise 13 - Compute the Determinant¶
Let $A = \begin{vmatrix} -3\ & -4\ \\ \ \ 2\ & -2\ \end{vmatrix}$. Compute $|A|$ using mental computations only.
View Solution
$$ \begin{vmatrix} -3\ & -4\ \\ \ \ 2\ & -2\ \end{vmatrix} = -3(-2) - (-4)(2) = 6 + 8 = 14 $$Exercise 14 - Compute the Determinant¶
Let $B = \begin{vmatrix}\ \ 2\ &\ \ 0\ &\ \ 2\ \\ \ \ 3\ &\ \ 0\ &\ \ 2\ \\ -3\ &\ \ 1\ & -2\ \end{vmatrix}$. Compute $|B|$ using mental computations only.
View Solution
$$ \begin{vmatrix}\ \ 2\ &\ \ 0\ &\ \ 2\ \\ \ \ 3\ &\ \ 0\ &\ \ 2\ \\ -3\ &\ \ 1\ & -2\ \end{vmatrix} = -0\left|M_{21}\right| + 0\left|M_{22}\right| - 1\begin{vmatrix}\ \ 2\ &\ \ 2\ \\ \ \ 3\ &\ \ 2\ \end{vmatrix} = -\left( 2(2) - 3(2) \right) = -\left( 4 - 6 \right) = 2 $$Exercise 15 - Compute the Determinant¶
Let $C = \begin{vmatrix} -3\ &\ \ 0\ &\ \ 0\ \\ -2\ &\ \ 3\ &\ \ 0\ \\ -3\ &\ \ 1\ &\ \ 3\ \end{vmatrix}$. Compute $|C|$ using mental computations only.
View Solution
$$ \begin{vmatrix} -3\ &\ \ 0\ &\ \ 0\ \\ -2\ &\ \ 3\ &\ \ 0\ \\ -3\ &\ \ 1\ &\ \ 3\ \end{vmatrix} = -3(3)(3) = -27 $$Exercise 16 - Compute the Determinant¶
Let $D = \begin{vmatrix}\ \ 2\ & -2\ &\ \ 2\ \\ -2\ &\ \ 3\ & -2\ \\ \ \ 1\ &\ \ 4\ &\ \ 1\ \end{vmatrix}$. Compute $|D|$ using mental computations only.
View Solution
Columns one and three are equal so $$ \begin{vmatrix}\ \ 2\ & -2\ &\ \ 2\ \\ -2\ &\ \ 3\ & -2\ \\ \ \ 1\ &\ \ 4\ &\ \ 1\ \end{vmatrix} = 0 $$Exercise 17 - Compute the Determinant¶
Let $E = \begin{vmatrix} -1\ &\ \ 4\ & -1\ \\ -4\ &\ \ 0\ &\ \ 4\ \\ \ \ 4\ &\ \ 0\ & -1\ \end{vmatrix}$. Compute $|E|$ using mental computations only.
View Solution
$$ \begin{vmatrix} -1\ &\ \ 4\ & -1\ \\ -4\ &\ \ 0\ &\ \ 4\ \\ \ \ 4\ &\ \ 0\ & -1\ \end{vmatrix} = -4\begin{vmatrix} -4\ &\ \ 4\ \\ \ \ 4\ & -1\ \end{vmatrix} + 0\left|M_{22}\right| - 0\left|M_{32}\right| = -4\left(\,(-4)(-1) - 4(4)\,\right) = -4\left( 4 - 16 \right) = 48 $$Exercise 18 - Compute the Determinant¶
Let $F = \begin{vmatrix} -3\ &\ \ 0\ & -3\ &\ \ 1\ \\ \ \ 1\ &\ \ 0\ &\ \ 0\ &\ \ 0\ \\ \ \ 2\ &\ \ 0\ & -3\ & -4\ \\ \ \ 1\ &\ \ 2\ & -1\ &\ \ 4\ \end{vmatrix}$. Use the Laplace expansion twice and compute $|F|$ by calculating one $2\times 2$ determinant only.
View Solution
$$ \begin{vmatrix} -3\ &\ \ 0\ & -3\ &\ \ 1\ \\ \ \ 1\ &\ \ 0\ &\ \ 0\ &\ \ 0\ \\ \ \ 2\ &\ \ 0\ & -3\ & -4\ \\ \ \ 1\ &\ \ 2\ & -1\ &\ \ 4\ \end{vmatrix} = (-1)(2)\begin{vmatrix} -3\ &\ \ 1\ \\ -3\ & -4\ \end{vmatrix} = -2\left( (-3)(-4) - (-3)(1) \right) = -2\left( 12 + 3 \right) = -30 $$Exercise 19 - Compute the Determinant¶
Let $G = \begin{vmatrix} -3\ & -2\ &\ \ 0\ & -1\ \\ \ \ 1\ &\ \ 1\ &\ \ 0\ & -2\ \\ \ \ 1\ & -2\ &\ \ 0\ &\ \ 4\ \\ \ \ 3\ &\ \ 3\ &\ \ 0\ & -4\ \end{vmatrix}$. Compute $|G|$ using mental computations only.
View Solution
This matrix has a column of zeros so $$ \begin{vmatrix} -3\ & -2\ &\ \ 0\ & -1\ \\ \ \ 1\ &\ \ 1\ &\ \ 0\ & -2\ \\ \ \ 1\ & -2\ &\ \ 0\ &\ \ 4\ \\ \ \ 3\ &\ \ 3\ &\ \ 0\ & -4\ \end{vmatrix} = 0 $$Exercise 20 - Compute the Determinant¶
Let $H = \begin{vmatrix} -3\ &\ \ 3\ & -1\ & -1\ \\ -3\ & -4\ &\ \ 1\ & -4\ \\ \ \ 0\ &\ \ 0\ &\ \ 0\ &\ \ 0\ \\ \ \ 3\ & -1\ & -2\ & -3\ \end{vmatrix}$. Compute $|H|$ using mental computations only.
View Solution
This matrix has a row of zeros so $$ \begin{vmatrix} -3\ &\ \ 3\ & -1\ & -1\ \\ -3\ & -4\ &\ \ 1\ & -4\ \\ \ \ 0\ &\ \ 0\ &\ \ 0\ &\ \ 0\ \\ \ \ 3\ & -1\ & -2\ & -3\ \end{vmatrix} = 0 $$Exercise 21 - Compute the Determinant¶
Let $M = \begin{vmatrix} -3\ &\ \ 4\ &\ \ 1\ \\ \ \ 3\ & -4\ & -1\ \\ \ \ 0\ & -1\ & -1\ \end{vmatrix}$. Compute $|M|$ using mental computations only.
View Solution
Row two is the negative of row 1 so $$ \begin{vmatrix} -3\ &\ \ 4\ &\ \ 1\ \\ \ \ 3\ & -4\ & -1\ \\ \ \ 0\ & -1\ & -1\ \end{vmatrix} = 0 $$Exercise 22 - Compute the Determinant¶
Let $N = \begin{vmatrix} -1\ &\ \ 0\ &\ \ 0\ &\ \ 0\ \\ \ \ 0\ &\ \ 3\ & -4\ &\ \ 0\ \\ \ \ 0\ &\ \ 2\ &\ \ 1\ &\ \ 0\ \\ \ \ 0\ &\ \ 0\ &\ \ 0\ & -2\ \end{vmatrix}$. Compute $|N|$ using mental computations only.
View Solution
$$ \begin{vmatrix} -1\ &\ \ 0\ &\ \ 0\ &\ \ 0\ \\ \ \ 0\ &\ \ 3\ & -4\ &\ \ 0\ \\ \ \ 0\ &\ \ 2\ &\ \ 1\ &\ \ 0\ \\ \ \ 0\ &\ \ 0\ &\ \ 0\ & -2\ \end{vmatrix} = -1\begin{vmatrix}\ \ 3\ & -4\ &\ \ 0\ \\ \ \ 2\ &\ \ 1\ &\ \ 0\ \\ \ \ 0\ &\ \ 0\ & -2\ \end{vmatrix} = -1(-2)\begin{vmatrix}\ \ 3\ & -4\ \\ \ \ 2\ &\ \ 1\ \end{vmatrix} = 2\left( 3(1) - 2(-4) \right) = 22 $$We will eventually be required to consider matrices and vectors of complex numbers. Recall that a complex number $a + bi$ has real part $\mathfrak{Re}(a+bi) = a$ and imaginary part $\mathfrak{Im}(a + bi) = b$.
Definition¶
Complex Conjugate
The conjugate of complex number $a + bi$ is given by
$$ \text{conj}(a + bi) = (a + bi)^* = a - bi $$
Often the complex conjugate of a number is denoted with an overbar for symbol, however if we use the overbar to denote a vector, we will need to use something else. Another common symbol is the asterisk or splat.
Example 5 - Conjugate¶
$$ A = \begin{bmatrix}\ 1-2i &\ 2+3i&\ 3+i\ \\ \ 4-i &\ 5+2i&\ 6-3i\ \\ \ 7+2i &\ 8-i &\ 9+2i\ \end{bmatrix} $$
then the conjugate matrix of matrix $A$ is given by
$$ A^* = \text{conj}(A) = \begin{bmatrix}\ 1+2i &\ 2-3i&\ 3-i\ \\ \ 4+i &\ 5-2i&\ 6+3i\ \\ \ 7-2i &\ 8+i &\ 9-2i\ \end{bmatrix} $$
Definition¶
Hermitian
The Hermitian operator or dagger $\dagger$ denotes the complex conjugate transpose of a complex-valued matrix. If $A$ is an $m\times n$ matrix; either complex-valued or real-valued, then
$$ A^H = A^{\dagger} = \left(A^*\right)^\mathsf{T} = \left(A^\mathsf{T}\right)^* = \left[ a_{ij} \right]^H = \left[ a_{ji}^* \right] $$
Since the transpose of a $1\times 1$ matrix or scalar is itself, we can denote complex conjugation using the Hermitian operator
$$ (a + bi)^H = (a + bi)^* = a - bi $$
Example 6 - Hermitian¶
If $A$ is the matrix in Example 5,
$$ A = \begin{bmatrix}\ 1-2i &\ 2+3i&\ 3+i\ \\ \ 4-i &\ 5+2i&\ 6-3i\ \\ \ 7+2i &\ 8-i &\ 9+2i\ \end{bmatrix}, $$
then the Hermitian of matrix $A$ is given by
$$ A^{H} = A^{\dagger} = \left(A^*\right)^\mathsf{T} = \left(A^\mathsf{T}\right)^* = \begin{bmatrix} \ 1+2i &\ 4+i &\ 7-2i\ \\ \ 2-3i &\ 5-2i &\ 8+i\ \\ \ 3-i &\ 6+3i &\ 9-2i\ \end{bmatrix} $$
Theorem 4¶
Properties of Complex Conjugate
If $z = a + bi$ and $w = c + di$ are complex numbers, then
$(z + w)^* = z^* + w^* = w^* + z^* = (w + z)^*$
$(zw)^* = z^*w^* = w^*z^* = (wz)^*$
$\left(\dfrac{z}{w}\right)^* = \dfrac{z^*}{w^*}$
$\left(z^*\right)^* = z$
$\mathfrak{Re}(z) = \dfrac{z + z^*}{2}$
$\mathfrak{Im}(z) = \dfrac{z - z^*}{2i}$
If $z = z^*$, then $z\in\mathbb{R}$; $z$ is a real number.
These same properties work for complex-valued matrices.
Theorem 5¶
Properties of Hermitian
If $A$ and $B$ are complex-valued $m\times n$ matrices, $C$ is a complex-valued $n\times p$ matrix, and $t\in\mathbb{C}$ is a complex scalar, then
$ \left(A+B\right)^H = A^H + B^H = B^H + A^H = (B + A)^H$
$ \left(AC\right)^H = C^HA^H$
$ \left(tA\right)^H = t^H\,A^H = t^*\,A^H$
$ \left(A^H\right)^H = A$
If $A = A^H$, then $A$ is $n\times n$, and for $1\le i\le n$, $a_{ii}\in\mathbb{R}$. We call a square matrix that is equal to its own Hermitian a Hermitian matrix.
If $A$ is a square, Hermitian, real-valued matrix, then $A$ is symmetric.
Example 7 - Hermitian Matrix¶
$$ B = \begin{bmatrix}\ 1 &\ 2-i &\ 3+2i\ \\ \ 2+i &\ 4 &\ 5-3i\ \\ \ 3-2i &\ 5+3i &\ 6 \end{bmatrix} $$
is an Hermitian matrix. Notice that this matrix is not symmetric.
Example 8 - 2x2 Determinant of Conjugate¶
Consider a $2\times 2$ matrix
$$ A = \begin{bmatrix}\ a\ &\ b\ \\ \ c\ &\ d\ \end{bmatrix} $$
and the conjugate matrix
$$ A^* = \begin{bmatrix}\ a^*\ &\ b^*\ \\ \ c^*\ &\ d^*\ \end{bmatrix} $$
The determinant
$$ \text{det}\left(A^*\right) = \begin{vmatrix}\ a^*\ &\ b^*\ \\ \ c^*\ &\ d^*\ \end{vmatrix} = a^*d^* - b^*c^* = (ad)^* - (bc)^* = (ad - bc)^* = |A|^* $$
Theorem 6¶
Conjugation Properties of Determinant
If matrix $A$ is $n\times n$, then
$\text{det}\left(A^*\right) = \text{det}(A)^*$
$\left|A^{\dagger}\right| = \left|A^H\right| = \left|\,\left(A^\mathsf{T}\right)^*\,\right| = \left|\,A^\mathsf{T}\,\right|^* = |A|^*$
$\textrm{adj}(A^*) = \left(\textrm{adj}(A)\right)^*$
$\textrm{adj}(A^H) = \left(\textrm{adj}(A)\right)^H$
If $A$ is symmetric, $A=A^\mathsf{T}$, then $\textrm{adj}(A)^\mathsf{T} = \textrm{adj}(A^\mathsf{T}) = \textrm{adj}(A)$, so $\textrm{adj}(A)$ is also symmetric.
If $A$ is Hermitian, $A=A^H$, then $\textrm{adj}(A)^H = \textrm{adj}(A^H) = \textrm{adj}(A)$, so $\textrm{adj}(A)$ is also Hermitian.
Compute each determinant.
Exercise 23 - Compute the Determinant¶
Let $A = \begin{vmatrix}\ \ 1+5i\ & -2-4i\ \\ -4+5i\ &\ \ 1+5i\ \end{vmatrix}$
View Solution
$$ \begin{align*} \begin{vmatrix}\ \ 1+5i\ & -2-4i\ \\ -4+5i\ &\ \ 1+5i\ \end{vmatrix} &= (1+5i)(1+5i) - (-2-4i)(-4+5i) \\ \\ &= (1 + 5i + 5i + 25i^2) - (8 -10i +16i - 20i^2) \\ \\ &= (-24 + 10i) - (28 + 6i) = -52 + 4i \end{align*} $$Exercise 24 - Compute the Determinant¶
Let $B = \begin{vmatrix}\ \ 1-i\ & -1+2i\ &\ \ 1+i\ \\ \ \ 4i\ & -1+3i\ &\ \ 5+2i\ \\ \ \ 3+5i\ &\ \ 5+3i\ &\ 2-4i\ \end{vmatrix}$
View Solution
$$ \begin{align*} \begin{vmatrix}\ \ 1-i\ & -1+2i\ &\ \ 1+i\ \\ \ \ 4i\ & -1+3i\ &\ \ 5+2i\ \\ \ \ 3+5i\ &\ \ 5+3i\ &\ 2-4i\ \end{vmatrix}\begin{array}{l} \\ \\ R_3+5R_1 \end{array} &= \begin{vmatrix}\ \ 1-i\ & -1+2i\ &\ \ 1+i\ \\ \ \ 4i\ & -1+3i\ &\ \ 5+2i\ \\ \ \ 8\ &\ \ 13i\ &\ 7+i\ \end{vmatrix}\begin{array}{l} \\ \\ R_3+2iR_2 \end{array} \\ \\ &= \begin{vmatrix}\ \ 1-i\ & -1+2i\ &\ \ 1+i\ \\ \ \ 4i\ & -1+3i\ &\ \ 5+2i\ \\ \ \ 0\ & -6+11i\ &\ 3+11i\ \end{vmatrix}\begin{array}{l} \\ R_2-2i(1+i)R_1 \\ \\ \end{array} \\ \\ &= \begin{vmatrix}\ \ 1-i\ & -1+2i\ &\ \ 1+i\ \\ \ \ 0\ &\ \ 1+9i\ &\ \ 9+2i\ \\ \ \ 0\ & -6+11i\ &\ 3+11i\ \end{vmatrix} \\ \\ &= (1-i)\begin{vmatrix}\ \ 1+9i\ &\ \ 9+2i\ \\ -6+11i\ &\ 3+11i\ \end{vmatrix} \\ \\ &= (1-i)\left( (1+9i)(3+11i) - (9+2i)(-6+11i) \right) \\ \\ &= (1-i)\left( -96 + 38i - (-76+87i) \right) \\ \\ &= (1-i)(-20-49i) = -20-49 -49i+20i = -69-29i \end{align*} $$Exercise 25 - Compute the Determinant¶
Let $C = \begin{vmatrix}\ \ 2-5i\ &\ \ 0\ &\ \ 0\ \\ -5-i\ & -4+3i\ &\ \ 0\ \\ -2-i\ &\ \ 4-3i\ &\ \ 5+2i\ \end{vmatrix}$
