Introdunction to Linear Algebra-第5章-Determinants

[TOC]
《Introdunction to Linear Algebra》的第 5 章: Determinants.

1. The determinant of $A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$ is $a d-b c$. Singular matrix $A=\left[\begin{array}{ll}a & x a \\ c & x c\end{array}\right]$ has $\text{det} =\mathbf{0}$.
2. $\begin{array}{l}\text { Row exchange } \\ \text { reverses signs }\end{array} P A=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=\left[\begin{array}{ll}c & d \\ a & b\end{array}\right] \text{has} \operatorname{det} P A=b c-a d=-\operatorname{det} \boldsymbol{A}$.
3. The determinant of $\left[\begin{array}{cc}x a+y A & x b+y B \\ c & d\end{array}\right]$ is $x(a d-b c)+y(A d-B c)$. $\begin{array}{l}\text { Det is linear in } \\ \text { row } \mathbf{1} \text { by itself}\end{array}$.
4. Elimination $E A=\left[\begin{array}{cc}a & b \\ 0 & d-\frac{c}{a} b\end{array}\right] \quad \operatorname{det} E A=a\left(d-\frac{c}{a} b\right)=$ product of pivots $=\operatorname{det} A$.
5. If $A$ is $n$ by $n$ then $1, 2, 3, 4$ remain true: $\operatorname{det}=\mathbf{0}$ when $A$ is singular, det reverses sign when rows are exchanged, det is linear in row $1$ by itself, det = product of the pivots. Always $\operatorname{det} B A=(\operatorname{det} B)(\operatorname{det} A)$ and $\operatorname{det} A^{\mathrm{T}}=\operatorname{det} A$. This is an amazing number.

秩的应用:

1. Determinants give $A^{- 1}$ and $A^{- 1} \boldsymbol{b}$ (this formula is called Cramer’s Rule).
2. When the edges of a box are the rows of $A$, the volume is $|\text{det} A|$.
3. For $n$ special numbers $\lambda$, called eigenvalues, the determinant of $A - \lambda I$ is zero.
   This is a truly important application and it fills Chapter 6.

5.1 The Properties of Determinants

The Properties of the Determinant

直接上秩的定义太难, 先从其性质说起. 下面介绍 10 个性质, 其中前 3 个性质能够确定秩的本质, 因此也能推导出其余 7 个性质.

1. The determinant of the $n$ by $n$ identity matrix is 1.

$$
\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|=1 \quad \text { and } \quad\left|\begin{array}{ccc}
1 & & \\
& \ddots & \\
& & 1
\end{array}\right|=1
$$

2. The determinant changes sign when two rows are exchanged (sign reversal), 对于列的交换也是一样的.

3. The determinant is a linear function of each row separately.

秩的运算规则:

• multiply row $1$ by any number $t$ det is multiplied by $t$: $ \left|\begin{array}{cc}t a & t b \\ c & d\end{array}\right|=t\left|\begin{array}{ll}a & b \\ c & d\end{array}\right| $.

• add row $1$ of $A$ to row $1$ of $A^{\prime}$: $\left|\begin{array}{cc}a+a^{\prime} & b+b^{\prime} \\ c & d\end{array}\right|=\left|\begin{array}{ll}a & b \\ c & d\end{array}\right|+\left|\begin{array}{cc}a^{\prime} & b^{\prime} \\ c & d\end{array}\right|$.

此性质的关键在于一行动, 其余行不动.
如果多行变化的话, 就成了幂乘.

$$
\left|\begin{array}{ll}
t & 0 \\
0 & t
\end{array}\right|=t^{2}
$$

因此秩其实就是体积(2 维时为面积).

4. If two rows of $A$ are equal, then $\text{det} A= 0$.

5. Subtracting a multiple of one row from another row leaves $\text{det} A$ unchanged.

推论: 对于 LU 分解: $\text{det} A= ± \text{det} U$.

6. $A$ matrix with a row of zeros has $\text{det} A= 0$.

7. If $A$ is triangular then $\text{det} A= a_{11} a_{22} \cdots a_{nn}$= product of diagonal entries.

$$
\text{Triangular} \left|\begin{array}{ll}a & b \\ 0 & d\end{array}\right|=a d \quad \text{and also} \quad\left|\begin{array}{ll}a & 0 \\ c & d\end{array}\right|=a d
$$

8. If $A$ is singular then $\text{det} A=0$. If $A$ is invertible then $\text{det} A \ne 0$.

Multiply pivots $\text{det} A= ± \text{det} U = ±$ (product of the pivots).

9. The determinant of $AB$ is $\text{det} A$ times $\text{det} B$: $|AB|= |A| |B|$.

$A \text{ times } A^{-1}$ : $AA^{-1} = I \text{ so } (\operatorname{det} A)\left(\operatorname{det} A^{-1}\right)=\operatorname{det} I=1$.

9. The transpose $A^T$ has the same determinant as $A$.

5.2 Permutations and Cofactors

本节概要

1. $\mathbf{2} \text{ by } \mathbf{2}$: $a d-b c$ has $2 !$ terms with $\pm$ signs. $\boldsymbol{n}$ by $\boldsymbol{n}: \text{det} A$ adds $\boldsymbol{n} !$ terms with $\pm$ signs.

2. For $n=3$, $\operatorname{det} A$ adds $3 !=6 $ terms. Two terms are $+a_{12} a_{23} a_{31} $ and $-a_{13} a_{22} a_{31}$ . Rows $1,2,3$ and columns $1,2,3$ appear once in each term.

3. That minus sign came because the column order $3,2,1$ needs one exchange to recover $1,2,3$ .

4. The six terms include $+a_{11} a_{22} a_{33}-a_{11} a_{23} a_{32}=a_{11}\left(a_{22} a_{33}-a_{23} a_{32}\right)=a_{11}\left(\right. \textbf{ cofactor } \left.C_{11}\right) $.

5. Always $\text{det} A=a_{11} C_{11}+a_{12} C_{12}+\cdots+a_{1 n} C_{1 n}$. Cofactors are determinants of size $n-1$.

The determinant of an $n \times n$ matrix can be found in three ways:

1. Multiply the $n$ pivots (times 1 or -1). This is the pivot formula.
2. Add up $n!$ terms (times 1 or -1). This is the “big” formula.
3. Combine $n$ smaller determinants (times 1 or -1). This is the cofactor formula.

the pivot formula

$A=LU$:
$$
\operatorname{det} A=(\operatorname{det} L)(\operatorname{det} U)=(1)\left(d_{1} d_{2} \cdots d_{n}\right)
$$
$PA=LU$:
$$
(\operatorname{det} P)(\operatorname{det} A)=(\operatorname{det} L)(\operatorname{det} U) \quad \text { gives } \quad \operatorname{det} A=\pm\left(d_{1} d_{2} \cdots d_{n}\right)
$$

$A = LU$ 可以理解为多个右下角低维度矩阵分解 $A_k = L_kU_k$. Each pivot is a ratio of determinants:

Pivots from determinants:
The $k$th pivot is

$$
d_{k}=\frac{d_{1} d_{2} \cdots d_{k}}{d_{1} d_{2} \cdots d_{k-1}}=\frac{\operatorname{det} A_{k}}{\operatorname{det} A_{k-1}}.
$$

The Big Formula for Determinants

有没有直接通过 $A$ 中元素算出秩的方法呢?
从每行中选出不同列的元素相乘, 然后乘以行交换正负符号系数, 最后其和为秩. 对于 $n$ 维方阵共有 $n!$ 项.

$$ \begin{array}{l|lll|} 3 \text { by } 3 & a_{11} & a_{12} & \boldsymbol{a_{13}} \\\\ \text { determinant } & \boldsymbol{a_{21}} & a_{22} & a_{23} \\\\ & a_{31} & \boldsymbol{a_{32}} & a_{33} \end{array} =\begin{array}{c} +a_{11} a_{22} a_{33}+a_{12} a_{23} a_{31}+\boldsymbol{a}_{\mathbf{1 3}} \boldsymbol{a}_{\mathbf{2 1}} \boldsymbol{a}_{\mathbf{3 2}} \\\\ -a_{11} a_{23} a_{32}-a_{12} a_{21} a_{33}-a_{13} a_{22} a_{31} \end{array} $$ $$ \begin{aligned} \left|\begin{array}{lll} a_{11} & a_{12} & a_{13} \\\\ a_{21} & a_{22} & a_{23} \\\\ a_{31} & a_{32} & a_{33} \end{array}\right| &=\left|\begin{array}{lll} a_{11} & & \\\\ & a_{22} & \\\\ & & a_{33} \end{array}\right|+\left|\begin{array}{lll} & a_{12} & \\\\ & & a_{23} \\\\ a_{31} & & \end{array}\right|+\left|\begin{array}{lll} & & \boldsymbol{a_{13}} \\\\ \boldsymbol{a_{21}} & & \\\\ & \boldsymbol{a_{32}} & \end{array}\right| \\\\ &+\left|\begin{array}{ccc} a_{11} & & \\\\ & & a_{23} \\\\ & a_{32} & \end{array}\right|+\left|\begin{array}{lll} & a_{12} & \\\\ a_{21} & & \\\\ & a_{32} & & a_{33} \end{array}\right|+\left|\begin{array}{lll} & & a_{13} \\\\ & a_{22} & \\\\ a_{31} & & \end{array}\right| \end{aligned} $$ $$ \operatorname{det} A = a_{11} a_{22} a_{33}\left|\begin{array}{lll}1 & & \\\\ & 1 & \\\\ & & 1\end{array}\right| +a_{12} a_{23} a_{31}\left|\begin{array}{lll} & 1 & \\\\ & & 1 \\\\ 1 & & \end{array}\right| +a_{13} a_{21} a_{32}\left|\begin{array}{lll} & & 1 \\\\ 1 & &\\\\ & 1 & \end{array}\right| \\\\ +a_{11} a_{23} a_{32}\left|\begin{array}{lll}1 & & \\\\ & & 1\\\\ & 1 & \end{array}\right| +a_{12} a_{21} a_{33}\left|\begin{array}{lll} & 1 & \\\\1 & & \\\\ & & 1\end{array}\right| +a_{13} a_{22} a_{31}\left|\begin{array}{lll} & & 1\\\\& 1 & \\\\ 1 & & \end{array}\right|. $$

$\operatorname{det} A=$ sum over all $\mathbf{n} ! $ column permutations $P=(\alpha, \beta, \ldots, \omega) $

$=\sum(\operatorname{det} P) a_{1 \alpha} a_{2 \beta} \cdots a_{n \omega}=\text { BIG FORMULA }$

例子:
$$
\text { The determinant of } Z=\left|\begin{array}{llll}
1 & 0 & \boldsymbol{a} & 0 \\
0 & 1 & \boldsymbol{b} & 0 \\
0 & 0 & \boldsymbol{c} & 0 \\
0 & 0 & \boldsymbol{d} & 1
\end{array}\right| \text { is } \boldsymbol{c}
$$

Determinant by Cofactors

矩阵可以分块, 那有没有分块嵌套求解秩的方法呢?

$\operatorname{det} A=a_{11}\left(a_{22} a_{33}-a_{23} a_{32}\right)+a_{12}\left(a_{23} a_{31}-a_{21} a_{33}\right)+a_{13}\left(a_{21} a_{32}-a_{22} a_{31}\right)$

即:

$$
\left|\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right|=\left|\begin{array}{lll}
a_{11} & & \\
& a_{22} & a_{23} \\
& a_{32} & a_{33}
\end{array}\right|+\left|\begin{array}{lll}
& a_{12} & \\
a_{21} & & a_{23} \\
a_{31} & & a_{33}
\end{array}\right|+\left|\begin{array}{lll}
& & a_{13} \\
a_{21} & a_{22} & \\
a_{31} & a_{32}
\end{array}\right|
$$

The cofactor expansion is $\text{det}A = a_{11}C_{11} + a_{12}C_{12} + · · · + a_{1n}C_{1n}$. 其中 $C_{1j} = ( -1)^{1+j} \text{det} M_{1j}$.

对于除了第一行的其他行也可以进行此操作:

The determinant is the dot product of any row $i$ of $A$ with its cofactors using other rows:

COFACTOR FORMULA $\operatorname{det} A=a_{i 1} C_{i 1}+a_{i 2} C_{i 2}+\cdots+a_{i n} C_{i n}$.
Each cofactor $C_{i j}$ (order $n-1$, without row $i$ and column $j$ ) includes its correct sign:
Cofactor: $C_{i j}=(-1)^{i+j} \operatorname{det} M_{i j} $

对于列也是可以如此操作:

Cofactors down column: $j \quad \operatorname{det} A=a_{1 j} C_{1 j}+a_{2 j} C_{2 j}+\cdots+a_{n j} C_{n j}$

Cofactors are useful when matrices have many zeros.

5.3 Cramer’s Rule, Inverses, and Volumes

本节概要

1. $A^{-1}$ equals $C^{\mathbf{T}} / \operatorname{det} A$. Then $\left(A^{-1}\right)_{i j}=\operatorname{cofactor} C_{j i}$ divided by the determinant of $A$.
2. Cramer’s Rule computes $\boldsymbol{x}=A^{-1} \boldsymbol{b}$ from $x_{j}=\operatorname{det}(A$ with column $j$ changed to $\boldsymbol{b}) / \operatorname{det} A$.
3. Area of parallelogram $=|a d-b c|$ if the four corners are $(0,0),(a, b),(c, d)$, and $(a+c, b+d)$.
4. Volume of box $=|\operatorname{det} \boldsymbol{A}|$ if the rows of $A$ (or the columns of $A$ ) give the sides of the box.
5. The cross product $\boldsymbol{w}=\boldsymbol{u} \times \boldsymbol{v} \text{ is } \operatorname{det}\left[\begin{array}{ccc}\boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \\\\ u_{1} & u_{2} & u_{3} \\\\ v_{1} & v_{2} & v_{3}\end{array}\right] . \quad \begin{array}{l}\text { Notice } \boldsymbol{v} \times \boldsymbol{u}=-(\boldsymbol{u} \times \boldsymbol{v}) \\\\ \boldsymbol{w}_{1}, \boldsymbol{w}_{2}, \boldsymbol{w}_{3} \text { are cofactors of row } 1 . \\\\ \text { Notice } \boldsymbol{w}^{\mathrm{T}} \boldsymbol{u}=0 \text { and } \boldsymbol{w}^{\mathrm{T}} \boldsymbol{v}=0 .\end{array} $

CRAMER’s RULE If $\operatorname{det} A$ is not zero, $A \boldsymbol{x} = \boldsymbol{b}$ is solved by determinants:

$x_{1}=\frac{\operatorname{det} B_{1}}{{\operatorname{det} A}} \quad x_{2}=\frac{\operatorname{det} B_{2}}{\operatorname{det} A} \quad \cdots \quad x_{n}=\frac{\operatorname{det} B_{n}}{{\operatorname{det} A}}$

The matrix $B_{j}$ has the $j$th column of $A$ replaced by the vector $\boldsymbol{b}$.

下面详细展开:
$$
\text { Key idea }A\left[\begin{array}{lll}
x_{1} & 0 & 0 \\
x_{2} & 1 & 0 \\
x_{3} & 0 & 1
\end{array}\right]=\left[\begin{array}{lll}
b_{1} & a_{12} & a_{13} \\
b_{2} & a_{22} & a_{23} \\
b_{3} & a_{32} & a_{33}
\end{array}\right]=B_{1} \\
\text { Product rule } \quad(\operatorname{det} A)\left(x_{1}\right)=\operatorname{det} B_{1} \quad \text { or } \quad x_{1}=\frac{\operatorname{det} B_{1}}{\operatorname{det} A} \\
\text { Same idea }\left[\begin{array}{lll}
a_{1} & a_{2} & a_{3}
\end{array}\right]\left[\begin{array}{lll}
1 & x_{1} & 0 \\
0 & x_{2} & 0 \\
0 & x_{3} & 1
\end{array}\right]=\left[\begin{array}{lll}
a_{1} & b & a_{3}
\end{array}\right]=B_{2}
$$

复杂度:
To solve an $n$ by $n$ system, Cramer’s Rule evaluates $n + 1$ determinants.

Cramer’s Rule is inefficient for numbers but it is well suited to letters. 例如如下方式求解逆矩阵:

$$
\begin{array}{l}
\text { Columns of } A^{-1} \\
\text { are } \boldsymbol{x} \text { and } \boldsymbol{y}
\end{array}\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\left[\begin{array}{l}
x_{1} \\
x_{2}
\end{array}\right]=\left[\begin{array}{l}
1 \\
0
\end{array}\right] \quad\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\left[\begin{array}{l}
y_{1} \\
y_{2}
\end{array}\right]=\left[\begin{array}{l}
0 \\
1
\end{array}\right]
$$

Those share the same matrix $A$ . We need $|A|$ and four determinants for $x_{1}, x_{2}, y_{1}, y_{2}$ :

$$
\left|\begin{array}{ll}
a & b \\
c & d
\end{array}\right| \text { and }\left|\begin{array}{ll}
1 & b \\
0 & d
\end{array}\right| \quad\left|\begin{array}{ll}
a & 1 \\
c & 0
\end{array}\right| \quad\left|\begin{array}{ll}
0 & b \\
1 & d
\end{array}\right| \quad\left|\begin{array}{ll}
a & 0 \\
c & 1
\end{array}\right|
$$

$$
x_{1}=\frac{d}{|A|}, x_{2}=\frac{-c}{|A|}, y_{1}=\frac{-b}{|A|}, y_{2}=\frac{a}{|A|} \text { and then } A^{-1}=\frac{1}{a d-b c}\left[\begin{array}{rr}
d & -b \\
-c & a
\end{array}\right]
$$

the determinant of each $B_j$ in Cramer’s Rule is a cofactor of $A$.

The $i, j$ entry of $A^{-1}$ is the cofactor $C_{j i}\left(\right. \text{ not } \left.C_{i j}\right)$ divided by $\operatorname{det} A$.

FORMULA FOR $A^{-1}$:

$$ \left(A^{-1}\right)_{i j}=\frac{C_{j i}}{\operatorname{det} A} \quad and \quad A^{-1}=\frac{C^{\mathrm{T}}}{\operatorname{det} A} $$

Direct proof of the formula $A^{-1}=C^{\mathrm{T}} / \operatorname{det} A$ This means $ A C^{\mathrm{T}}=(\operatorname{det} A) I$:
$$
\left[\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right]\left[\begin{array}{lll}
C_{11} & C_{21} & C_{31} \\
C_{12} & C_{22} & C_{32} \\
C_{13} & C_{23} & C_{33}
\end{array}\right]=\left[\begin{array}{ccc}
\operatorname{det} A & 0 & 0 \\
0 & \operatorname{det} A & 0 \\
0 & 0 & \operatorname{det} A
\end{array}\right]
$$

$$
a_{11} C_{11}+a_{12} C_{12}+a_{13} C_{13}=\operatorname{det} A \quad \text { This is exactly the cofactor rule! }
$$

Area of a Triangle

对三角形的面积可以不使用开方而是使用秩解出:

The triangle with corners $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ has area $=\frac{\text { determinant }}{2} $:
Area of triangle :

$$
\frac{1}{2}\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{array}\right| \quad Area =\frac{1}{2}\left|\begin{array}{ll}x_{1} & y_{1} \\ x_{2} & y_{2}\end{array}\right| \quad when \left(x_{3}, y_{3}\right)=(0,0)
$$

为什么求三角形面积可以使用求秩公式: 因为求面积满足秩的前 3 个特性, 使得面积就是秩.

1. When $A= I$, the parallelogram becomes the unit square. Its area is $\text{det} I= 1$.
2. When rows are exchanged, the determinant reverses sign. The absolute value (positive area) stays the same-it is the same parallelogram.
3. If row 1 is multiplied by $t$, 下图左侧 shows that the area is also multiplied by $t$. Sup­pose a new row ($x_1^{\prime}, y_1^{\prime}$ is added to ($x_1, y_1$) (keeping row 2 fixed). 下图右侧 shows that the solid parallelogram areas add to the dotted parallelogram area (because the two triangles completed by dotted lines are the same).
   

拓展维度, 高维度的体积本质上也是秩.

The Cross Product

DEFINITION The cross product of $\boldsymbol{u}=\left(u_{1}, u_{2}, u_{3}\right)$ and $\boldsymbol{v}=\left(v_{1}, v_{2}, v_{3}\right)$ is a vector:
$$
\boldsymbol{u} \times \boldsymbol{v}=\left|\begin{array}{ccc}\boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \\ u_{1} & u_{2} & u_{3} \\ v_{1} & v_{2} & v_{3}\end{array}\right|=\left(u_{2} v_{3}-u_{3} v_{2}\right) \boldsymbol{i}+\left(u_{3} v_{1}-u_{1} v_{3}\right) \boldsymbol{j}+\left(u_{1} v_{2}-u_{2} v_{1}\right) \boldsymbol{k} .
$$

This vector $\boldsymbol{u} \times \boldsymbol{v}$ is perpendicular to $\boldsymbol{u}$ and $\boldsymbol{v}$. The cross product $\boldsymbol{v} \times \boldsymbol{u}$ is $-(\boldsymbol{u} \times \boldsymbol{v})$.

叉积性质如下:

1. $\boldsymbol{v} \times \boldsymbol{u}$ reverses rows 2 and 3 in the determinant so it equals $-(\boldsymbol{u} \times \boldsymbol{v})$.
2. The cross product $\boldsymbol{u} \times \boldsymbol{v}$ is perpendicular to $\boldsymbol{u}$ (and also to $\boldsymbol{v}$). The direct proof is to watch terms cancel, producing a zero dot product:

$$
\boldsymbol{u} \cdot(\boldsymbol{u} \times \boldsymbol{v})=u_{1}\left(u_{2} v_{3}-u_{3} v_{2}\right)+u_{2}\left(u_{3} v_{1}-u_{1} v_{3}\right)+u_{3}\left(u_{1} v_{2}-u_{2} v_{1}\right)=0
$$

3. The cross product of any vector with itself (two equal rows) is $\boldsymbol{u} \times \boldsymbol{u} = 0$.

4. 长度与方向
   The length of $\boldsymbol{u} \times \boldsymbol{v}$ equals the area of the parallelogram with sides $\boldsymbol{u}$ and $\boldsymbol{v}$.

$$
|\boldsymbol{u} \times \boldsymbol{v}|=|\boldsymbol{u}||\boldsymbol{v}||\sin \theta| \quad \text { and } \quad|\boldsymbol{u} \cdot \boldsymbol{v}|=|\boldsymbol{u}||\boldsymbol{v}||\cos \theta|
$$

方向符合右手准则, 例子:

The cross product of $i=(1,0,0)$ and $j=(0,1,0)$ obeys the right hand rule. That cross product $k=i \times j$ goes up not down:

Thus $i \times j =k$. The right hand rule also gives $j \times k = i$ and $k \times i= j$. Note the cyclic order. In the opposite order (anti-cyclic) the thumb is reversed and the cross product goes the other way: $k \times j = -i$ and $i \times = -j $ and $j \times i = -k$.

DEFINITION The cross product is a vector with length $|\boldsymbol{u}||\boldsymbol{v}||\sin \theta|$. Its direction is perpendicular to $\boldsymbol{u}$ and $\boldsymbol{v}$. It points “up” or “down” by the right hand rule.

用力学举例
在 $(u_1,u_2,u_3)$ 的质心处施加 $F=(F_x,F_y,F_z)$的力,$\boldsymbol{u} \times F = 0$,没有旋转,不为0则有旋转,$||\boldsymbol{u}|| ||F|| sin \theta$ 为转矩动量.

Triple Product = Determinant = Volume

Triple product:

$$
\begin{array}{l}
(\boldsymbol{u} \times \boldsymbol{v}) \cdot \boldsymbol{w}=\left|\begin{array}{lll}
w_{1} & w_{2} & w_{3} \\
u_{1} & u_{2} & u_{3} \\
v_{1} & v_{2} & v_{3}
\end{array}\right|=\left|\begin{array}{ccc}
u_{1} & u_{2} & u_{3} \\
v_{1} & v_{2} & v_{3} \\
w_{1} & w_{2} & w_{3}
\end{array}\right|
\end{array}
$$

$( \boldsymbol{u} \times \boldsymbol{u=v}) · \boldsymbol{w} = 0$ exactly when the vectors $\boldsymbol{u}, \boldsymbol{v}, \boldsymbol{w}$ lie in the same plane.