[MIT OCW] 18.02 Multivariable Calculus - 3 笔记

MIT OCW

18.02 Multivariable Calculus

Matrices

Linear Transformation

If we want apply some operation on a vector $\mathbf{x}=(x_1,x_2)$ to change it into $\mathbf{x'}=(-x_2,x_1)$, the process can be described as a matrix.
Given a vector $\mathbf{x}=\begin{bmatrix}x_1\\x_2\end{bmatrix}$ and a matrix $\mathbf{R}=\begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}$, a new vector will be generated after we apply the matrix $\mathbf{R}$ to $\mathbf{x}$ that is $$\mathbf{R}\mathbf{x}=\begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}-x_2\\x_1\end{bmatrix}.$$

A more concrete example is that if we know how the target vector $\mathbf{u}=\begin{bmatrix}u_1\\u_2\\u_3\end{bmatrix}$ is derived from $\mathbf{x}=\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}$, for instance $$\left\{\begin{aligned}u_1&=2x_1+3x_2+3x_3\\u_2&=2x_1+4x_2+5x_3\\u_3&=x_1+x_2+2x_3\end{aligned}\right.,$$ then we can intermediately write such operation as a matrix that is $$\mathbf{A}=\begin{bmatrix}2&3&3\\2&4&5\\1&1&2\end{bmatrix}.$$
We now have the following equation $$\mathbf{A}\mathbf{x}=\mathbf{u}.$$

Minor and Cofactor

For a matrix $\mathbf{A}$, delete the row and the column where a specific element $a_{ij}$ exactly lies, and we get a smaller matrix. Treat it as a determinant, and that's called the minor of $a_{ij}$.
The cofactor is almost the same as the minor, but have something to do with signs, which means that for the element $a_{ij}$, we get the cofactor of it by multiplying $(-1)^{i+j}$ before the minor.
Write the cofactor of some element $a_{ij}$ as $A_{ij}$, and compose a new matrix $$\mathbf{C}=\begin{bmatrix} A_{11} & A_{12} & A_{13} & \cdots \\ A_{21} & A_{22} & A_{23} & \cdots \\ A_{31} & A_{32} & A_{33} & \cdots \\ \vdots & \cdots & \cdots & \cdots \end{bmatrix},$$ which is called the cofactor matrix, namely comatrix.

Transpose

For a matrix $\mathbf{A}$, the transpose matrix of it $\mathbf{A}^{\mathsf{T}}$ is derived from $\mathbf{A}$ where one's rows are the other's columns in the same order, vice versa.
Concretely, take $\mathbf{A}=\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$ as an example, the tranpose matrix $\mathbf{A}^{\mathsf{T}}=\begin{bmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \end{bmatrix}$.

Adjugate

For a matrix $\mathbf{A}$, the adjugate matrix of it $\operatorname{adj}(\mathbf{a})=\mathbf{C}^{\mathsf{T}}$, where $\mathbf{C}$ is the comatrix.

Inverse

The inverse of a matrix $\mathbf{A}$ with the notation $\mathbf{A}^{-1}$ fulfills $$\mathbf{A}\mathbf{A}^{-1}=\mathbf{I},\mathbf{A}^{-1}\mathbf{A}=\mathbf{I},$$ where $\mathbf{I}$ is an identity matrix.
For an equation $\mathbf{A}\mathbf{X}=\mathbf{B}$, if we want to solve $\mathbf{X}$, we just apply $\mathbf{A}^{-1}$ to the both sides thus get $$\mathbf{A}^{-1}(\mathbf{A}\mathbf{X})=\mathbf{A}^{-1}\mathbf{B},$$ namely $\mathbf{X}=\mathbf{A}^{-1}\mathbf{B}$.
To calculate the inverse of a matrix $A$, we need to know the adjugate matrix $\operatorname{adj}(\mathbf{A})$ and the determinant of $\mathbf{A}$ to in the following formula $$\mathbf{A}^{-1}=\frac{\operatorname{adj}(\mathbf{A})}{\det(\mathbf{A})}.$$
e.g. Calculate the inverse of $\mathbf{A}=\begin{bmatrix}2&3&3\\2&4&5\\1&1&2\end{bmatrix}$.
Solution
First, we calculate the comatrix
$$\mathbf{C}=\begin{bmatrix} +\begin{vmatrix}4&5\\1&2\end{vmatrix}& -\begin{vmatrix}2&5\\1&2\end{vmatrix}& +\begin{vmatrix}2&4\\1&1\end{vmatrix}\\ -\begin{vmatrix}3&3\\1&2\end{vmatrix}& +\begin{vmatrix}2&3\\1&2\end{vmatrix}& -\begin{vmatrix}2&3\\1&1\end{vmatrix}\\ +\begin{vmatrix}3&3\\4&5\end{vmatrix}& -\begin{vmatrix}2&3\\2&5\end{vmatrix}& +\begin{vmatrix}2&3\\2&4\end{vmatrix} \end{bmatrix}=\begin{bmatrix} +3 & -(-1) & +(-2)\\ -3 & +1 & -(-1)\\ +3 & -4 & +(-2) \end{bmatrix}=\begin{bmatrix} 3 & 1 & -2\\ -3 & 1 & 1\\ 3 & -4 & 2 \end{bmatrix},$$
and tranpose it
$$ \mathbf{C}^{\mathsf{T}}=\begin{bmatrix} 3 & -3 & 3\\ 1 & 1 & -4\\ -2 & 1 & 2 \end{bmatrix}. $$
Easy to get $\det(\mathbf{A})=3$, therefore the inverse can be calculated
$$ \mathbf{A}^{-1}=\frac13\begin{bmatrix} 3 & -3 & 3\\ 1 & 1 & -4\\ -2 & 1 & 2 \end{bmatrix}=\begin{bmatrix} 1 & -1 & 1\\ \frac13 & \frac13 & -\frac43\\ -\frac23 & \frac13 & \frac23 \end{bmatrix}. $$

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