MIT18.06 Linear Algebra (20)

lec20

2x2 inverse

$$\left[\begin{array}{ll}a & b \\c & d\end{array}\right]^{-1}=\frac{1}{a d-b c}\left[\begin{array}{cc}d & -b \\-c & a\end{array}\right]$$

inverse

$$A^{-1}=\frac{1}{\det A} C^{\top}$$

for cofactor $C$ , see cofactor

proof

check:

$$A C^\top=(\det A) I$$

$$\left[\begin{array}{ccc} a_{11} & \cdots & a_{1 n} \\ \vdots & & \vdots \\ a_{n 1} & \cdots & a_{n n} \end{array}\right]\left[\begin{array}{ccc} C_{11} & \cdots & C_{n 1} \\ \vdots & & \vdots \\ C_{1 n} & \cdots & C_{nn} \\ \end{array}\right] = \left[\begin{array}{cccc} \det A & 0 & 0 \\ 0 & \ddots & 0 \\ 0 & 0 & \det A \end{array}\right]$$

why get 0?

in $A=\left[\begin{array}{ll}a & b \\c & d\end{array}\right]^{-1},C^\top = \left[\begin{array}{cc}d & -b \\-c & a\end{array}\right]$ :

$$a(-b)+ba$$

means:

$$\begin{vmatrix} a & b\\ a & b\end{vmatrix} = 0$$

Cramer’s rule

Solve $Ax=b$

$$x=A^{-1}b=\frac{1}{\det A}C^\top b$$

then

$$x_j=\frac{\det B_j}{\det A}$$

where

$$B_j = \left[\begin{array}{cccccc} a_{11} & \cdots & a_{1\,j-1} & b_{1} & a_{1 \,j+1} & \cdots & a_{1 n} \\ a_{21} & \cdots & a_{2\,j-1} & b_{2} & a_{2 \, j+1} & \cdots & a_{2 n} \\ \vdots & & \vdots & \vdots & \vdots & & \vdots \\ a_{n 1} & \cdots & a_{n\,j-1} & b_{n} & a_{n\,j+1} & \cdots & a_{n n} \end{array}\right]$$

Because:

$$x_1 \det A= \det B_1=\begin{vmatrix} \vdots & A_{12} & \cdots & A_{1n}\\ b & \vdots & & \vdots \\ \vdots & A_{n2} & \cdots & A_{nn} \end{vmatrix} = b_1C_{11} + \cdots + b_nC_{n1} = C^\top b$$

determinate and volume

find area of $\triangle OAB$ :

$$\frac{1}{2} \begin{vmatrix} x_A & y_A \\ x_B & y_B\end{vmatrix}$$

find area of $\triangle ABC$ :

$$\frac{1}{2} \begin{vmatrix} x_A & y_A & 1\\ x_B & y_B & 1\\ x_C & y_C & 1\end{vmatrix}$$