91影视

Matrixis aset of numbers arranged inrowsandcolumnsso as to form a rectangular array.

The numbers are called the elements, or entries, of the matrix.

If there are $m$rows and $n$columns, the matrix is said to be a 鈥� $m$by $n$鈥� matrix, written 鈥� $m脳n$.鈥�

Consider a $2脳2$matrix,

$A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

To find: $T\left(\stackrel{鈫�}{x}\right)=\left[\begin{array}{c}\det \left[\begin{array}{cc}\stackrel{鈫�}{x}& \stackrel{鈫�}{w}\end{array}\right]\\ det\left[\begin{array}{cc}\stackrel{鈫�}{v}& \stackrel{鈫�}{x}\end{array}\right]\end{array}\right]$

$T\left(\stackrel{鈫�}{x}\right)=\left[\begin{array}{cc}{x}_1& b\\ x_2& d\\ a& x_1\\ c& x_2\end{array}\right]$

$T\left(\stackrel{鈫�}{x}\right)=\left[\begin{array}{c}d{x}_1-b{x}_2\\ -c{x}_1+a{x}_2\end{array}\right]$

This means the standard matrix of $T$is,

$B=\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$

To find,$\det B=ad-bc$

The relationship of the determinant is,

$\det B=\det A$

To show,

$$\begin{gathered} BA=\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]\left[\begin{array}{cc}a& b\\ c& d\end{array}\right] \\ BA=\left[\begin{array}{cc}ad-bc& 0\\ 0& ad-bc\end{array}\right] \\ BA=(ad-bc)\left[\begin{array}{cc}a& b\\ c& d\end{array}\right] \end{gathered}$$

If is non-invertible (but non-zero), then and are collinear, so .

$鈭�伪鈭�\mathrm{鈩�},(c,d)=伪(a,b)$

Then, forrole="math" localid="1660402432979" $\stackrel{鈫�}{x}$role="math" localid="1660402437557" $ker$role="math" localid="1660402428720" $B$, we have:

$$\begin{gathered} \left[\begin{array}{cc}伪b& -b\\ -伪a& a\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=0 \\ b\left(伪x_1-x_2\right)=0, \\ a\left(-伪x_1+x_2\right)=0 \\ 伪x_1-x_2=0 \\ {x}_2=伪x_{1}n \end{gathered}$$

So,

$\ker B=span\left(\left[\begin{array}{c}1\\ 伪\end{array}\right]\right)$

On the other hand,

$$\begin{gathered} imA=\left\{\left[\begin{array}{cc}a& b\\ 伪a& 伪b\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]:\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]鈭�{\mathrm{鈩�}}^2\right\} \\ imA=\left\{\left[\begin{array}{c}a{x}_1+b{x}_2\\ 伪\left(a{x}_1+b{x}_2\right)\end{array}\right]:\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]鈭�{\mathrm{鈩�}}^2\right\} \\ imA=span\left(\left[\begin{array}{c}1\\ 伪\end{array}\right]\right) \end{gathered}$$

Thus, ker $B=imA$. By similar computing, we can see that ker $A=imB$, too.

If$A$is invertible, then so is$B$, obviously. In that case,

We have,

$$\begin{gathered} A^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right] \\ {A}^{-1}=\frac{1}{\det B}B. \end{gathered}$$