求解矩阵方程

例子1:

求解矩阵$X$：$X+［\begin{array}{cc}\hfill 3.& \hfill 2\\ \hfill 1& \hfill 0\end{array}］=［\begin{array}{rr}\hfill 6& \hfill 3.\\ \hfill 7& \hfill -1\end{array}］$

$\begin{array}{r}X+［\begin{array}{cc}\hfill 3.& \hfill 2\\ \hfill 1& \hfill 0\end{array}］-［\begin{array}{cc}\hfill 3.& \hfill 2\\ \hfill 1& \hfill 0\end{array}］=［\begin{array}{cc}\hfill 6& \hfill 3.\\ \hfill 7& \hfill -1\end{array}］-［\begin{array}{cc}\hfill 3.& \hfill 2\\ \hfill 1& \hfill 0\end{array}］\\ X+［\begin{array}{cc}\hfill 0& \hfill 0\\ \hfill 0& \hfill 0\end{array}］=［\begin{array}{cc}\hfill 6-3.& \hfill 3.-2\\ \hfill 7-1& \hfill -1-0\end{array}］\\ X=［\begin{array}{cc}\hfill 3.& \hfill 1\\ \hfill 6& \hfill -1\end{array}］\end{array}$

例子2:

求解矩阵$X$：$X-［\begin{array}{cc}\hfill -9& \hfill -3.\\ \hfill 6& \hfill 0\end{array}］=［\begin{array}{cc}\hfill 4& \hfill 0\\ \hfill 12& \hfill -10\end{array}］$

$\begin{array}{r}X-［\begin{array}{cc}\hfill -9& \hfill -3.\\ \hfill 6& \hfill 0\end{array}］=［\begin{array}{cc}\hfill 4& \hfill 0\\ \hfill 12& \hfill -10\end{array}］\\ X-［\begin{array}{cc}\hfill -9& \hfill -3.\\ \hfill 6& \hfill 0\end{array}］+［\begin{array}{cc}\hfill -9& \hfill -3.\\ \hfill 6& \hfill 0\end{array}］=［\begin{array}{cc}\hfill 4& \hfill 0\\ \hfill 12& \hfill -10\end{array}］+［\begin{array}{cc}\hfill -9& \hfill -3.\\ \hfill 6& \hfill 0\end{array}］\\ X-［\begin{array}{cc}\hfill 0& \hfill 0\\ \hfill 0& \hfill 0\end{array}］=［\begin{array}{cc}4+（-9）\hfill & 0+（-3.）\hfill \\ 12+6\hfill & -10+0\hfill \end{array}］\\ X=［\begin{array}{cc}\hfill -5& \hfill -3.\\ \hfill 18& \hfill -10\end{array}］\end{array}$

例子3:

用矩阵求解方程组:$\{\begin{array}{l}7x+5y=3.\\ 3.x-2y=22\end{array}$

$\begin{array}{c}7x+5y=3.\\ 3.x-2y=22\end{array}\to ［\begin{array}{c}7x+5y\\ 3.x-2y\end{array}］=［\begin{array}{c}\hfill 3.\\ \hfill 22\end{array}］$

把左边的矩阵写成系数和变量的乘积。

$［\begin{array}{cc}7& 5\\ 3.& -2\end{array}］［\begin{array}{c}x\\ y\end{array}］=［\begin{array}{c}3.\\ 22\end{array}］$

$\uparrow \uparrow \uparrow$

$\begin{array}{l}\text{系数}\text{变量}\text{常数}\\ \text{矩阵}\text{矩阵}\text{矩阵}\end{array}$

首先，求系数矩阵的逆。的逆$［\begin{array}{rr}\hfill 7& \hfill 5\\ \hfill 3.& \hfill -2\end{array}］$是

$\begin{array}{l}\hfill \frac{1}{7（-2）-（3.）（5）}［\begin{array}{rr}\hfill -2& \hfill -5\\ \hfill -3.& \hfill 7\end{array}］\\ =-\frac{1}{29}［\begin{array}{rr}\hfill -2& \hfill -5\\ \hfill -3.& \hfill 7\end{array}］\\ =［\begin{array}{rr}\hfill \frac{2}{29}& \hfill \frac{5}{29}\\ \hfill \frac{3.}{29}& \hfill -\frac{7}{29}\end{array}］\end{array}$

接下来，将矩阵方程的两边乘以逆矩阵．因为矩阵乘法是不交换，逆矩阵应该在左边每一个矩阵方程的一边。

$［\begin{array}{cc}\hfill \frac{2}{29}& \hfill \frac{5}{29}\\ \hfill \frac{3.}{29}& \hfill -\frac{7}{29}\end{array}］［\begin{array}{rr}\hfill 7& \hfill 5\\ \hfill 3.& \hfill -2\end{array}］［\begin{array}{c}\hfill x\\ \hfill y\end{array}］=［\begin{array}{rr}\hfill \frac{2}{29}& \hfill \frac{5}{29}\\ \hfill \frac{3.}{29}& \hfill -\frac{7}{29}\end{array}］［\begin{array}{c}\hfill 3.\\ \hfill 22\end{array}］$

$［\begin{array}{cc}\hfill 1& \hfill 0\\ \hfill 0& \hfill 1\end{array}］［\begin{array}{c}\hfill x\\ \hfill y\end{array}］=［\begin{array}{r}\hfill 4\\ \hfill -5\end{array}］$

的单位矩阵左图验证了逆矩阵的正确计算。

$［\begin{array}{c}\hfill x\\ \hfill y\end{array}］=［\begin{array}{r}\hfill 4\\ \hfill -5\end{array}］$

解决方案是$（4，-5）$．