当一个矩阵和不可逆矩阵相乘，怎么求解这个未知矩阵？

一、题目

已知 $\left[\begin{array}{ll}1& 1\\ 2& 2\end{array}\right]\mathit{A}=\left[\begin{array}{ll}2& 3\\ 4& 6\end{array}\right]$, 则 $\mathit{A}=?$

难度评级：

二、解析

由于 $\left[\begin{array}{ll}1& 1\\ 2& 2\end{array}\right]$ 不可逆，因此，只能将矩阵 $A$ 每一个元素都设出来，然后列方程求解：

$$A=\left[\begin{array}{ll}{x}_1& y_1\\ x_2& y_2\end{array}\right]\Rightarrow$$

$$\left[\begin{array}{ll}1& 1\\ 2& 2\end{array}\right]A=\left[\begin{array}{ll}2& 3\\ 4& 6\end{array}\right]\Rightarrow$$

$$\left[\begin{array}{ll}1& 1\\ 2& 2\end{array}\right]\left[\begin{array}{ll}{x}_1& y_1\\ x_2& y_2\end{array}\right]=\left[\begin{array}{ll}2& 3\\ 4& 6\end{array}\right]\Rightarrow$$

$$\{\begin{array}{l}{x}_1+x_2=2\\ y_1+y_2=3\\ 2x_1+2x_2=4\\ 2y_1+2y_2=6\end{array}\Rightarrow \{\begin{array}{l}{x}_1+x_2=2\\ y_1+y_2=3\end{array}\Rightarrow$$

令：

$$x_1=t,y_1=u$$

则：

$$\{\begin{array}{l}{x}_1=t\\ x_2=2-t\\ y_1=u\\ y_2=3-u\end{array}\Rightarrow$$

$$A=\left[\begin{array}{ll}t& u\\ 2-t& 3-u\end{array}\right]$$

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