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

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：

$$
A=\left[\begin{array}{ll}x_{1} & y_{1} \\ x_{2} & y_{2}\end{array}\right] \Rightarrow
$$

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

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

$$
\left\{\begin{array}{l}x_{1}+x_{2}=2 \\ y_{1}+y_{2}=3 \\ 2 x_{1}+2 x_{2}=4 \\ 2 y_{1}+2 y_{2}=6\end{array} \Rightarrow \left\{\begin{array}{l}x_{1}+x_{2}=2 \\ y_{1}+y_{2}=3\end{array} \Rightarrow\right.\right.
$$

令：

$$
x_{1}=t, y_{1}=u
$$

则：

$$
\left\{\begin{array}{l}x_{1}=t \\ x_{2}=2-t \\ y_{1}=u \\ y_{2}=3-u\end{array} \Rightarrow\right.
$$

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

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