矩阵乘法和求逆运算都在这道题里了

一、题目

已知 $X A + 2 E = X + B$ , 其中 $A = [\begin{array}{ll} 1 & 2 \\ 1 & 1 \end{array}], B = [\begin{array}{cc} 2 & 2 \\ 1 & - 1 \end{array}]$ , 则 $X = ?$

难度评级：

二、解析

$X A + 2 E = X + B \Rightarrow$

$X (A - E) = B - 2 E \Rightarrow$

$x = (B - 2 E) (A - E)^{- 1}$

$A - E = [\begin{array}{ll} 0 & 2 \\ 1 & 0 \end{array}] \Rightarrow$

$[\begin{array}{llll} 0 & 2 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{array}] \Rightarrow [\begin{array}{llll} 1 & 0 & 0 & 1 \\ 0 & 1 & \frac{1}{2} & 0 \end{array}] \Rightarrow$

$(A - E)^{- 1} = [\begin{array}{ll} 0 & 1 \\ \frac{1}{2} & 0 \end{array}]$

$B - 2 E = [\begin{array}{ll} 0 & 2 \\ 1 & - 3 \end{array}]$

于是：

$X = [\begin{array}{cc} 0 & 2 \\ 1 & - 3 \end{array}] [\begin{array}{cc} 0 & 1 \\ \frac{1}{2} & 0 \end{array}] = [\begin{array}{cc} 1 & 0 \\ \frac{- 3}{2} & 1 \end{array}]$