Question

19．已知矩阵$A=[\begin{array}{l}2\\ 1\end{array}\right.$$\left.\begin{array}{l}1\\ 3\end{array}]$，$B=[\begin{array}{l}1\\ 0\end{array}\right.$$\left.\begin{array}{l}1\\-1\end{array}]$．求矩阵C，使得AC=B．

Answer 1

分析 求出A^-1，由AC=B，得（A^-1A）C=A^-1B，即可得出结论．

解答 解：因为|A|=2×3-1×1=5，
所以${A^{-1}}=[{\begin{array}{l}{\frac{3}{5}}&{-\frac{1}{5}}\\{-\frac{1}{5}}&{\frac{2}{5}}\end{array}}]=[{\begin{array}{l}{\frac{3}{5}}&{-\frac{1}{5}}\\{-\frac{1}{5}}&{\frac{2}{5}}\end{array}}]$，
由AC=B，得（A^-1A）C=A^-1B，
所以$C={A^{-1}}B=[{\begin{array}{l}{\frac{3}{5}}&{-\frac{1}{5}}\\{-\frac{1}{5}}&{\frac{2}{5}}\end{array}}][{\begin{array}{l}1&1\\ 0&{-1}\end{array}}]=[{\begin{array}{l}{\frac{3}{5}}&{\frac{4}{5}}\\{-\frac{1}{5}}&{-\frac{3}{5}}\end{array}}]$．

点评 本题考查矩阵的乘法，考查逆矩阵的求法，考查学生的计算能力，属于中档题．