Question

2．设线性方程组的增广矩阵为$（\begin{array}{l}{2}&{3}&{{t}_{1}}\\{0}&{1}&{{t}_{2}}\end{array}）$，解为$\left\{\begin{array}{l}{x=3}\\{y=5}\end{array}\right.$，则三阶行列式$[\begin{array}{l}{1}&{-1}&{{t}_{1}}\\{0}&{1}&{-1}\\{-1}&{{t}_{2}}&{-6}\end{array}]$的值为19．

Answer 1

分析 $\left\{\begin{array}{l}{x=3}\\{y=5}\end{array}\right.$，是方程$\left\{\begin{array}{l}{2x+3y={t}_{1}}\\{y={t}_{2}}\end{array}\right.$的解，代入即可求得t1和t2的值，代入行列式，按第一列展开，即可求得行列式的值．

解答 解：由题意可知：$\left\{\begin{array}{l}{x=3}\\{y=5}\end{array}\right.$，是方程$\left\{\begin{array}{l}{2x+3y={t}_{1}}\\{y={t}_{2}}\end{array}\right.$的解，
解得：$\left\{\begin{array}{l}{{t}_{1}=21}\\{{t}_{2}=5}\end{array}\right.$，
∴$|\begin{array}{l}{1}&{-1}&{21}\\{0}&{1}&{-1}\\{-1}&{5}&{-6}\end{array}|$=1×$|\begin{array}{l}{1}&{-1}\\{5}&{-6}\end{array}|$+（-1）×$|\begin{array}{l}{-1}&{21}\\{1}&{-1}\end{array}|$=-6-（-1）×5+（-1）×（1-1×21）=19，
故答案为：19．

点评 本题主要考查增广矩阵的求解，考查行列式的展开式，考查计算能力，属于基础题．