题目
设行列式|({a)_(11)}&({a)_(12)}{a)_(21)}&({a)_(22)}|等于（ ）A. m+nB. -（m+n）C. n-mD. m-n
设行列式$|\begin{array}{l}{{a}_{11}}&{{a}_{12}}\\{{a}_{21}}&{{a}_{22}}\end{array}|=m，|\begin{array}{l}{{a}_{13}}&{{a}_{11}}\\{{a}_{23}}&{{a}_{21}}\end{array}|$=n，则行列式$|\begin{array}{l}{{a}_{11}}&{{a}_{12}+{a}_{13}}\\{{a}_{21}}&{{a}_{22}+{a}_{23}}\end{array}|$等于（ ）
A. m+n
B. -（m+n）
C. n-m
D. m-n
题目解答
答案
D. m-n
解析
步骤 1：计算行列式$|\begin{array}{l}{{a}_{11}}&{{a}_{12}}\\{{a}_{21}}&{{a}_{22}}\end{array}|$
根据行列式的定义，行列式$|\begin{array}{l}{{a}_{11}}&{{a}_{12}}\\{{a}_{21}}&{{a}_{22}}\end{array}|$的值为$a_{11}a_{22}-a_{21}a_{12}$，题目中给出这个值为$m$，即$m=a_{11}a_{22}-a_{21}a_{12}$。
步骤 2：计算行列式$|\begin{array}{l}{{a}_{13}}&{{a}_{11}}\\{{a}_{23}}&{{a}_{21}}\end{array}|$
同样根据行列式的定义，行列式$|\begin{array}{l}{{a}_{13}}&{{a}_{11}}\\{{a}_{23}}&{{a}_{21}}\end{array}|$的值为$a_{13}a_{21}-a_{23}a_{11}$，题目中给出这个值为$n$，即$n=a_{13}a_{21}-a_{23}a_{11}$。
步骤 3：计算行列式$|\begin{array}{l}{{a}_{11}}&{{a}_{12}+{a}_{13}}\\{{a}_{21}}&{{a}_{22}+{a}_{23}}\end{array}|$
根据行列式的定义，行列式$|\begin{array}{l}{{a}_{11}}&{{a}_{12}+{a}_{13}}\\{{a}_{21}}&{{a}_{22}+{a}_{23}}\end{array}|$的值为$a_{11}(a_{22}+a_{23})-a_{21}(a_{12}+a_{13})$，即$a_{11}a_{22}+a_{11}a_{23}-a_{21}a_{12}-a_{21}a_{13}$。根据步骤1和步骤2，可以将这个表达式重写为$m-(a_{21}a_{13}-a_{23}a_{11})$，即$m-n$。
根据行列式的定义，行列式$|\begin{array}{l}{{a}_{11}}&{{a}_{12}}\\{{a}_{21}}&{{a}_{22}}\end{array}|$的值为$a_{11}a_{22}-a_{21}a_{12}$，题目中给出这个值为$m$，即$m=a_{11}a_{22}-a_{21}a_{12}$。
步骤 2：计算行列式$|\begin{array}{l}{{a}_{13}}&{{a}_{11}}\\{{a}_{23}}&{{a}_{21}}\end{array}|$
同样根据行列式的定义，行列式$|\begin{array}{l}{{a}_{13}}&{{a}_{11}}\\{{a}_{23}}&{{a}_{21}}\end{array}|$的值为$a_{13}a_{21}-a_{23}a_{11}$，题目中给出这个值为$n$，即$n=a_{13}a_{21}-a_{23}a_{11}$。
步骤 3：计算行列式$|\begin{array}{l}{{a}_{11}}&{{a}_{12}+{a}_{13}}\\{{a}_{21}}&{{a}_{22}+{a}_{23}}\end{array}|$
根据行列式的定义，行列式$|\begin{array}{l}{{a}_{11}}&{{a}_{12}+{a}_{13}}\\{{a}_{21}}&{{a}_{22}+{a}_{23}}\end{array}|$的值为$a_{11}(a_{22}+a_{23})-a_{21}(a_{12}+a_{13})$，即$a_{11}a_{22}+a_{11}a_{23}-a_{21}a_{12}-a_{21}a_{13}$。根据步骤1和步骤2，可以将这个表达式重写为$m-(a_{21}a_{13}-a_{23}a_{11})$，即$m-n$。