题目
单选题-|||-令 =((alpha )_(1),(alpha )_(2),beta ) =((alpha )_(1),(alpha )_(2),gamma ) 为三阶方阵,且-|||-|A|=1 |B|=2 ,则 |2(alpha )_(1),-(alpha )_(2),gamma -beta |= __-|||-A. -2-|||-B.2-|||-C. -8-|||-D.8
题目解答
答案
解析
步骤 1：拆分行列式
将行列式 $|2{\alpha }_{1},-{\alpha }_{2},\gamma -\beta |$ 拆分为 $|2{\alpha }_{1},-{\alpha }_{2},\gamma |$ 和 $|2{\alpha }_{1},-{\alpha }_{2},\beta |$ 的差。
步骤 2：应用行列式性质
根据行列式的性质，将行列式的某一行(列)乘以一个数k，行列式的值变为原来的k倍。因此，$|2{\alpha }_{1},-{\alpha }_{2},\gamma | = 2 \times (-1) \times |{\alpha }_{1},{\alpha }_{2},\gamma | = -2|{\alpha }_{1},{\alpha }_{2},\gamma |$。同理，$|2{\alpha }_{1},-{\alpha }_{2},\beta | = 2 \times (-1) \times |{\alpha }_{1},{\alpha }_{2},\beta | = -2|{\alpha }_{1},{\alpha }_{2},\beta |$。
步骤 3：代入已知条件
已知 $|A|=|{\alpha }_{1},{\alpha }_{2},\beta |=1$ 和 $|B|=|{\alpha }_{1},{\alpha }_{2},\gamma |=2$。因此，$|2{\alpha }_{1},-{\alpha }_{2},\gamma -\beta | = -2|{\alpha }_{1},{\alpha }_{2},\gamma | - (-2|{\alpha }_{1},{\alpha }_{2},\beta |) = -2 \times 2 - (-2 \times 1) = -4 + 2 = -2$。
将行列式 $|2{\alpha }_{1},-{\alpha }_{2},\gamma -\beta |$ 拆分为 $|2{\alpha }_{1},-{\alpha }_{2},\gamma |$ 和 $|2{\alpha }_{1},-{\alpha }_{2},\beta |$ 的差。
步骤 2：应用行列式性质
根据行列式的性质，将行列式的某一行(列)乘以一个数k，行列式的值变为原来的k倍。因此，$|2{\alpha }_{1},-{\alpha }_{2},\gamma | = 2 \times (-1) \times |{\alpha }_{1},{\alpha }_{2},\gamma | = -2|{\alpha }_{1},{\alpha }_{2},\gamma |$。同理，$|2{\alpha }_{1},-{\alpha }_{2},\beta | = 2 \times (-1) \times |{\alpha }_{1},{\alpha }_{2},\beta | = -2|{\alpha }_{1},{\alpha }_{2},\beta |$。
步骤 3：代入已知条件
已知 $|A|=|{\alpha }_{1},{\alpha }_{2},\beta |=1$ 和 $|B|=|{\alpha }_{1},{\alpha }_{2},\gamma |=2$。因此，$|2{\alpha }_{1},-{\alpha }_{2},\gamma -\beta | = -2|{\alpha }_{1},{\alpha }_{2},\gamma | - (-2|{\alpha }_{1},{\alpha }_{2},\beta |) = -2 \times 2 - (-2 \times 1) = -4 + 2 = -2$。