Question

9．把行列式$|{\begin{array}{l}{a_1}&3&{b_1}\\{{a_2}}&2&{b_2}\\{{a_3}}&{-2}&{b_3}\end{array}}|$按照第二列展開(kāi)，則-3×$|\begin{array}{l}{{a}_{2}}&{_{2}}\\{{a}_{3}}&{_{3}}\end{array}|$+2×$|\begin{array}{l}{{a}_{1}}&{_{1}}\\{{a}_{3}}&{_{3}}\end{array}|$+2×$|\begin{array}{l}{{a}_{1}}&{_{1}}\\{{a}_{2}}&{_{2}}\end{array}|$．

Answer 1

分析 利用行列式展開(kāi)的方法，即可得出結(jié)論．

解答 解：把行列式$|{\begin{array}{l}{a_1}&3&{b_1}\\{{a_2}}&2&{b_2}\\{{a_3}}&{-2}&{b_3}\end{array}}|$按照第二列展開(kāi)得到-3×$|\begin{array}{l}{{a}_{2}}&{_{2}}\\{{a}_{3}}&{_{3}}\end{array}|$+2×$|\begin{array}{l}{{a}_{1}}&{_{1}}\\{{a}_{3}}&{_{3}}\end{array}|$+2×$|\begin{array}{l}{{a}_{1}}&{_{1}}\\{{a}_{2}}&{_{2}}\end{array}|$．
故答案為：-3×$|\begin{array}{l}{{a}_{2}}&{_{2}}\\{{a}_{3}}&{_{3}}\end{array}|$+2×$|\begin{array}{l}{{a}_{1}}&{_{1}}\\{{a}_{3}}&{_{3}}\end{array}|$+2×$|\begin{array}{l}{{a}_{1}}&{_{1}}\\{{a}_{2}}&{_{2}}\end{array}|$．

點(diǎn)評(píng) 本題考查行列式展開(kāi)的方法，考查學(xué)生的計(jì)算能力，比較基礎(chǔ)．