题目
设D=|(3)&(1)&(-1)&(2)-5)&(1)&(3)&(-4)2)&(0)&(1)&(-1)1)&(-5)&(3)&(-3)|，D的（i，j）元的代数余子式记作Aij，求A31+3A32-2A33+2A34．
设D=$|\begin{array}{l}{3}&{1}&{-1}&{2}\\{-5}&{1}&{3}&{-4}\\{2}&{0}&{1}&{-1}\\{1}&{-5}&{3}&{-3}\end{array}|$，D的（i，j）元的代数余子式记作Aij，求A31+3A32-2A33+2A34．
题目解答
答案
解：∵D=$|\begin{array}{l}{3}&{1}&{-1}&{2}\\{-5}&{1}&{3}&{-4}\\{2}&{0}&{1}&{-1}\\{1}&{-5}&{3}&{-3}\end{array}|$，D的（i，j）元的代数余子式记作Aij，
∴A31+3A32-2A33+2A34
=$|\begin{array}{l}{1}&{-1}&{2}\\{1}&{3}&{-4}\\{-5}&{3}&{-3}\end{array}|$-3$|\begin{array}{l}{3}&{-1}&{2}\\{-5}&{3}&{-4}\\{1}&{3}&{-3}\end{array}|$-2$|\begin{array}{l}{3}&{1}&{2}\\{-5}&{1}&{-4}\\{1}&{-5}&{-3}\end{array}|$-2$|\begin{array}{l}{3}&{1}&{-1}\\{-5}&{1}&{3}\\{1}&{-5}&{3}\end{array}|$
=-9+6-20+30+12-3-3（-27-30+4-6-36-15）-2（-9+50-4-2-60-15）-2（9+3-25+1+45+15）
=330．
∴A31+3A32-2A33+2A34
=$|\begin{array}{l}{1}&{-1}&{2}\\{1}&{3}&{-4}\\{-5}&{3}&{-3}\end{array}|$-3$|\begin{array}{l}{3}&{-1}&{2}\\{-5}&{3}&{-4}\\{1}&{3}&{-3}\end{array}|$-2$|\begin{array}{l}{3}&{1}&{2}\\{-5}&{1}&{-4}\\{1}&{-5}&{-3}\end{array}|$-2$|\begin{array}{l}{3}&{1}&{-1}\\{-5}&{1}&{3}\\{1}&{-5}&{3}\end{array}|$
=-9+6-20+30+12-3-3（-27-30+4-6-36-15）-2（-9+50-4-2-60-15）-2（9+3-25+1+45+15）
=330．
解析
步骤 1：计算A_31
根据代数余子式的定义，A_31是去掉第3行和第1列后剩余的3阶行列式的值，且其符号由(-1)^(3+1)决定。计算如下：
A_31 = (-1)^(3+1) * $|\begin{array}{l}{1}&{-1}&{2}\\{1}&{3}&{-4}\\{-5}&{3}&{-3}\end{array}|$
= $|\begin{array}{l}{1}&{-1}&{2}\\{1}&{3}&{-4}\\{-5}&{3}&{-3}\end{array}|$
= 1*(3*(-3)-(-4)*3) - (-1)*(1*(-3)-(-4)*(-5)) + 2*(1*3-3*(-5))
= 1*(-9+12) - (-1)*(-3+20) + 2*(3+15)
= 1*3 - (-1)*17 + 2*18
= 3 + 17 + 36
= 56
步骤 2：计算3A_32
A_32是去掉第3行和第2列后剩余的3阶行列式的值，且其符号由(-1)^(3+2)决定。计算如下：
A_32 = (-1)^(3+2) * $|\begin{array}{l}{3}&{-1}&{2}\\{-5}&{3}&{-4}\\{1}&{3}&{-3}\end{array}|$
= - $|\begin{array}{l}{3}&{-1}&{2}\\{-5}&{3}&{-4}\\{1}&{3}&{-3}\end{array}|$
= - (3*(3*(-3)-(-4)*3) - (-1)*(-5*(-3)-(-4)*1) + 2*(-5*3-3*1))
= - (3*(-9+12) - (-1)*(-15+4) + 2*(-15-3))
= - (3*3 - (-1)*(-11) + 2*(-18))
= - (9 - 11 - 36)
= - (-38)
= 38
3A_32 = 3 * 38 = 114
步骤 3：计算-2A_33
A_33是去掉第3行和第3列后剩余的3阶行列式的值，且其符号由(-1)^(3+3)决定。计算如下：
A_33 = (-1)^(3+3) * $|\begin{array}{l}{3}&{1}&{2}\\{-5}&{1}&{-4}\\{1}&{-5}&{-3}\end{array}|$
= $|\begin{array}{l}{3}&{1}&{2}\\{-5}&{1}&{-4}\\{1}&{-5}&{-3}\end{array}|$
= 3*(1*(-3)-(-4)*(-5)) - 1*(-5*(-3)-(-4)*1) + 2*(-5*(-5)-1*1)
= 3*(-3+20) - 1*(-15+4) + 2*(25-1)
= 3*17 - 1*(-11) + 2*24
= 51 + 11 + 48
= 110
-2A_33 = -2 * 110 = -220
步骤 4：计算2A_34
A_34是去掉第3行和第4列后剩余的3阶行列式的值，且其符号由(-1)^(3+4)决定。计算如下：
A_34 = (-1)^(3+4) * $|\begin{array}{l}{3}&{1}&{-1}\\{-5}&{1}&{3}\\{1}&{-5}&{3}\end{array}|$
= - $|\begin{array}{l}{3}&{1}&{-1}\\{-5}&{1}&{3}\\{1}&{-5}&{3}\end{array}|$
= - (3*(1*3-3*(-5)) - 1*(-5*3-3*1) + (-1)*(-5*(-5)-1*1))
= - (3*(3+15) - 1*(-15-3) + (-1)*(25-1))
= - (3*18 - 1*(-18) + (-1)*24)
= - (54 + 18 - 24)
= - 48
2A_34 = 2 * (-48) = -96
步骤 5：计算A_31+3A_32-2A_33+2A_34
A_31+3A_32-2A_33+2A_34 = 56 + 114 - 220 - 96 = 170 - 316 = -146
根据代数余子式的定义，A_31是去掉第3行和第1列后剩余的3阶行列式的值，且其符号由(-1)^(3+1)决定。计算如下：
A_31 = (-1)^(3+1) * $|\begin{array}{l}{1}&{-1}&{2}\\{1}&{3}&{-4}\\{-5}&{3}&{-3}\end{array}|$
= $|\begin{array}{l}{1}&{-1}&{2}\\{1}&{3}&{-4}\\{-5}&{3}&{-3}\end{array}|$
= 1*(3*(-3)-(-4)*3) - (-1)*(1*(-3)-(-4)*(-5)) + 2*(1*3-3*(-5))
= 1*(-9+12) - (-1)*(-3+20) + 2*(3+15)
= 1*3 - (-1)*17 + 2*18
= 3 + 17 + 36
= 56
步骤 2：计算3A_32
A_32是去掉第3行和第2列后剩余的3阶行列式的值，且其符号由(-1)^(3+2)决定。计算如下：
A_32 = (-1)^(3+2) * $|\begin{array}{l}{3}&{-1}&{2}\\{-5}&{3}&{-4}\\{1}&{3}&{-3}\end{array}|$
= - $|\begin{array}{l}{3}&{-1}&{2}\\{-5}&{3}&{-4}\\{1}&{3}&{-3}\end{array}|$
= - (3*(3*(-3)-(-4)*3) - (-1)*(-5*(-3)-(-4)*1) + 2*(-5*3-3*1))
= - (3*(-9+12) - (-1)*(-15+4) + 2*(-15-3))
= - (3*3 - (-1)*(-11) + 2*(-18))
= - (9 - 11 - 36)
= - (-38)
= 38
3A_32 = 3 * 38 = 114
步骤 3：计算-2A_33
A_33是去掉第3行和第3列后剩余的3阶行列式的值，且其符号由(-1)^(3+3)决定。计算如下：
A_33 = (-1)^(3+3) * $|\begin{array}{l}{3}&{1}&{2}\\{-5}&{1}&{-4}\\{1}&{-5}&{-3}\end{array}|$
= $|\begin{array}{l}{3}&{1}&{2}\\{-5}&{1}&{-4}\\{1}&{-5}&{-3}\end{array}|$
= 3*(1*(-3)-(-4)*(-5)) - 1*(-5*(-3)-(-4)*1) + 2*(-5*(-5)-1*1)
= 3*(-3+20) - 1*(-15+4) + 2*(25-1)
= 3*17 - 1*(-11) + 2*24
= 51 + 11 + 48
= 110
-2A_33 = -2 * 110 = -220
步骤 4：计算2A_34
A_34是去掉第3行和第4列后剩余的3阶行列式的值，且其符号由(-1)^(3+4)决定。计算如下：
A_34 = (-1)^(3+4) * $|\begin{array}{l}{3}&{1}&{-1}\\{-5}&{1}&{3}\\{1}&{-5}&{3}\end{array}|$
= - $|\begin{array}{l}{3}&{1}&{-1}\\{-5}&{1}&{3}\\{1}&{-5}&{3}\end{array}|$
= - (3*(1*3-3*(-5)) - 1*(-5*3-3*1) + (-1)*(-5*(-5)-1*1))
= - (3*(3+15) - 1*(-15-3) + (-1)*(25-1))
= - (3*18 - 1*(-18) + (-1)*24)
= - (54 + 18 - 24)
= - 48
2A_34 = 2 * (-48) = -96
步骤 5：计算A_31+3A_32-2A_33+2A_34
A_31+3A_32-2A_33+2A_34 = 56 + 114 - 220 - 96 = 170 - 316 = -146