线性代数第二版答案篇一：工程数学线性代数课后同济第五版12345篇二：线性代数第二章答案第二章矩阵及其运算1已经明白线性变换x12y12y2y3x23y1y25y3x33y12y23y3求从变量x1x2x3到变量y1y2y3的线性变换解由已经明白x1221y1x2315y2x323y23y1221x1749y1故y2315x2637y2y323x3243y32y17x14x29x3y26x13x27x3y33x12x24x32已经明白两个线性变换1y13z1z2x12y1y3x22y13y22y3y22z1z3y3z23z3x34y1y25y3求从z1z2z3到x1x2x3的线性变换解由已经明白x1201y120221x2232y223220x415y4150123613z11249z210116z30z11z2z33x16z1z23z3因此有x212z14z29z3x310z1z216z31111233设A111B124求3AB2A及ATB111051111123111解3AB2A311112421111110511110581112132230562111217202901114292111123058ATB1111240561110512904计算以下乘积4317(1)12325701431747321135解123217(2)23165701577202293(2)(123)213解(123)2(132231)(10)12(3)1(12)32(1)22242解1(12)1(1)121233(1)32361310122140(4)13111344021310126782140解13120561134402a11a12a13x1(5)(x1x2x3)a12a22a23x2aaa132333x3解a11a12a13x1(x1x2x3)a12a22a23x2aaa132333x3x1(a11x1a12x2a13x3a12x1a22x2a23x3a13x1a23x2a33x3)x2x35设A22a11x12a22x2a33x32a12x1x22a13x1x32a23x2x3112B1130征询2(1)ABBA吗解ABBA由于AB344BA1362因此ABBA8(2)(AB)2A22ABB2吗解(AB)2A22ABB2由于AB222522252(AB)22但28141429538681A22ABB241181230101615274因此(AB)2A22ABB2(3)(AB)(AB)A2B2吗解(AB)(AB)A2B2由于AB222AB00521220226(AB)(AB)250109102838而A2B24113417故(AB)(AB)A2B26举反列说明以下命题是错误的（也可参考书上的答案）(1)假设A20那么A0解取A00101那么A20但A001那么A2A但A0且AE0(2)假设A2A那么A0或AE解取A(3)假设AXAY且A0那么XY解取1A00X11Y1110011那么AXAY且A0但XY7设A解10求A2A3Ak1101010A21121101010A3A2A21131011Akk108设A01求Ak00解首先观察101A20220000332A3A2A0300443A4A3A0400554A5A4A0500kkA00kk102211022202232362434103545k0k(k1)k22kk1k用数学归纳法证明篇三：线性代数第二章答案
