线性代数科学出版社课后习题答案.pdf

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1、工程数学（线性代数与概率统计）习题一一、211.221(1)5；12x1122322.(x1)(xx1)xxx1；22xxx1ab223.abab22ab1114.314532278153658950a05.b0c0,(第一行与第三行对应成比例）0d01236.312182766618。231二．求逆序数1.34215即5221002.4312即53200n(n1)3.n(n1)21即(n1)(n2)212(n1)(

2、n2)1013(2n1)(2n)(2n2)4201n1n1n2104.n(n1)[12(n1)][(n1)(n2)21]2*2三．四阶行列式中含有aa的项为aaaaaaaa11231123324411233442四．计算行列式值41240724009451202r14r21202r17r412021.010520r10r015220r15r00178532340117011701172.01113111111111111011301110110100ccc

3、c333123411013101110100101110311011100001abacae1113.bdcddeadfbce1114abcdefbfcfef111a100b101101b104.按第一行展开a1c10c101c101d01d001dc111c1aba(ab1)(cd1)ad1d0d1d5.abc2a2a2bbac2b2c2ccaba2a2ab2a2ac2a2a2bbac2b0bac2b0bac2b2c2ccab02ccab

4、02ccab其中a2a2aa2a2aa02aa02a2bbac2b2bb2b2ba2b2bc2b2c2ccab2c2ccab2c0cab2c0caba2a2aa2a0a2a0a2a2bb2b2bb02bb0(ac)2ccab2c2cc2c2ca2c2cb122a2aa2aabc212(ab)(ac)2bb2ccab22125abc3ab(ab)(ac)(aab3ac)其余同法可求。方法2:abc2a2a2bbac2b2c2ccabbac2b2b2b2bba

5、c(abc)2a2a2ccab2ccab2c2c2240r2r2240214135303556.rr270313123204832051r4r1021112221000222222227.2232rr(i2)00102(n2)!i2222n000n28.a001a0000a000a000a000000n1按第一行展开a(1)00a000a0000a100a000a1000n1n1nn1

6、nn2nn2a(1)(1)aaa五．证明下列等式1.222222aabb0a(ba)ba0a(ba)ba232aab2brar2aab2br2ar0ba2(ba)(ab)13231111111112222222a(a1)(a2)(a3)a2a1(a2)(a3)2222222b(b1)(b2)(b3)b2b1(b2)(b3)2.cc222221222c(c1)(c2)(c3)c2c1(c2)(c3)2222222d(d1)(d2)(d3)d2d1(d2)(d3

7、)22a2a14a46a9a2a12322c3c1b2b14b46b9c32c2b2b123022c4c1c2c14c46c9c42c2c2c12322d2d14d46d9d2d1233.x1000x10000x1000x10000x0000x00D按第一行展开xn000x1000x1aaaaxaaaaaxann1n221n1n2n321010000x10000x00n11n2xDa(1)(1)xDa

8、n1nn1n000x1aaaaxann2n321同理DxDa，返回代入得k1kk1nn1DxD