线性代数行列按行(列)展开设D=| 3 -5 2 1 || 1 1 0 -5 || -1 3 1 3 | | 2 -4 -1 -3 |,D的(i,j)元的余子式和代数余子式依次记作Mij和Aij,求A11+A12+A13+A14及 M11+M21+M31+M41.A11+A12+A13+A14等于用1,1,1,1代替D的第1行
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![线性代数行列按行(列)展开设D=| 3 -5 2 1 || 1 1 0 -5 || -1 3 1 3 | | 2 -4 -1 -3 |,D的(i,j)元的余子式和代数余子式依次记作Mij和Aij,求A11+A12+A13+A14及 M11+M21+M31+M41.A11+A12+A13+A14等于用1,1,1,1代替D的第1行](/uploads/image/z/12516060-12-0.jpg?t=%E7%BA%BF%E6%80%A7%E4%BB%A3%E6%95%B0%E8%A1%8C%E5%88%97%E6%8C%89%E8%A1%8C%EF%BC%88%E5%88%97%EF%BC%89%E5%B1%95%E5%BC%80%E8%AE%BED%3D%7C+3+-5+2+1+%7C%7C+1+1+0+-5+%7C%7C+-1+3+1+3+%7C+%7C+2+-4+-1+-3+%7C%2CD%E7%9A%84%EF%BC%88i%2Cj%EF%BC%89%E5%85%83%E7%9A%84%E4%BD%99%E5%AD%90%E5%BC%8F%E5%92%8C%E4%BB%A3%E6%95%B0%E4%BD%99%E5%AD%90%E5%BC%8F%E4%BE%9D%E6%AC%A1%E8%AE%B0%E4%BD%9CMij%E5%92%8CAij%2C%E6%B1%82A11%2BA12%2BA13%2BA14%E5%8F%8A+M11%2BM21%2BM31%2BM41.A11%2BA12%2BA13%2BA14%E7%AD%89%E4%BA%8E%E7%94%A81%2C1%2C1%2C1%E4%BB%A3%E6%9B%BFD%E7%9A%84%E7%AC%AC1%E8%A1%8C)
线性代数行列按行(列)展开设D=| 3 -5 2 1 || 1 1 0 -5 || -1 3 1 3 | | 2 -4 -1 -3 |,D的(i,j)元的余子式和代数余子式依次记作Mij和Aij,求A11+A12+A13+A14及 M11+M21+M31+M41.A11+A12+A13+A14等于用1,1,1,1代替D的第1行
线性代数行列按行(列)展开
设D=| 3 -5 2 1 |
| 1 1 0 -5 |
| -1 3 1 3 |
| 2 -4 -1 -3 |,
D的(i,j)元的余子式和代数余子式依次记作Mij和Aij,求
A11+A12+A13+A14及 M11+M21+M31+M41.
A11+A12+A13+A14等于用1,1,1,1代替D的第1行
所得的行列式,即
A11+A12+A13+A14=| 1 1 1 1 |=...=| 2 -5 |=4
| -1 1 0 -5 | | 0 2 |
| -1 3 1 3 |
| 2 -4 -1 -3 |
M11+M21+M31+M41=A11-A21+A31-A41
| 1 -5 2 1 |
= | -1 1 0 -5 |=...=0
| 1 3 1 3 |
| -1 -4 -1 -3 |
问一下这一步是怎么得到的:M11+M21+M31+M41=A11-A21+A31-A41
线性代数行列按行(列)展开设D=| 3 -5 2 1 || 1 1 0 -5 || -1 3 1 3 | | 2 -4 -1 -3 |,D的(i,j)元的余子式和代数余子式依次记作Mij和Aij,求A11+A12+A13+A14及 M11+M21+M31+M41.A11+A12+A13+A14等于用1,1,1,1代替D的第1行
M11+M21+M31+M41是行列式第一列所有元素的余子式的和.
注意的行列式的余子式与对应的代数余子式的关系:Aij=(-1)^(i+j)Mij
所以A11=M11,A21=(-1)^(2+1)M21=-M21,A31=M31,A41=-M41
所以M11=A11 ,M21=-A21,M31=A31,M41=-A41
故有M11+M21+M31+M41=A11-A21+A31-A41.
展开没错,但展开后第二个行列式计算错了啊。