作业帮线性代数行列式,齐次线性方程组的题目~
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线性代数行列式,齐次线性方程组的题目~
画圈的两道 .希望有具体的步骤,
画圈的两道 .希望有具体的步骤,
1 2 3 ...n
2 1 3 ...n
3 2 1 ...n
......
n 2 3 ...1
c1+c2+...+cn
n(n+1)/2 2 3 ...n
n(n+1)/2 1 3 ...n
n(n+1)/2 2 1 ...n
......
n(n+1)/2 2 3 ...1
ri-r1,i=2,3,...,n
n(n+1)/2 2 3 ...n
0 -1 3 ...0
0 0 -2 ...0
......
0 0 0 ...-(n-1)
= n(n+1)/2 * (-1)^(n-1) * (n-1)!
= (-1)^(n-1)* (n+1)!/2
系数行列式 =
5-λ 2 2
2 6-λ 0
2 0 4-λ
= (4-λ)(5-λ)(6-λ)-4(6-λ)-4(4-λ) --直接对角线法则得
= (4-λ)(5-λ)(6-λ)-4(10-2λ) --后两项合并得因子(5-λ)
= (4-λ)(5-λ)(6-λ)-8(5-λ)
= (5-λ)[(4-λ)(6-λ)-8]
= (5-λ)[λ^2-10λ+16] -- 十字相乘法分解
= (5-λ)(2-λ)(8-λ)
所以 λ = 2 或 5,8
注:这个行列式的计算比较特殊,不能用行列式的性质化简提出λ的因子
2 1 3 ...n
3 2 1 ...n
......
n 2 3 ...1
c1+c2+...+cn
n(n+1)/2 2 3 ...n
n(n+1)/2 1 3 ...n
n(n+1)/2 2 1 ...n
......
n(n+1)/2 2 3 ...1
ri-r1,i=2,3,...,n
n(n+1)/2 2 3 ...n
0 -1 3 ...0
0 0 -2 ...0
......
0 0 0 ...-(n-1)
= n(n+1)/2 * (-1)^(n-1) * (n-1)!
= (-1)^(n-1)* (n+1)!/2
系数行列式 =
5-λ 2 2
2 6-λ 0
2 0 4-λ
= (4-λ)(5-λ)(6-λ)-4(6-λ)-4(4-λ) --直接对角线法则得
= (4-λ)(5-λ)(6-λ)-4(10-2λ) --后两项合并得因子(5-λ)
= (4-λ)(5-λ)(6-λ)-8(5-λ)
= (5-λ)[(4-λ)(6-λ)-8]
= (5-λ)[λ^2-10λ+16] -- 十字相乘法分解
= (5-λ)(2-λ)(8-λ)
所以 λ = 2 或 5,8
注:这个行列式的计算比较特殊,不能用行列式的性质化简提出λ的因子