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本人自己做的,不一定对,仅供参考!不正确的欢迎大家指正.
1.a.
[0 0 1]
[0 0 0]
[0 1 0]
b. A is not invertible, because detA=0
2.a. x=-1, y=-3, z=0
b.
[4 -1 -3]
[3 -1 0]
[3 -1 -1]
3.a. t=1,-2,5, or 0
b. Because it's not invertible => det=0
also, detA=detA^T
Therefore, t=1,-2,5, or 0
4.略
5.a. dimH=3
b. dimH=2
c. dimH=3
d. dimH=3
6.略
7.a. Basis for row B: {(0,1,2,3,4),(0,0,0,-1,-2)}
b. Basis for ColB:
{[ 1] [ 3]}
{[-1], [-4]}
{[ 0] [ 4]}
c. Basis for NulB:
{[1] [ 0] [ 0]}
{[0] [-2] [ 2]}
{[0], [ 1], [ 0]}
{[0] [ 0] [-2]}
{[0] [ 0] [ 1]}
d. RankB=dimColB=The number of pivot columns=2
8.略
1.a.
[0 0 1]
[0 0 0]
[0 1 0]
b. A is not invertible, because detA=0
2.a. x=-1, y=-3, z=0
b.
[4 -1 -3]
[3 -1 0]
[3 -1 -1]
3.a. t=1,-2,5, or 0
b. Because it's not invertible => det=0
also, detA=detA^T
Therefore, t=1,-2,5, or 0
4.略
5.a. dimH=3
b. dimH=2
c. dimH=3
d. dimH=3
6.略
7.a. Basis for row B: {(0,1,2,3,4),(0,0,0,-1,-2)}
b. Basis for ColB:
{[ 1] [ 3]}
{[-1], [-4]}
{[ 0] [ 4]}
c. Basis for NulB:
{[1] [ 0] [ 0]}
{[0] [-2] [ 2]}
{[0], [ 1], [ 0]}
{[0] [ 0] [-2]}
{[0] [ 0] [ 1]}
d. RankB=dimColB=The number of pivot columns=2
8.略