Question

急问一道简单的线性代数题目设 A= [ 4 2 3 ],且矩阵X满足AX=A+2X,求X.[ 1 1 0 ][-1 2

Answer 1

搞错了 重来

因为 AX=A+2X

所以 (A-2E)X = A

(A-2E,A) =

2 2 3 4 2 3

1 -1 0 1 1 0

-1 2 1 -1 2 3

r1-2r2,r3+r2

0 4 3 2 0 3

1 -1 0 1 1 0

0 1 1 0 3 3

r1-3r3,r2+r3

0 0 -1 2 -12 -9

1 0 1 1 4 3

0 1 1 0 3 3

r2+r1,r3+r1,r1*(-1)

0 0 1 -2 12 9

1 0 0 3 -8 -6

0 1 0 2 -9 -6

交换行

1 0 0 3 -8 -6

0 1 0 2 -9 -6

0 0 1 -2 12 9

X =

3 -8 -6

2 -9 -6

-2 12 9