Question

设矩阵A为3阶方阵,|A|=1,把A按列分块A=(A1,A2,A3),其中Aj(j=1,2,3)为A的第j列,求|A1+

Answer 1

(A1+A2+A3,A1+2A2,9A3) = (A1,A2,A3) K

K=

1 1 0

1 2 0

1 0 9

等式两边取行列式得 |A1+A2+A3,A1+2A2,9A3| = |A||K| = 1*9 = 9.