Let a be a square matrix of order 3, | a | = - 4, let AI be the i-th column vector of a, then a = (A1, A2, A3), then the determinant | A3 + 3A1, A2, 4A1 | = | Math enthusiast | Mathematical art

Let a be a square matrix of order 3, | a | = - 4, let AI be the i-th column vector of a, then a = (A1, A2, A3), then the determinant | A3 + 3A1, A2, 4A1 | =

4*4=16

Let a be a 3-order matrix, a1a2a3 be a 3-dimensional linearly independent column vector, and Aa1 = 4a1-4a2 + 3a3, aa2 = negative 6a1-a2 + a3, aa3 = 0

A (a1a2a3) = (a1a2a3) B,
Matrices A and B have the same eigenvalues

Given that A1 and A2 are column vectors, matrix A = (2A1 + A2. A1-a2) B = (A1, A2) if determinant | a | = 6, then | B | =?

A=(2a1+a2,a1-a2)=(a1,a2)P=BP
P=2 1
1 -1
Then B = AP ^ - 1
P^-1=1/3 1/3
1/3 -2/3
|B|=|A||P^-1|=6×(-1/3)=-2

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