It is known that A1, A2, B1, B2, y are all three-dimensional column vectors, and the determinants | A1, B1, y | = | A1, B2, y | = | A2, B1, y | = | A2, B2, y | = 3 So | - 2Y, a1 + A2, B1 + 2B2 | =?

It is known that A1, A2, B1, B2, y are all three-dimensional column vectors, and the determinants | A1, B1, y | = | A1, B2, y | = | A2, B1, y | = | A2, B2, y | = 3 So | - 2Y, a1 + A2, B1 + 2B2 | =?

|-This is the first time that we are going to be able to + A2 + A2 + A2, a1 + 2B2 + 2b2b2 1242; | - 2Y, a1,2b2 + | - 2Y, A2, B1 + 124; -2y, A2, B1 + + | - 2y-2y, A1, B1 | + -2;-2y-2y, A1, A1, B1 | | -2;- 2y-2y-2y, A1, A1, A1, B1, b1-2-2-2-2-y-y, A2, A2, A2, b1-12-4-4-four-4-4-4-4-12-this is the public factor = -2-2-the factor = -2-2-2-2-the public factor for the public factor = - 2-2-2-2-the public factor = - 2 | A2, B1, y | - 4 | A2, B2, y | -- commutative column = - 2 * 3

Let A1, A2, A3, B1, B2 all be 4 * 1 column vectors, and the fourth-order determinant A1, A2, A3, B1 = m, A1, A2, B2, A3 = n, then the determinant A3, A2, A1, B1 + B2=

It is known that | A1 A2 A3 B1 | = m, | A1 A2 B2 A3 | = n
Exchange two columns of determinant, determinant sign change
So | A3 A2 A1 B1 | = - M
| a3 a2 b2 a1| = -n
| a3 a2 a1 b2| =n
|a3,a2,a1,b1+b2| =| a1 a2 a3 b1 | + | a3 a2 a1 b2| =n-m

It is known that A1, A2, B1, B2, R are all three-dimensional column vectors, and the determinant | A1, B1, R | = | A1, B2, R | = | A2, B1, R | = | A2, B2, R | = 3, then | - 3R, a1 + A2, B1 + 2B2 | =?

|-3r,a1+a2,b1+2b2|
= -3 (|r,a1,b1| + 2|r,a1,b2| + |r,a2,b1| + 2|r,a2,b2|)
= -3 ( |a1,b1,r| + 2|a1,b2,r| + |a2,b1,r| + 2|a2,b2,r|)
= -3 * 6 * 3
= - 45.

How to calculate the new vector after the three-dimensional vector a (A1, A2, A3) rotates x degrees counterclockwise around the three-dimensional vector b (B1, B2, B3)?