The third order square matrix A = (A1, A2, A3), AJ (J = 1,2,3) is the j-th column of a, and the determinant of a / A / = 2, If B = (A1, A2 + 2A3, 3a3), then the determinant of B / =? A.16 b.12 c.54 d.6

The third order square matrix A = (A1, A2, A3), AJ (J = 1,2,3) is the j-th column of a, and the determinant of a / A / = 2, If B = (A1, A2 + 2A3, 3a3), then the determinant of B / =? A.16 b.12 c.54 d.6

|B| = |a1,a2+2a3,3a3|
=| A1, A2, 3a3 | + | A1, 2A3, 3a3 | [determinant property]
=3 | A1, A2, A3 | + 0 [the two columns are proportional]
= 3*2 = 6
Select (d)

A (a4-2a3-3a2-a-1) - A2 (1-a-a2) + (a4-a + 1) where a = - radical 3

a=-√3
So A2 = (- √ 3) (- √ 3) = 3
a3=a×a2=-3√3
a4=a×a3=-3√3(-√3)=3(√3)²=9
a5=a×a4=-9√3
a(a4-2a3-3a2-a-1)-a2(1-a-a2)+(a4-a+1)
=a5-2a4-3a3-a2-a-a2+a3+a4+a4-a+1
=a5-2a3-2a2-2a+1
=-9√3+6√3-6+2√3+1
=-√3-5

A. 5a-4a = 1 B. (A2) 3 = A5 c.2a2 × 3a3 = 6A5 d.3a2 + 2A3 = 5A3 which operation is correct

c,
Your math needs to be improved
A a
B a6
D is not operational

Given the set M = {x | x = 1 + A2, a belongs to natural number set}, P = {x | x = a2-4a = 5, a belongs to positive integer}, try to judge the relationship between M and P?

∵M={x|x=1+a^2,a∈N}
∵P={y|y=a^2-4a+5,a∈Z+}
∴P={y|y=(a-2)^2+1,a∈Z+}
When a ∈ Z +,
Both a ^ 2 and (A-2) ^ 2 represent the squares of all integers
∴M=P

If M = {x | x = 1 + A2, a ∈ n *}, P = {x | x = a2-4a + 5, a ∈ n *}, then the relation between M and P is______ ．

P = {x | x = a2-4a + 5, a ∈ n *} = {x | x = (A-2) 2 + 1, a ∈ n *} ∵ a ∈ n *} A-2 ≥ - 1, and A-2 ∈ Z, that is, (A-2) 2 ∈ {0, 1, 2 }And M = {x | x = A2 + 1, a ∈ n *}, ⊊ m ⊊ P. so the answer is m ⊊ P

Given the set M = {x | x = 1 + a square, a ∈ r}, P = {x | x = a square - 4A + 5, a ∈ r}, try to judge the relationship between M and P

x=1+a^2≥1
M={x|x=1+a^2,a∈R}={x|x≥1}
x=a^2-4a+5=(a-2)^2+1≥1
P={x|x=a^2-4a+5,a∈R}={x|x≥1}
So m = P
If you don't understand, please hi me, I wish you a happy study!

If M = {x | x = 1 + A2, a ∈ n *}, P = {x | x = a2-4a + 5, a ∈ n *}, then the relation between M and P is______ ．

P = {x | x = a2-4a + 5, a ∈ n *} = {x | x = (A-2) 2 + 1, a ∈ n *} ∵ a ∈ n *} A-2 ≥ - 1, and A-2 ∈ Z, that is, (A-2) 2 ∈ {0, 1, 2 }And M = {x | x = A2 + 1, a ∈ n *}, ⊊ m ⊊ P. so the answer is m ⊊ P

Let m = {x | x = 5-4a + A2, a ∈ r}, n = {y | y = 4a2 + 4A + 2, a ∈ r} the following is true: a.m = n, B.N is the proper subset of M, C.M is the proper subset of n
Need process

This is a simple exercise after class
First of all, remove the set of M and N. in fact, you just need to draw the graph of x = 5-4a + A2 and y = 4a2 + 4A + 2 to see if there is intersection and coincidence, and then you can choose the answer
Or, even if the value range of X and Y is OK, x = (A-2) 2 + 1, then x is greater than or equal to 1;
If y = 4 (a + 1 / 2) 2 + 1, then y is greater than or equal to 1
So m and N are equal

Set M = {x | x = 1 + A2, a ∈ n *}, P = {y | y = x2-4x + 5, X ∈ n *}, the correct relation in the following is ()
A. M ⊊ Pb. P ⊊ MC. M = PD. M ⊈ P and P ⊈ M

P = {y | y = x2-4x + 5, X ∈ n *} = {y | y = (X-2) 2 + 1, X ∈ n *}, we can see that the element in P set is the square of all natural numbers plus 1, and M = {x | x = 1 + A2, a ∈ n *}, where the element is the square of all positive integers plus 1, so m ⊊ P is selected: a

Given that M = {x | x = a ^ 2 + 1, a belongs to Z}, n = {y | y = a ^ 2-4a + 5, a belongs to Z}, then the relationship between M and N is?
M=N
If "a belongs to a and 8-A belongs to a", and a belongs to N, what is the number of sets a with two elements?
Let m = {x | - 1 be satisfied

M = {x | x = a ^ 2 + 1, a belongs to Z} n = {y | y = a ^ 2-4a + 5, a belongs to Z} = {y | y = (A-2) ^ 2 + 1, a belongs to Z} because when a belongs to Z, both a ^ 2 and (A-2) ^ 2 represent the squares of all integers, so n = ma belongs to N and there are only 2 elements, a belongs to N, so 8-A belongs to N and there are only 2 elements, so 8 - (8-A) = a gets 0? That is