How to multiply matrices 1 1 1 1 1 6 6 Why 2 * 2 = 12 3 3 3 3 3 18 18 How is this matrix multiplied?

How to multiply matrices 1 1 1 1 1 6 6 Why 2 * 2 = 12 3 3 3 3 3 18 18 How is this matrix multiplied?

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This is the definition of matrix multiplication~
Multiplication of two matrices:
1,1,1 1,1
2,2,2 * 2,2
3,3,3 3,3
The elements of row a and column B of the new matrix are equal to the elements of row a of the first matrix, which are multiplied and added by the elements of column B of the second matrix
For example, the value of the third row and the second column of the new matrix in this question is:
3*1+3*2+3*3=18
among
3 (the third row and first column of the first matrix) * 1 (the first row and second column of the second matrix) + 3 (the third row and second column of the first matrix) * 2 (the second row and second column of the second matrix) + 3 (the third row and third column of the first matrix)) * 3 (the third row and second column of the second matrix)
So the new matrix is:
1*1+1*2+1*3,1*1+1*2+1*3
2*1+2*2+2*3,2*1+2*2+2*3
3*1+3*2+3*3,3*1+3*2+3*3
Namely:
6,6
12,12
18,18
Matrix multiplication requires that the specifications of the two matrices should be the same, that is, when the first matrix is row a and column B, the second matrix is row B and column C
That is, the number of columns of the first matrix should be equal to the number of rows of the second matrix, otherwise it cannot be multiplied

Seeking: cross multiplication and matrix multiplication
Well, to solve the examples of these two algorithms, we only need to solve them,
I don't want to see a lot of analysis. It's a headache. It's much easier to understand with examples

2
X －3X＋2＝0
Cross multiplication is (X-2) (x-1) = 0
Need more practice, no way

Given vector M = (COSA, Sina), vector n = (2, - 1), and vector m × vector n = 0, find the range of function f (x) = cos2x + tanasinx (x ∈ R)

M = (COSA, Sina), vector n = (2, - 1), and vector m × vector n = 0
∴2cosA-sinA=0
∴tanA=sinA/cosA=2
∴f(x)=cos2x+tanAsinx
=cos2x+2sinx
=1-2(sinx)^2+2sinx
∵sinx∈[-1,1]
∴f(x)∈[-3,3/2]

Through the point P (2,1), make the positive half axis of X, Y axis at a, B, and find the equation of the line when the area of triangle AOB reaches the minimum

Let the equation of the straight line be Y-1 = K (X-2) (K0
Then s △ AOB ≥ (1 / 2) [2 √ (- 4K) (- 1 / k) + 4]
=(1/2)×(2×2+4)
=4
If and only if - 4K = - 1 / K, that is, k = - 1 / 2, the equal sign holds
When k = - 1 / 2, the area of triangle AOB is the smallest
The equation of the straight line is y = (- 1 / 2) x + 2

The range y = - 2x ^ 2-x + 6 x belongs to [- 1,1] y = (3x-1) / (x ^ 2 + 2) f (x) = x + √ (x + 1)

Y = - 2x ^ 2 - x + 6 = - 2 (x ^ 2 + X × 1 / 2) + 6 = - 2 (x + 1 / 4) ^ 2 + 49 / 8, maximum x = - 1 / 4, y max = 49 / 8, minimum x = 1, y Min = 3Y = (3x-1) / [(3x-1) ^ 2 / 9 + 2x / 3] = (3x-1) / [(3x-1) ^ 2 / 9 +

The length of a rectangle is the fourth power of 4.2 × 10, and the width is the fourth power of 2 × 10
 

Area of rectangle = length * width = = 4.2 * 10 ^ 4 * 2 * 10 ^ 4 = 8.4 * 10 ^ 8
Perimeter = 2 (length + width) = 2 (4.2 * 10 ^ 4 + 2 * 10 ^ 4) = 2 * 6.2 * 10 ^ 4 = 1.24 * 10 ^ 5

It is known that a, B and C are three real numbers which are not all zero, then the root of the equation x2 + (a + B + C) x + A2 + B2 + C2 = 0 is ()
A. There are two negative roots B. There are two positive roots C. There are two positive roots and one negative root d. There are no real roots

∵ △ = (a + B + C) 2-4 (A2 + B2 + C2) = - 3a2-3b2-3c2 + 2Ab + 2BC + 2Ac = - (A-C) 2 - (B-C) 2 - (a-b) 2-a2-b2-c2, and a, B and C are three real numbers which are not all zero, ∵ (A-C) 2 - (B-C) 2 - (a-b) 2 - ≤ 0, a2-b2-c2 < 0, ∵ △ 0, ∵ the original equation has no real root

If M and N are opposite to each other, then m-2 + n is absolute=

Easy to get: M + n = 0
|m-2+n|
=|m+n-2|
=|-2|
=2

The function f (x) = x * x (ax-3) defined on R is known, where a is a constant
The function f (x) = x * x (ax-3) defined on R is known, where a is a constant. (1) if x = 1, the function f (x) obtains the extremum, find the value of A. (2) if f (x) is an increasing function in the interval (- 1,0), find the value range of A

1
f(x)=x*x(ax-3)=ax^3-3x^2
f'(x)=3ax^2-6x
f'(1)=3a-6=0
a=2
two
f'(x)=3ax^2-6x=3a(x^2-2x/a+1/a^2)-3/a=3a(x-1/a)^2-3/a
When a > 0
Then x = 1 / a = - 1
therefore
a>0
When a = 0, there is no solution

Given that the line L: y = KX + 2 and the circle x ^ 2 + (Y-1) ^ 2 = 4, let the line and the circle intersect at two points a and B, the absolute value of AB is the square of 14 and K is greater than 0
(1) Finding the equation of straight line
(2) Let circle D and circle C be symmetric with respect to line L, and the equation of circle D is obtained

（1） It is easy to know that a straight line l always passes through a fixed point (0,2), and the distance between the fixed point and the center of a circle (0,1) is 1 ＜ 2 (radius of a circle). Therefore, a straight line and a circle always intersect at two points. According to the vertical diameter theorem and Pythagorean theorem, the distance from a straight line L to the center of a circle = √ 2 / 2. Then from the "distance formula from a point to a straight line", the distance from a straight line L: kx-y + 2 = 0 to the center of a circle (0,1) can be obtained as 1 / √ (K & sup2; + 1); +1) (K ＞ 0) = = = = > k = 1. Straight line L: y = x + 2. (2) from the proposition, we can know that the radius of circle D is 2. If the center of circle D is (a, b), then point (a, b) and point (0,1) are symmetric with respect to the straight line y = x + 2. By combining numbers and shapes, we can get a = - 1, B = 2. That is to say, the center of circle D is (- 1,2), ‖ circle D: (x + 1) & sup2; + (Y-2) & sup2; = 4