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What is the multiplication formula of second order matrix
Cross multiplication
What is the multiplication formula of matrix
If a, B and C represent three matrices with C = AB, a is n rows and m columns, B is m rows and Q columns, then C is n rows and Q columns
Then for any element CIJ of C matrix, there is a
Cij=ai1*b1j+ai2*b2j+ai3*b3j+...+ain*bnj
i=1,2,3,...,n,j=1,2,3,...q
As shown in the figure, the area of a right triangle is 20 square centimeters, and the area of a circle is () square centimeters
A. 80B. 62.8C. 125.6
Let the radius of the circle be r cm. From the meaning of the question, we can get: R2 △ 2 = 20, & nbsp; & nbsp; R2 = 40, the area of the circle = π R2 = 3.14 × 40 = 125.6 (square cm); answer: the area of the circle is 125.6 square cm
1. If the inequality f (x) = LG (x + A / X - 2), if for any x ∈ [2, + ∞] f (x) > 0, find the value range of real number a
2. When x ∈ (1,2), the inequality (x-1) 2 ＜ logax holds and the range of a is obtained
1. LG (x + A / X - 2). If f (x) > 0 is constant for any x ∈ [2, + ∞], then x + A / X - 2 > 1. If f (x) > 0 is constant for any x ∈ [2, + ∞], then x & sup2; - 3x + a > 0 is constant, then a ＞ 3x-x & sup2; if f (x) > 0 is constant, then 3x-x & sup2 should be obtained. The maximum value of X ∈ [2, + ∞] can be 2, (x-1) 2 ＜ logax
Finding the partial derivative of Z = (1 + XY) ^ y at (1,1)
z = (1+xy)^y
∂z/∂x = y²(1+xy)^(y-1)
lnz = yln(1+xy)
∂z/∂y /z = ln(1+xy) + xy/(1+xy)
∂z/∂y = [ln(1+xy) + xy/(1+xy)] (1+xy)^y
∂z/∂x (1,1) = 1²(1+1*1)^(1-1)=1
∂z/∂y (1,1) = [ln(1+1*1)+1*1/(1+1*1)](1+1*1)^1 = 2(ln2 + ½) = 1+ ln4
Use a black chess piece of the same size to place the figure in the way shown in the figure. According to this rule, the nth figure needs a chess piece______ (expressed by an algebraic expression containing n)
The first graph needs chessboard 4; the second graph needs chessboard 4 + 3 = 7; the third graph needs chessboard 4 + 3 + 3 = 10 The nth graph needs 4 + 3 (n-1) = 3N + 1 pieces, so the answer is: 3N + 1
1. It is known that the image of the quadratic function y = ax ^ 2 + BX + C of X passes through the intersection of the straight line y = - 3x + 3 and the X axis, Y axis and the point (- 3,0), and the analytic expression of the quadratic function is obtained
In the function y = - 3x + 3, let x = 0, then y = 3, that is, the intersection of the line y = - 3x + 3 and the Y axis is (0,3). Then let y = 0, then - 3x + 3 = 0, and the solution is x = 1, that is, the intersection of the line y = - 3x + 3 and the X axis is (1,0), because the image of the quadratic function y = ax & sup2; + BX + C passes through the intersection of the line y = - 3x + 3 and the X axis, Y axis, and the point (- 3,0), that is
The known function y = - 2x-2x-5
When x takes what value, y < 0?
Y = - 1 / 2 x 2 - 3x - 50 / 2
x-1
Given that the real number A.B satisfies the following relation a2-5a + 2 = 0, 2b2-5b + 1 = O, find the value of (AB + 1) / b
2B ^ 2-5b + 1 = 0 can be changed into (1 / b) ^ 2-5 * 1 / B + 2 = 0
The equation x ^ 2-5x + 2 = 0 can be constructed by a ^ 2-5A + 2 = 0 and (1 / b) ^ 2-5 * 1 / B + 2 = 0, then a and 1 / b are two of the equations
a+1/b=5
a*1/b=2
Then (AB + 1) / b = a + 1 / b = 5
There is a triangle ABC with an angle E in the middle of it to prove that ab + AC is greater than EB + EC
Extend be to AC to f
In triangle ABF, AB + AF > BF = be + EF
In triangle EFC, EF + FC > EC
So AB + AF + EF + FC > EB + EF + EC
So AB + AF + FC > EB + EC
So AB + AC > EB + EC
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