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A simple matrix exercise, Matrix a [1,0] m matrix [a, b], a and M matrix are interchangeable, determine the relationship between a, B, C, D, and write matrix M [ 2,-1] [ c,d] In addition, I want to ask the next matrix 2 * 3, will it have the reciprocal of the matrix
Directly substituting am = ma, (a, b) = (a + 2B, C + 2D) (2a-c, 2b-d) (- B - D), then B = 0, C + 2D = 2a-c = 0, d = - A, C = 2A, so (10) M = a (2 - 1) = AA. A matrix which is not a square matrix can also have "inverse". A typical "inverse" is Moore Penrose inverse
Matrix exercises,
In R ^ 4, find the transition matrix A from base X1 to xn to base Y1 to YN, and find the coordinates of vector a under the specified base
(1) [X1 = (1,2, - 1,0) ^ t, X2 = (1, - 1,1,1) ^ t, X3 = (- 1,2,1,1) ^ t, X4 = (- 1, - 1,0,1) ^ t,] [Y1 = (2,1,0,1) ^ t, y2 = (0,1,2,2) ^ t, Y3 = (- 2,1,1,2) ^ t, Y4 = (1,3,1,2) ^ t], a = (1,0,0) ^ t under the base x1, X2, X3, X4;
（2） 【X1=(1,1,1,1)^T ,X2=(1,1,-1,-1)^T,X3=(1,-1,1,-1)^T,X4=(1,-1,-1,1)^T】【Y1=(1,1,0,1)^T,Y2=（2,1,3,1)^T,Y3=(1,1,0,0)^T,Y4=(0,1,-1,-1)^T】,
The coordinates of a = (1,0,0, - 1) ^ t under the basis Y1, Y2, Y3, Y4
The problem you asked didn't use the transition matrix. It only requires the coordinates of the vector under the base. Then you can use the elementary row transformation. See the figure below
Some matrix problems
1.det(A)det(A*)=?
2. Inverse matrix (inverse matrix A + inverse matrix B) =?
3. A is symmetric matrix and B is antisymmetric matrix. It is proved that ABA is antisymmetric
Hope to give the general process, which properties are often used, talent matrix feel more flexible transformation, what good ideas
Hi me
Let f (x) = ax + LNX (a) belong to R. (1) find the monotone interval of F (x); (2) Let G (x) = x * 2-2x + 2, if any x 1 belongs to (0, positive infinity), all of them are
Let f (x) = ax + LNX (a) belong to R. (1) find the monotone interval of F (x); (2) Let G (x) = x * 2-2x + 2, if x 1 belongs to (0, positive infinity), x 2 belongs to [0,1], so that f (x 1) is less than g (x 2), find the value range of real number a
f'(x)=a+1/x=a(x+1/a)/x
When a > 0, the solution of - 1 / A0 is 0
Two simple mathematical problems of initial rise
（1+1/2）（1+1/2²）（1+1/2^4）(1+1/2^8)+1/2^15
(6X+7)²(3X+4)(X+1)=6
The calculation process should be specified
（1+1/2）（1+1/2²）（1+1/2^4）(1+1/2^8)+1/2^15
=2*(1-1/2)（1+1/2）（1+1/2²）（1+1/2^4）(1+1/2^8)+1/2^15
=2*(1-1/2^2)（1+1/2²）（1+1/2^4）(1+1/2^8)+1/2^15
=2*(1-/2^4)（1+1/2^4）(1+1/2^8)+1/2^15
=2*(1-1/2^8)(1+1/2^8)+1/2^15
=2*(1-1/2^16)+1/2^15
=2-1/2^15+1/2^15
=2
What's the second question for? To solve the equation? Please write the question clearly
Y = 1 / (2cosx-1) derivation
y=1/(2cosx-1)
y'=[1/(2cosx-1)]'
=[(1')*(2cosx-1)-1*(2cosx-1)']/[(2cosx-1)^2]
=2sinx/[(2cosx-1)^2]
Be interesting
Go to the bookstore and buy a test paper linking kindergarten and primary school
If there is no solution for the system of equations {ax + 3Y = 9} about X and y, find the solution of A. {2x-y = 1
2x-y=1, y=2x-1
Ax + 3Y = 9, ax + 6y-3 = 9, (6 + a) x = 12, only when a = - 6, there is no solution
Select nine numbers from the ten numbers of 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 to form 2-digit and 3-digit respectively, and then the sum of these three digits must be equal to 2010,
Get the right answer
957+843+210=2010
Given that the hyperbola x ^ 2 / 3-y ^ 2 = 1, the straight line y = KX + m intersects the hyperbola at two points c and D, and the vertical bisector of C and D passes B (0, - 1), the value range of M is obtained
Y = KX + m (k, M is not equal to 0)
Given that hyperbola x2 / 3-y2 = 1, straight line y = KX + m (k, m ≠ 0) intersects hyperbola at two points c and D, and the vertical bisector of CD passes through point B (0, - 1), try to find the value range of M
If the slope of CD = k, then the slope of the vertical bisector = - 1 / K
Let C and d be (x1, Y1), (X2, Y2), and let m be (a, b),
Let the bisector be l: y = - X / K + B2
Because l goes through (0, - 1), B2 = - 1
L is y = - X / k-1
Because (x12-x22) / 3 = (Y12 + 1) - (Y22 + 1)
=>(x1+x2)/3(y1+y2)=(y1-y2)/(x1-x2)=k
Then a / 3B = k,
If M is also on the line L, then B = - A / k-1 (substituting k = A / 3b)
B = - 1 / 4, k = - 4A / 3
Obviously, point m is also on the line y = KX + m, then B = Ka + M
Then - 1 / 4 = - 3k2 / 4 + M
3k2=4m+1
Substituting y = KX + m into hyperbolic equation to eliminate y
X2 / 3-k2x2-2kmx-m2-1 = 0 make the equation have two real roots
Then 4m2k2-4 (- M2-1) (1 / 3-k2) > 0
=>m2/3-k2+1/3>0
=>m2+1>3k2=4m+1
The solution is m > 4 or m
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