How to solve the matrix [34 - 641241 - 12 - 70] and change the matrix into ladder matrix and simplest matrix

How to solve the matrix [34 - 641241 - 12 - 70] and change the matrix into ladder matrix and simplest matrix

r1-3r2,r2+r3
0 -2 -18 1
0 4 -3 1
-1 2 -7 0
r2+2r1
0 -2 -18 1
0 0 -39 3
-1 2 -7 0
Exchange bank
-1 2 -7 0
0 -2 -18 1
0 0 -39 3
--This is the ladder matrix
r1*(-1),r2*(-1/2),r3*(-1/39)
1 -2 7 0
0 1 9 -1/2
0 0 1 -1/13
r1-7r3,r2-9r3
1 -2 0 7/13
0 1 0 5/26
0 0 1 -1/13
r1+2r2
1 0 0 12/13
0 1 0 5/26
0 0 1 -1/13
--This is the simplest form of a row (or row simplified ladder matrix)

In this paper, [2, - 3,7,4,3; 1,2,0, - 2, - 4; - 1,5, - 7, - 6,7; 3, - 2,8,3,0] is transformed into row simplest matrix

r3-r1-r2,r1-2r2,r3+r1
0 -7 7 8 11
1 2 0 -2 -4
0 7 -7 -8 3
0 -1 1 1 1
r1-7r4,r2+2r4,r3+7r4
0 0 0 1 4
1 0 2 0 -2
0 0 0 -1 10
0 -1 1 1 1
r3+r1,r4-r1
0 0 0 1 4
1 0 2 0 -2
0 0 0 0 14
0 -1 1 0 -3
Sequence: R3 * (1 / 14), r1-4r3, R2 + 2r3, R4 + 3r3, R4 * (- 1)
0 0 0 1 0
1 0 2 0 0
0 0 0 0 1
0 1 -1 0 0
Exchange will do
1 0 2 0 0
0 1 -1 0 0
0 0 0 1 0
0 0 0 0 1
Pay attention to the method when doing exercises

It is known that the x square of 16 plus the Y square of 12 is equal to 1, and the fixed-point coordinates and focus coordinates of the ellipse are written out

Vertex coordinates are (4,0), (- 4,0), (0,2 √ 3), (0, - 2 √ 3)
Focus coordinates (- 2,0), (2,0)

Given the vector M = (COSA, Sina), n = (2, − 1), and m · n = 0. (1) find the value of Tana; (2) find the range of function f (x) = cos2x + tanasinx (x ∈ R)

(1) (2) from (1) we know that Tana = 2 leads to f (x) = cos2x + 2sinx = 1 − 2sin2x + 2sinx = - 2 (SiNx − 12) 2 + 32. (6) because x ∈ R, so SiNx ∈ [- 1, 1]. (7) when SiNx = 12, f (x) has the maximum value of 32; (9) when SiNx = - 1, f (x) has the minimum value of - 3; (11) when SiNx = - 1, f (x) has the maximum value of 32; (9) when SiNx = - 1, f (x) has the minimum value of - 3; (11) So the range of F (x) is [− 3, 32]. (12 points)

Make a straight line L through point P (1,2) and cross the positive half axis of X axis. The positive half axis of Y lies at two points a and B. find the equation of the straight line l when the area of △ AOB reaches the minimum

Let AB: X / A + Y / b = 1, a > 0, b > 0,
It passes P (1,2),
∴1=1/a+2/b>=2√[2/(ab)],ab>=8,
When 1 / a = 2 / b = 1 / 2, i.e. a = 2, B = 4, take the equal sign,
At this time, the minimum area of △ AOB is obtained, and the equation of AB is
x/2+y/4=1,
That is, 2x + y-4 = 0

Find the range of y = (X & # 178; + 3x-2) / (X & # 178; - x + 1)
Please answer as soon as possible

It's too busy to use the discriminant method to multiply y by the denominator, and then merge the similar terms about X. then Δ is greater than 0 to get the inequality solution about y, and the inequality can get the range of Y
The method is correct and can't be typed out

Enlarge a rectangle of 4cm in length and 3cm in width by 3:1, and the area of the figure is the second power of () cm

Enlarge a rectangle of 4cm in length and 3cm in width by 3:1, and the area of the figure is the second power of (108) cm
4 × 3 × 3 × 3 = 108 square centimeter

If the system of equations ax + by = 1 x ^ 2 + y ^ 2 = 10 with respect to XY has solutions and all solutions are integers, then the number of ordered real number pairs (a, b) is
If the system of equations ax + by = 1 x ^ 2 + y ^ 2 = 10 has solutions and all solutions are integers, then the number of ordered real number pairs is?
Correct answer: 32

There are eight groups of (1,3) (- 1,3) (- 1, - 3), (1, - 3) (3,1) (3, - 1) (- 3, - 1), (- 3,1) of integer solutions of x ^ 2 + y ^ 2 = 10. All the tangents passing through any point satisfy the following conditions: all the eight lines passing through any two points satisfy the following conditions: the equations {ax + by = 1, x ^ 2 + y ^ 2 = 10 have solutions, and all the solutions are integers C (8,2)

If M and N are opposite numbers and the absolute value of a is 2, find the value of a + 2 (M + n)

From the meaning of the title:
m+n=0
A = 2 or - 2
When a = 2,
a+2（m+n）=2+2×0=2
When a = - 2,
a+2（m+n）=-2+2×0=-2

Given the function f (x) = x2-ax + 2a-1 (a is a real constant); (1) if a = 0, find the monotone increasing interval of function y = | f (x) | (2) let the minimum value of F (x) in interval [1,2] be g (a), find the expression of G (a); (3) let H (x) = f (x) x, if function H (x) is an increasing function in interval [1,2], find the value range of real number a

(1) When a = 0, f (x) = x2-1, then combining with the image of y = | f (x) | we can get that the function increases monotonically on (- 1, 0), (1, + ∞). (2) the symmetry axis of the function f (x) = x2-ax + 2a-1 is x = A2, when A2 ≤ 1, i.e. a ≤ 2, G (a) = f (1) = a; when 1 < A2 < 2, i.e. 2 < a < 4