Using determinant definition to calculate determinant The answer doesn't matter, it's the process

Using determinant definition to calculate determinant The answer doesn't matter, it's the process

The last number of the first column is n, expanded by the first column, and the determinant = (- 1) to the power of 2N-1*
1 0 ...0
0 2.0
0 0 3..0
.
0 ... n-1
=2N-1 power of (- 1) * n!
Using determinant definition to calculate determinant
0 0 0 a1
0 0 a2 0
0 a3 0 0
a4 0 0 0
According to the definition, take the positions (1,4) (2,3) (3,2) (4,1) of A1, A2, A3 and A4 to get that n (1234) = 0 and n (4321) = 6 are even numbers, so they are positive; other items contain at least one zero element, so other items are 0, so d = a1a2a3a4
Use determinant definition to calculate the following determinants
1 1 1 0
0 1 0 1
0 1 1 1
0 0 1 0
Non zero items:
A11a22a34a43 = 1, with negative sign -
A11a24a32a43 = 1, with positive sign +
So, d = - 1 + 1 = 0
| 1 1 1 0 |
| 0 1 0 1 |
| 0 1 1 1 |
| 0 0 1 0 |
=
Chinese style |1 0 1|__ _ |1 1 0|_____ |1 1 0|_____ |1 1 0|
1 * |1 1 1| - 0 * |1 1 1| + 0 * |1 0 1| - 0 * |1 0 1|
Chinese style |0 1 0|__ _ |... unfold
| 1 1 1 0 |
| 0 1 0 1 |
| 0 1 1 1 |
| 0 0 1 0 |
=
Chinese style |1 0 1|__ _ |1 1 0|_____ |1 1 0|_____ |1 1 0|
1 * |1 1 1| - 0 * |1 1 1| + 0 * |1 0 1| - 0 * |1 0 1|
Chinese style |0 1 0|__ _ |0 1 0|_____ |0 1 0|_____ |1 1 1|
=
|1 0 1|
|1 1 1|
|0 1 0|
=
Wei |1 1|_____ |0 1|_____ |0 1|
1 * |1 0| - 1 * |1 0| + 0 * |1 1|
=(1*0-1*1)-(0*0-1*1)+0
=0
| 1 1 1 0 |
|0 1 0 1 | (for determinant)
| 0 1 1 1 |
| 0 0 1 0 |
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Use the definition of determinant to calculate the following determinants
a11 a12 a13 a14 a15
a21 a22 a23 a24 a25
a31 a32 0 0 0
a41 a42 0 0 0
a51 a52 0 0 0
Solution: by the definition of determinant, each term in the definition is obtained by multiplying exactly one number in each row and column of determinant
Because the elements of columns 3, 4 and 5 in rows 3, 4 and 5 are all 0
So every item in the determinant definition is equal to 0
So the determinant is equal to 0
Given a = {y | y = log2 (x), X ＞ 1}, B = {y | y = (1 / 2) ^ x, X ＞ 2}, then a ∩ B is equal to?
A∩B={y|0
It is known that the solution set of (3a + 2b) x ＞ 4A + 3b is (- some number, 2), and the solution inequality (a + b) x ＞ 3A + B is obtained
What is the slope? The formula?
Slope is the degree of inclination. The slope is generally expressed by K. The value of slope k is the tangent of the angle between the straight line and the positive direction of X axis. If any two points on the straight line are (x1, Y1), (X2, Y2), then the slope of the straight line k = (y2-y1) / (x2-x1). The straight line is parallel to the Y axis, and the slope does not exist. It is parallel to the X axis, and the slope is 0
The slope, also known as the "angle coefficient", represents the degree of inclination of a straight line relative to the abscissa axis. The tangent of the angle between a line and the positive half axis of the abscissa axis of a plane rectangular coordinate system is the slope of the line relative to the coordinate system. If the line is perpendicular to the X axis, the tangent of the right angle is infinite, so there is no slope for this line. When the slope of the line L exists, for the linear function y = KX + B, (oblique section) k is the slope of the function image.
For details: http://baike.baidu.com/view/271319.htm... open
The slope, also known as the "angle coefficient", represents the degree of inclination of a straight line relative to the abscissa axis. The tangent of the angle between a line and the positive half axis of the abscissa axis of a plane rectangular coordinate system is the slope of the line relative to the coordinate system. If the line is perpendicular to the X axis, the tangent of the right angle is infinite, so there is no slope for this line. When the slope of the line L exists, for the linear function y = KX + B, (oblique section) k is the slope of the function image.
For details: http://baike.baidu.com/view/271319.htm Put it away
The slope is the slope of the line
Not equal to (yx2-2) / (yx1)
2X + 5Y + 4Z = 6,3x + y-7z = - 4,4x + 3Y + 12z = - 2 find x + y + Z
Divide the left of the three expressions by 9 to get x + y + Z = 0
2X + 5Y + 4Z = 6 (1) 3x + Y - 7z = 4 (2) (1) formula × 2: 4x + 10Y + 8Z = 12 (3) (2) formula × 3: 9x + 3Y - 21z
Add the three equations: 2x + 3x + 4x + 5Y + y + 3Y + 4z-7z + 12z = 6-4-2
We get: 9x + 9y + 9z = 9x (x + y + Z) = 0
That is x + y + Z = 0
x＋y＋z＝0
x=-2 y=2 z=0;x+y+z=0
(2x+5y+4z)+(3x+y-7z）+(4x+3y+12z)=6+(-4)+(-2)
(2+3+4)x+(5+1+3)y+(4-7+12)z=0
9x+9y+9z=0
9(x+y+z)=0
(x+y+z)=0
If the image of the first-order function y = KX + B (K ≠ 0) passes through point a (0,2) and intersects with the image of the positive scale function y = - x at point B, and the abscissa of point B is - 1, then the analytic expression of the first-order function is ()
A. y=x+2B. y=-x+2C. y=x-2D. y=-x-2
The abscissa of point B is - 1, y = - (- 1) = 1, the coordinate of point B is (- 1, 1), B = 2 − K + B = 1, the solution is k = 1b = 2, and the analytic expression of this linear function is y = x + 2
Given the set a = {x | y = √ (x ^ 2 + 2x-3)}, B = {y | y = log2 & X, x > 1}, then a intersects B =?
The element a is x, so it's the domain
x^2+2x-3>=0
(x+3)(x-1)>=0
X=1
So a is the set of real numbers greater than or equal to 1 and less than or equal to - 3
The B element is y, so it's a range
x> 1, base 2 > 1, increasing function
So Y > log2 (1) = 0
So the set of B is greater than 0
So a ∩ B = {Z | z > = 1}
{Y|Y>O}
A={x|y=√(x^2+2x-3)},
x^2+2x-3≥0,
X ≥ 1, or X ≤ - 3
B={y|y=log2&x,x>1},
X>1.
Then a cross B =? X > 1
A={x|y=√(x^2+2x-3)},
x^2+2x-3≥0,
X ≥ 1, or X ≤ - 3
B={y|y=log2&x,x>1},
X>1.
Then a intersects B =? X > 1