Properties of determinant in linear algebra

Properties of determinant in linear algebra

Property 1: the determinant is equal to its transposed determinant. Property 2: if two rows of the determinant are the same, the determinant is 0. Property 3: if two rows of the determinant are proportional, the determinant is 0. Property 4: add the multiple of one row of the determinant to another row, and the value of the determinant remains unchanged. Property 5: for the position of two rows in the newline determinant, the determinant is inversed

In the non right triangle ABC, the lines of high BD and CE intersect at O. when the angle a = 60 degrees, the angle BOC is zero=

Use the sum of internal angles of quadrilateral to answer 120 degrees for 360 degrees

It is known that in non RT △ ABC, a = 45 ° and the straight line of high BD and CE intersects at point h. draw a graph and find out the degree of BHC

① As shown in Figure 1, when △ ABC is an acute triangle, ∵ BD and CE are the high lines of △ ABC, ∵ ADB = 90 °, ∵ BEC = 90 °, in △ abd, ∵ a = 45 °, ∵ abd = 90 ° - 45 ° = 45 °, ∵ BHC = ∵ abd + ∵ BEC = 45 ° + 90 ° = 135 °; when △ ABC is an obtuse triangle, ∵ BD and CE are △ ab

It is known that in non RT △ ABC, a = 45 ° and the straight line of high BD and CE intersects at point h. draw a graph and find out the degree of BHC

① As shown in Figure 1, when △ ABC is an acute triangle, ∵ BD and CE are the high lines of △ ABC, ∵ ADB = 90 °, ∵ BEC = 90 °, in △ abd, ∵ a = 45 °, ∵ abd = 90 ° - 45 ° = 45 °, ∵ BHC = ∵ abd + ∵ BEC = 45 ° + 90 ° = 135 °; ② when △ ABC is an obtuse triangle, ∵ BD and CE are the high lines of △ ABC, ∵ a + ∵ ace = 90 °, ∵ BHC + ∵ HCD = 90 ° and ∵ ace = ∵ HCD (the vertex angles are equal) ）In conclusion, the degree of ∠ BHC is 135 ° or 45

It is known that in the non right triangle ABC, ∠ a = 45 ° where the straight line of height AD and CE intersects h, what is the degree of ∠ BHC?

1) When △ ABC is an acute triangle,
∵ BD, CE is the height of △ ABC, ∠ a = 45 °
∴∠ADB=∠BEH=90°
In △ abd, ∠ abd = 180 ° - 90 ° - 45 ° = 45 °
∵∠ BHC is the outer angle of △ bhe
∴∠BHC=90°+45°=135°
(2) When △ ABC is an obtuse triangle
∵ h is the intersection point of two straight lines with the height of △ ABC, ∠ a = 45 °
∴∠ABD=180°-90°-45°=45°
In RT △ beh, ∠ BHC = 180 ° - 90 ° - 45 ° = 45 °
The degree of ∠ BHC is 135 ° or 45 °

As shown in the figure, the bisector of triangle ABC, angle ABC and angle ACD intersects at point O. try to explore the relationship between angle BOC and angle A and prove the conclusion

Angle a minus angle BCD = 1 / 2 (angle ACD angle b)

As shown in the figure, the circle O is the circumscribed circle of △ ABC, BC = 4 cm, a = 30 degree_ Connect ob, OC, find the degree of BOC. (2) find the area of circle o
(3) Do you have another way to find the area of circle O without constructing △ OBC?

(1) Because the circle angle is equal to the shift of the center angle, so,

If 4x-3y = 0 and X and y are not equal to zero, then the value of 4x-5y / 4x = 5Y is () a.1/31 b.31 C. - 1 / 4 d.32

It's for 4x-5y / 4x + 5Y. Yes, choose C
4x-3y=0
4x=3y
therefore
Original formula = 4x-5y / 4x + 5Y
=(3y-5y)/(3y+5y)
=-2y/8y
=-1/4

If the equations 4x-3y and X ≠ o, y ≠ o, then what is the value of 4x-5y / 4x + 5Y

Is 4x-3y = 0?
4x=3y
y=(4/3)x
So the original formula = [4x-5 (4 / 3) x] / [4x + 5 (4 / 3) x]
=(-8/3)x/(32/3)x
=-1/4

The trajectory equation of a point in space which is equal to the distance between a (- 1,2,3) and B (0,0,5) is
The trajectory equation of a point in space which is equal to the distance between a (- 1,2,3) and B (0,0,5) is