Change the following determinants into triangular determinants -2 2 -4 0 4 -1 3 5 3 1 -2 -3 2 0 5 1

Change the following determinants into triangular determinants -2 2 -4 0 4 -1 3 5 3 1 -2 -3 2 0 5 1

-2 2 -4 0
4 -1 3 5
3 1 -2 -3
2051 column 2 plus column 1, column 3 minus column 1 * 2
=
-2 0 0 0
4 3 -5 5
3 4 -8 -3
2 2 1 1 column 3 plus column 4, column 4 minus column 2 * 5 / 3
=
-2 0 0 0
4 3 0 0
3 4 -11 -29/3
2 - 7 / 3 line 3 minus line 4 * 2
=
-2 0 0 0
4 3 0 0
-1 0 -15 -5
2 - 7 / 3 column 4 minus column 3 * 1 / 3
=
-2 0 0 0
4 3 0 0
-1 0 -15 0
2 2 2 -3
So we get the triangle determinant
D= -2 *3 * -15 *-3= 270

Change the following determinants into triangles
2 -1 11 3
-1 4 -7 0
3 8 9 -5
-3 1 0 4

r3+r4,r1+2r2,r4-3r2
0 7 -3 3
-1 4 -7 0
0 9 9 -1
0 -11 21 4
r1+3r3,r4+4r3
0 34 24 0
-1 4 -7 0
0 9 9 -1
0 25 57 0
(if you have learned the determinant expansion theorem, expand by the first column, and then expand by the last column.)
r1r2,c2c3,r2r3
-1 0 -7 4
0 -1 9 9
0 0 24 34
0 0 57 25
So d = (- 1) ^ 3 (24 * 25 - 34 * 57) = 1338

Use the reduced order method to calculate the following determinant, the first line 1 + x 11, the second line 1 1-x 11, the third line 1 11 + Y 1
The first line 1 + x 11, the second line 1 1-x 11, the third line 11 + Y 1, the fourth line 11-y are used to calculate the following determinants,

The determinant property can be calculated as shown in the figure. The economic mathematics team will help you to solve it, please adopt it in time. Thank you!

As shown in the figure, ABC is an equilateral triangle, BD is the middle line, extend BC to e, so that CE = CD

It is proved that: ∵ ABC is an equilateral triangle, BD is the middle line, ∵ ABC = ∠ ACB = 60 °, DBC = 30 ° (isosceles triangle three lines in one), and ∵ CE = CD, ∵ CDE = ∠ CED. And ∵ BCD = ∠ CDE + ∠ CED, ∵ CDE = ∠ CED = 12 ∠ BCD = 30 °. ∵ DBC = ∠ Dec. ∵ DB = de (equiangular to equilateral)

As shown in the figure, in the triangle ABC, the angle c = 90 degrees, the angle BAC is equal to 60 degrees, and the vertical bisector of side AB intersects AB at D, and BC at E. if de = 2cm, find
As shown in the figure, in the triangle ABC, the angle c = 90 degrees, the angle BAC is equal to 60 degrees, and the vertical bisector of side AB intersects AB at D and BC at E. if de = 2cm, find the lengths of De and be

Then be = 2DE = 4cm
When de bisects AB vertically, AE = be = 4cm;
If ∠ EAB = ∠ B = 30 °, then ∠ CAE = ∠ BAC - ∠ EAB = 30 °, CE = AE / 2 = 2cm

As shown in the figure, in the triangle ABC, the angle c is equal to 90 degrees, the angle B is equal to 30 degrees, De is the vertical bisector of AB, BC is equal to 12 cm, then CE is equal to 0

4 pieces of 3 (is e on AB line? Because you didn't send pictures, I can only guess roughly)

As shown in the figure, in △ ABC, the vertical bisector De of AB intersects AB at D and BC at E. if CE = 3cm, calculate the length of be

∵ - C = 90 °, ∵ BAC = 60 °, ∵ B = 90 ° - 60 ° = 30 °, ∵ De is the vertical bisector of AB, ∵ AE = be, ∵ BAE = ∵ B = 30 °, ∵ CAE = ∵ BAE, ∵ de = CE = 3cm, and ∵ B = 30 °, ∵ be = 2DE = 2 × 3 = 6cm

As shown in Figure + in the triangle ABC, ad ⊥ BC, CE ⊥ AB, connected ed, F.H are ac.ed The verification of the midpoint of: FH ⊥

Connect DF, EF ⊥ ad ⊥ BC, then △ ACD is the right triangle f at the midpoint of the hypotenuse AC ⊥ DF = 1 / 2Ac ⊥ CE ⊥ AB, then △ ace is the right triangle f at the midpoint of the hypotenuse AC ⊥ EF = 1 / 2Ac ⊥ EF = DF ⊥ h is the midpoint of De, then eh = dhfh = FH ≌ EFH ≌ DFH (SSS) ⊥ fhe = ∠ FHD ⊥ fhe + ∠ FHD = 180 °

In the triangle ABC, EC = 2be, CD = 2bd, the area of the triangle BDE is 14 square centimeters, find the area of the triangle ABC?

Please see the picture above

In the triangle ABC, ad is perpendicular to BC, D is perpendicular to the foot, EC is perpendicular to AB, e is perpendicular to the foot

The triangle Abe and CEB are right triangles,
And angle B is the common angle,
So the triangle Abe is similar to CEB
We can know be / BD = BC / ab
Be / BC = BD / AB
In the triangle Abe and ABC,
Angle B is the common angle,
Plus be / BC = BD / AB
There is an angle between the two sides
We can see that BDE is similar to BAC