If the eigenvalues of the third-order matrix A are 1, - 1,2, what is the determinant of a ^ 3-5a ^ 2 | Math enthusiast | Mathematical art

If the eigenvalues of the third-order matrix A are 1, - 1,2, what is the determinant of a ^ 3-5a ^ 2

Because all the eigenvalues of a are 1,2, - 1
So the eigenvalues of a ^ 3-5a ^ 2 are - 4, - 12, - 6
So | a ^ 3-5a ^ 2 | = (- 4) (- 12) (- 6) = - 288

The first line 21000, the second line 12100, the third line 01210, the fourth line 00121, the fifth line 00012 of the determinant is transformed into the upper triangular form and the result is calculated?

2 1 0 0 0
1 2 1 0 0
0 1 2 1 0
0 0 1 2 1
0 0 0 1 2
R1r5
1 2 1 0 0
0 1 2 1 0
0 0 1 2 1
0 0 0 1 2
2 1 0 0 0
R5-2r1 was obtained
1 2 1 0 0
0 1 2 1 0
0 0 1 2 1
0 0 0 1 2
0 -3 -2 0 0
R5 + 3R2 is obtained as follows:
1 2 1 0 0
0 1 2 1 0
0 0 1 2 1
0 0 0 1 2
0 0 4 3 0
R5-4r3 was obtained as follows
1 2 1 0 0
0 1 2 1 0
0 0 1 2 1
0 0 0 1 2
0 0 0 -5 -4
R5 + 5r4 is obtained as follows:
1 2 1 0 0
0 1 2 1 0
0 0 1 2 1
0 0 0 1 2
0 0 0 0 6
So determinant = 1 × 1 × 1 × 6 = 6

Find the value of determinant in the first line, the second line, the third line and the fourth line

3 1 1 1
1 3 1 1
1 1 3 1
1 1 1 3 = 3^4 + 1 + 1 + 1 - 3 - 3 - 3 -3 = 81 - 9 = 72

The transformation of determinant into triangle determinant, and its value (need process): - 22 - 404 - 13531 - 2 - 32051

The first line extracts (- 2) to get (- 2) * {1, - 1,2,0; 4, - 1,3,5; 3,1, - 2, - 3; 2,0,5,1};
The first line is multiplied by (- 4), (- 3), (- 2) respectively and added to the second, third and fourth lines to get (- 2) * {1, - 1,2,0; 0,3, - 5,5; 0,4, - 8, - 3; 0,2,1,1};
The fourth line is multiplied by (- 1) and added to the second line to get (- 2) * {1, - 1,2,0; 0,1, - 6,4; 0,4, - 8, - 3; 0,2,1,1};
The second line is multiplied by (- 4) and (- 2) respectively and added to the third and fourth line to get (- 2) * {1, - 1,2,0; 0,1, - 6,4; 0,0,16, - 19; 0,0,13, - 7};
Finally, the original formula = (- 2) * (16 * (- 7) - 13 * (- 19)) = - 270

As shown in the figure, it is known that ∠ 1, 2 and 3 are the three outer angles of triangle ABC, and ∠ 1: ∠ 2: ∠ 3 = 2:3:4. Calculate the degree of each inner angle of triangle ABC
The basis of every step,

Because the sum of the external angles of the triangle is 360
Let ∠ 1 = 2x
2x+3x+4x=360
x=40
The three outer angles of the triangle are 80, 120 and 160 respectively
If the three internal angles of a triangle are a, B and C
A+B+C=180
A+B=80
B+C=120
A+C=160
A = 60, B = 20, C = 100 (unit degree)

As shown in the figure, the bisectors of the two outer angles of △ ABC intersect at point D, ∠ a = 50 °, then ∠ D=______ ．

According to the theorem of sum of internal angles, the definition of bisector of angles and the properties of external angles of triangles, it is obtained that ∠ d = 180 ° - (∠ 1 + ∠ 2) = 180 ° - 12 (∠ CBE + ∠ BCF) = 180 ° - 12 (180 ° - ∠ ABC + 180 ° - ∠ BCA) = 180 ° - 12 (180 ° + ∠ a) = 90 ° - 12 ∠ a = 65 °

It is known that the ratio of the degrees of the three outer angles of the triangle ABC is 2:3:4, and the degrees of the three inner angles are calculated

Triangle external angle and 360
360/（2+3+4）=40
40*2=80
40*3=120
40*4=160
180-80=100
180-120=60
180-160=20
It's 100 / 60 / 20 degrees

As shown in the figure, in △ ABC, ∠ B = 48 ° and the bisector of the triangle ∠ DAC and ∠ ACF intersects at point E, then ∠ AEC

∵ FA bisection ∠ DAC
∴∠1=∠DAC/2
∵ FC bisection ∠ ACF
∴∠2=∠ACF/2
∴∠1+∠2=(∠DAC+∠ACF)/2
∵∠B+∠3+∠4=180°
∠B=48°
∴∠3+∠4=132°
∵∠3+∠DAC=180
∠4+∠ACF=180
∴∠3+∠4+∠DAC+∠ACF=360
∴∠DAC+∠ACF=228
∴∠1+∠2=114
∴∠AEC=180-∠1-∠2=66°

As shown in the figure, ∠ B = 47 ° in △ ABC, the bisector of the triangle's outer angle ∠ DAC and ∠ ACF intersects at point E, then ∠ Abe=______ °．

As shown in the figure: make em ⊥ BD, en ⊥ BF, EO ⊥ AC vertical feet as m, N, O respectively, ∵ AE, CE are bisectors of ∠ DAC and ∠ ACF, ∵ EM = EO, EO = en, ∵ EM = en, ∵ be are angular bisectors of ∠ ABC, ∵ Abe = 12 ∠ ABC = 23.5 °

If the eigenvalues of the third-order matrix A are 1, - 1,2, what is the determinant of a ^ 3-5a ^ 2