Prove: the determinant of n-order matrix (C B next to a 0) is equal to deta * DETB

Prove: the determinant of n-order matrix (C B next to a 0) is equal to deta * DETB

See P14, 5th edition of linear algebra, edited by Department of mathematics, Tongji University

Let a be a matrix of order 3, and the determinant of a is equal to 1 / 2. Find the adjoint matrix and inverse matrix of A

A * and a ^ - 1 cannot be obtained from the determinant of A
Because the matrix a satisfying | a | = 1 / 2 is not unique

Is there a simple algorithm for (46.5 * 50% - 3) dividing by 5 / 6?

(46.5x50%-3)÷5/6
=46.5x5/10x6/5-3x6/5
=4.65x6-18/5
=6x(4.65-0.6)
=6x4.05
=24.3

Let number a be x and number B be y. (1) the difference between two times of number a and three times of number B; (2) the sum of two times of the square of number a and the cube of number B
The square difference of the reciprocal of a and B

2x－3y
2x²＋y³
(1／x)²－(1／y)²

49 times 3.14 = how much

Equal to 153.86

The right side of a triangle is 2.32 meters long, the left side is 3.8 meters long, and the lower side is 4 meters long,

The ancient Chinese had a theorem for a long time
s=√[p*(p-a)*(p-b)*(p-c)]
Where p = (a + B + C) / 2 = 2.32 + 3.8 + 4 = 5.06
So s = √ [5.06 * (5.06-2.32) * (5.06-3.8) * (5.06-4)] = 4.303172393

The solution equation is: X-2 / 7 x + 12: x = 13:14
The problem-solving process is a little more detailed, urgently needed!
By the way, help me solve this equation: x + (3x-16) △ 25% = 10 (x + 1)
O(∩_ Thank you

(x-2/7x＋12)：x=13:14
13X=(5/7X+12)x14
13X-10X=168
3X=168
X=56

A semicircle with a circumference of 15.42 decimeters has an area of () square decimeters

Circumference of semicircle = half of circumference + diameter = 3.14 × radius × 2 △ 2 + radius × 2 = 3.14 × radius + radius × 2 = (3.14 + 2) × radius = 15.42
So the radius of the circle is 15.42 △ 3.14 + 2 = 3 decimeters
Area of circle = 3.14 × 3 & # 178; = 28.26 square decimeters
I hope I can help you. Happy New Year

The simple operation of 0.78 × 201 is 8.5 × 10.1, 1.96 × 0.25-0.25 × 0.96

Analysis,
0.78×201
=(0.78)×(200+1)
=0.78×200+0.78
=156.78.
8.5×10.1
=8.5×(10+0.1)
=8.5×10+0.85
=85.85
1.96×0.25-0.25×0.96
=0.25×(1.96-0.96)
=0.25×1
=0.25

BCD is on the same line, and BC = 2CD, triangle ABC and triangle CDE are equilateral triangles. Connect AE and calculate the degree of angle CAE

Let EF / / BD intersect AC with F, then △ CEF is an equilateral triangle, EF = AF = CF ∧ CAE = 30 °