How to define matrix in MATLAB

How to define matrix in MATLAB

You can use the command to define an array
cells(m,n)
The definition is m * n-dimensional matrix
You can also define a matrix directly, such as a 3 × 3 matrix
A=(x11,x12,x13;x21,x22,x23;x31,x32,x33)

How to define special matrix in MATLAB

You can use the command to define an array
cells(m,n)
The definition is m * n-dimensional matrix
You can also define a matrix directly, such as a 3 × 3 matrix
A=(x11,x12,x13;x21,x22,x23;x31,x32,x33)

How to define matrix in MATLAB and how to solve matrix
For example, there is a matrix A, a B, and a C. given a * b = C, and the elements in a and C, find the B matrix

You have a wide range of questions. Let's make a concrete analysis of specific topics. Now I will briefly introduce some knowledge points of matrix involved in MATLAB. Since you ask how to define matrix, let's talk about the most basic, such as a = [1, 2; 3, 4; 5, 4]; & nbsp; & nbsp; & nbsp; & nbsp; note: matrix is generally represented by [], that is, bracket, and single element can be separated by comma or space, The semicolon in the brackets means to change the line, and the semicolon after it means not to display. There is still a lot of knowledge about the matrix, if you want to ask. The solution of the matrix: the solution of your problem can be expressed in two ways: 1. & nbsp; & nbsp; B = A / C & nbsp; & nbsp; 2. & nbsp; & nbsp; b = C * inv (a) & nbsp; & nbsp; note: the matrix can't be divided by left, divided by right, or multiplied by the inverse

The last question on the third page of the sixth grade primary school mathematics Evaluation Handbook published by Jiangsu Education Press
It's the one in stock in warehouse A and warehouse B, the one in 2010

Set: warehouse a transported x tons to warehouse B
（108—X）/2=120+X
360-2X=120+X
360-120=X+2X
240=3X
X=80
A: 80 tons were transported from warehouse A to warehouse B

[e ^ (x + y) - e ^ x] DX + [e ^ (x + y) + e ^ y] dy = 0, find the general solution of I

∵[e^(x+y)-e^x]dx+[e^(x+y)+e^y]dy=0
==>(e^y-1)e^xdx+(e^x+1)e^ydy=0
==>e^xdx/(e^x+1)+e^ydy/(e^y-1)=0
==>d(e^x)/(e^x+1)+d(e^y)/(e^y-1)=0
==>Ln (e ^ x + 1) + ln │ e ^ Y-1 │ = ln │ C (C is an integral constant)
==>(e^x+1)(e^y-1)=C
The general solution of the original differential equation is (e ^ x + 1) (e ^ Y-1) = C (C is an integral constant)

A large square of n * n can count several squares (expressed in n)

There are 178 squares with side length n
There are two squares with side length (n-1)
There are 3 squares with side length (n-2)
…………………………
There are 178 squares with side length 1
in total:
1²+2²+3²+…… +N & # 178; = n (n + 1) (2n + 1) / 6 (pieces)

There are two cuboid pillars in the children's palace, which are 4 meters high. The ground is square and the perimeter of the bottom is 1.2 meters. What is the total area of the two pillars painted
Square meter

The side area is: 1.2 × 4 = 4.8 square meters
The required paint area is 4.8 × 2 = 9.6 square meters
If you don't understand, please ask

The middle addition of four three symbols equals eight

3×3-3÷3=8

Find the general solution of dy / DX = 2x + y

Let me introduce a method
Let 2x + y = U
If we take the derivative of X on both sides, we get the following result
2+dy/dx=du/dx
therefore
du/dx-2=u
du/dx=u+2
1/(u+2)du=dx
∫1/(u+2)du=∫dx
ln|u+2|=x+ln|c|
u+2=ce^x
2x+y+2=ce^x
y=ce^x-2x-2

If the cube of 3x times the square of Y + the square of ay times the cube of x = the cube of x times the square of Y, then a = []

When XY is not zero, it is - 2, otherwise it can take any value