Matrix transposition

[Obeliks2]

Hello, I need to make a code for two transposition matrices, and then they are counted according to the task in the picture. How can I do that? I just don't understand at all

Matrix A
|1|
|2|
|4|

Matrix B
|0|
|3|
|2|

Task
C=2A*B

transposition over the letter A or multiplication * (T)

[MikeyBoy]

Is the problem that you don't understand the maths - i.e. that you don't know what to do to get the result?

Or is it that you don't understand how to write the code that implements the maths that you do understand?

[Obeliks2]

I don't understand the matrix math and how the code can be done with transposition matrices and task like this

[lastchance]

@Obeliks2,
Your written description of the problem makes no sense at all.

transposition over the letter A or multiplication * (T)

What exactly is that supposed to say?

If you can't state the problem correctly then people can't help you.

Perhaps you could provide a picture or link to the original exercise.

[seeplus]

Task
C=2A*B

You need to specify what is meant by '*' in the context of array multiplication. In matrix algebra, you can't multiply a 3x1 matrix by a 3x1 matrix.

The usual rule of matrix multiplication of 2 matrices of dimensions r1, c1 and r2, c2 is that r2 = c1 resulting in a matrix of dimensions r1, c2

[Ganado]

I think I get it. The * means transposition.
So it's saying transpose A, then multiple by with B.

OP, you should practice transposing/multiplying 2x1 and 1x2 matrices first, and see the patterns there, and then see the pattern in 3x1 and 1x3. Show us what what result should be in your example problem.

https://www.mathsisfun.com/algebra/matrix-multiplying.html

[seeplus]

So using the original matrices,
2A gives

2
4
8

which transposed gives

2 4 8

then multiply by

0
3
2

which gives

28

Or transposing B then gives

2
4
8

multiply 0 3 2 which gives

0 6 4
0 12 8
0 24 16

???

[Ganado]

Yes (assuming that's what it means). But it's Obeliks that needs to ask that and try it out him/herself.

Edit: You edited your post. I was saying yes to the 28.

[lastchance]

#include <iostream>
#include <valarray>
using namespace std;

int main()
{
   valarray<double> A{ 1, 2, 4 }, B{ 0, 3, 2 };
   cout << 2 * ( A * B ).sum() << '\n';
}

[Obeliks2]

https://imgur.com/a/YozrsK8
Here is a picture that shows more which matrices are in the task and which task is under them



Also, I found out code, but whether it does transpose matrices I still don't understand. And which code makes the task easier and more correct? This one or which made by lastchance?

#include <iostream>
 
int main()
{
    int A[] = { 1, 2, 4 };
    int B[] = { 0, 3, 2 };
    std::cout << "C = " << (2 * (A[0] * B[0] + A[1] * B[1] + A[2] * B[2]));
}

[Ganado]

For anyone that doesn't want to click on the screenshot, it's just saying: 2 A^TB

Both your 'found' code and lastchance's do essentially the same thing.
Both are written in such a way that the A array is "pre-transposed" so you are essentially just doing a dot product and then scaling it by 2.

[Obeliks2]

Does this mean that these two codes are already fully executed task or not yet?

[Ganado]

We have no idea what the requirements of your actual assignment are. The code in question produces the answer to the "problem" in the screenshot you have shown. If that's all you need to do, then you're done.

The code executes the multiplication/addition at run-time (assuming the compiler isn't aggressively optimizing).
Not sure what you're asking.