# Given a Matrix A, create a row vector of 1s that has same number of elements as A has rows

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Gaurav Srivastava il 4 Giu 2019
Commentato: Voss il 2 Feb 2024
Given a Matrix A, Create a row vector of 1's that has same number of elements as A has rows. Create a column vector of 1's that has the same number of elements as A has columns. Using matrix multiplication, assign the product of the row vector, the matrix A, and the column vector (in this order) to the variable result. A = [1:5; 6:10; 11:15; 16:20];
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Alex Mcaulley il 4 Giu 2019

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### Risposta accettata

Manali Gupta il 20 Mag 2021
Modificato: MathWorks Support Team il 20 Mag 2021
R_vector = ones(1,size(A,1)); C_vector=ones(size(A,2),1); result = R_vector*A*C_vector;
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can you explain the syntax row and column
Mariam Aldeepa il 2 Gen 2021
Why when you calculate the result multiplied the matrix "A" with another two matrices ?

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### Più risposte (11)

Alex Mcaulley il 4 Giu 2019
I think the goal of this forum is not to do anyone's homework, but to solve generic questions about Matlab that can help more users. This exercise is very simple, anyone who has ever used Matlab could do it, so, I think the best thing we can do is help @Gaurav to know how to learn the basics of Matlab.
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Carlo Romagnolo il 19 Apr 2021
Hi! I know this might be a long a shot, as this thread has been inactive for quite some time, but I was wondering if you could help me with a general question on this.
Here is my code to solve this problem:
A = [1:5; 6:10; 11:15; 16:20];
format compact
row = [1,1,1,1]
column = [1;1;1;1;1]
result = row * A * column
I was wondering if there is a better/more elegant way to answer the question which does not involve hardcoding the row and column vectors but instead uses functions.
Walter Roberson il 19 Apr 2021
ones(1,size(A, 1))

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Khom Raj Thapa Magar il 28 Ago 2020
Modificato: DGM il 21 Feb 2023
% Given
A = [1:5; 6:10; 11:15; 16:20];
row_vec = ones(1,size(A,1))
row_vec = 1×4
1 1 1 1
column_vec = ones(length(A),1)
column_vec = 5×1
1 1 1 1 1
result = row_vec * A * column_vec
result = 210
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DGM il 21 Feb 2023
The length() function does not return the number of columns in an array. It returns the size of the longest dimension. If A has more rows than columns, this code will fail. Use size() to get the size of a specific dimension.

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Sudhanshu Rasal il 5 Mag 2020
Modificato: DGM il 21 Feb 2023
Simple way to do this question is
X=[1 1 1 1]
Y=[1;1;1;1;1]
result=X*A*Y
????????
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el mehdi latrach il 14 Giu 2020
no, it does not work ,
it shows error
DGM il 21 Feb 2023
Modificato: DGM il 21 Feb 2023
The point of writing a program is to have the computer do the work. Literally writing out the vectors manually makes as much sense as calculating the inner product on paper and writing
result = 210; % this is my entire program
Both examples will fail for obvious reasons if the size of A changes.

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Tushar Parmar il 11 Mag 2020
A = [1:5; 6:10; 11:15; 16:20]
B(1:4)=1;
f=B*A
C(1:5)=1;
C=C'
result=f*C
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Tharindu Priyankara il 29 Ago 2020
THANKS
DGM il 21 Feb 2023
The point of writing a program is to have the computer do the work. Use size() to find the size of A; use ones() to generate the vectors programmatically.

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Arakala Gautham il 4 Apr 2020
R_vector = ones(1,size(A,1));
C_vector=ones(size(A,2),1);
result = R_vector*A*C_vector;
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DGM il 21 Feb 2023
Abhishek il 2 Lug 2023
A = [1:5; 6:10; 11:15; 16:20];
R = (A(:,1))';
R(:)=1;
C = (A(1,:))';
C(:)=1;
result = R*A*C
%@DGM bro is this code ookay?

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Rishabh Nirala il 20 Mag 2020
Modificato: DGM il 21 Feb 2023
A = [1:5; 6:10; 11:15; 16:20];
C = [1;1;1;1;1]
R = [1 1 1 1 ]
P = R * A
result = P*C
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DGM il 21 Feb 2023
Again, the point of writing a program is to have the computer do the work. Use size() to find the size of A; use ones() to generate the vectors programmatically.

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Sneham Shrikant Vaidya il 27 Mag 2020
Modificato: DGM il 21 Feb 2023
A = [1:5; 6:10; 11:15; 16:20];
A
x = [1 1 1 1 ]
y = [1;1;1;1;1]
z = A*y
result =x*z
you can also perform this way as we know z =(lxm)*(mxn) so we first multiply A*y as their inner dimension ara same
then we obtain result matrix z that has inner dimension equal to x so now we can multiply x*z to get final ans
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DGM il 21 Feb 2023
Use size() to find the size of A; use ones() to generate the vectors programmatically. Otherwise, this fails if the size of A changes.

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VISHWA D il 22 Giu 2020
A = [1:5; 6:10; 11:15; 16:20];
row_vector=[1 1 1 1 1]
col_vector=[1; 1; 1; 1]
result=(row_vector*(A'))*(col_vector)
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lijuan wang il 27 Ago 2020
I thought the assignment has a wrong requirement, right?
we should create a vector which has the some rows as A. So it should be row_vector=[1 1 1 1]'?
DGM il 21 Feb 2023
Modificato: DGM il 21 Feb 2023
@lijuan wang is correct. This answer creates the vectors incorrectly. Not only are they created manually in a way that would break if the size of A changes, they do not correspond to the specified dimensions of A. The result is correct, but the vectors are wrong, and the method is poor practice.

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Chintan il 21 Feb 2023
row_vector=ones(size(A(:,1)))'
coloumn_vector=ones(size(A(1,:)))'
result=row_vector*A*coloumn_vector
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DGM il 21 Feb 2023
I suppose that's one way, but size() supports dimension specification, which would avoid the need to address vectors of A or to transpose anything.
Also note that the ctranspose, ' operator is the complex conjugate transpose. If you just want to reorient a vector or matrix, use transpose, .' instead.

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Waseem Zafar il 17 Set 2023
Modificato: Waseem Zafar il 17 Set 2023
A = [1:5; 6:10, 11:15; 16:20]
R = [1 1 1 1]
C = [1;1;1;1;1]
Z = A*C
result = R*Z
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DGM il 17 Set 2023
This example is identical to at least two other bad examples in this thread, differing only in the variable names.
You are told to use the size of A to construct two vectors. This example, and at least four others define the vectors literally instead of programmatically creating them as told. Let me be clear. You literally wrote two of the intermediate results into the script instead of making your script calculate them. If I were grading this, I would accept it no more than if you had just written
result = 210;
... because it neglects the entire point of the exercise to presuppose the input and bake the results into the script as literals.
An answer can earn its place if it demonstrates something either good or bad, but there is no value in duplicate, unformatted, unexplained answers with obvious, openly-discussed flaws. It doesn't add anything to the discussion and only serves to confuse readers who are looking for help.
Is there always room left in a thread for another answer that's not a near-duplicate? No. Sometimes it's not. Is this question completely exhausted? Not sure, but with extremely simple direct questions, it doesn't take long. If you want to answer questions, pick a thread where there is still room for a new answer.

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asmaa il 2 Feb 2024
Create a row vector named d that contains sqrt(10) as the first element and pi^2 (
π2
) as the second elemen
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Voss il 2 Feb 2024
Create a row vector named d that contains -3.1 as the first element and 2.43 as the second elemen[t].
Solution:
d = [-3.1, 2.43];

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