# Expressing Matrix Product Ax as Linear Combination of Column Vectors in A

• mitch_1211
In summary, the conversation discusses how to express the matrix product Ax as a linear combination of the column vectors in A. The method involves multiplying the columns of A by constants and adding them together to form the matrix product b. The conversation also includes a clarification on the purpose of finding constants a, b, and c, and a discussion on how to approach similar problems.
mitch_1211
I want to express the matrix product Ax as a linear combination of the column vectors in A.

I know for that for (2,3;-1,4)(1;2) i just row reduce the 2 previous matrices and then the solutions give the multiples needed for each column.

How would i go about solving a similar problem but then the matrices are
(2,1,5;6,3,-8)(3;0;-5) the first matrix is 2x3 and the second one is 3x1, i can't augment these and solve for constants can I?

i'm also not entirely sure if the second matrix is x from Ax=b or if it the b matrix, my guess would be that it is the x matrix.

I've tested if the solutions i get multiplied by the appropriate rows of the first matrix gives the second matrix, and it works, but not sure if that was by change or if i am following a valid procedure.

Some explanation on best way to approach these problems would be much appreciated..

SHIVAN

Hi mitch,

First of all, if this is a homework question it should go in the homework section of the forum.

I think you may have misunderstood the question. I think the question is actually referring to the columns of A. Writing out the example of a 2x2 matrix will probably make this clearer.

$$\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} x_1 a_{11} + x_2 a_{12} \\ x_1 a_{21} + x_2 a_{22} \end{pmatrix} = x_1 \begin{pmatrix} | \\ \vec{a_1} \\ | \end{pmatrix} + x_2 \begin{pmatrix} | \\ \vec{a_2} \\ | \end{pmatrix}$$

Where $\vec{a_1}$ and $\vec{a_2}$ are the columns of A. Can you see the general rule?

spamiam said:
Can you see the general rule?

yes i think i can see what is happening now.. i have attached a photo of the matrices i am working with, C is what I am talking about.

i think from your explanation that, a(2;6)=3 b(1;3)=0 c(5;-8)=-5
but how would i easily find a,b,c?

in a) and b) i just augmented the 2 matrices and row reduced finding the constants, was i correct in doing that?

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mitch_1211 said:
i think from your explanation that, a(2;6)=3 b(1;3)=0 c(5;-8)=-5
but how would i easily find a,b,c?

I don't think this is what the question is asking... Could you post the exact wording? From what you said before, I think that they are looking for constants a,b, and c such that
$$\begin{pmatrix} 2 & 1& 5 \\ 6 & 3 & -8 \end{pmatrix} \begin{pmatrix} 3 \\ 0 \\ -5 \end{pmatrix} = a\begin{pmatrix} 2 \\ 6 \end{pmatrix} + b\begin{pmatrix} 1 \\ 3 \end{pmatrix}+ c\begin{pmatrix} 5 \\ -8 \end{pmatrix}$$

Do you know how to multiply matrices?
in a) and b) i just augmented the 2 matrices and row reduced finding the constants, was i correct in doing that?

No, I don't think that works. Take another look at my previous post.

spamiam said:
Could you post the exact wording?

Express the given matrix product Ax (=b) as a linear combination of the column vectors in A. (NOTE: this shows that the matrix product b belongs to the column space of A)

spamiam said:
Do you know how to multiply matrices?

Yea i know how to do that, and i think i can see what you did with the x and a example

spamiam said:
I think that they are looking for constants a,b, and c such that

Sorry, that is what i was trying to say with a(2;6)=3 b(1;3)=0 c(5;-8)=-5 , but i just realized, I am supposed to calculate the matrix multiplication first and call the result b for example and then find constants a,b,c that multiply the column vectors of A to give the numbers in b.

That seems like the long way around. Look at spamiam's post, and notice that the three vectors on the right side of the equation are the three columns in the left matrix on the left side of the equation. The column matrix on the left side of the equation has a lot to do with the coefficients a, b, and c.

## 1. What is a matrix product?

A matrix product is a mathematical operation in which two matrices are multiplied together to produce a new matrix. It is denoted by the symbol "A*B" and is defined as the dot product of rows of matrix A and columns of matrix B.

## 2. What is the significance of expressing matrix product Ax as a linear combination of column vectors in A?

Expressing matrix product Ax as a linear combination of column vectors in A allows us to understand the relationship between the two matrices and how they interact with each other. It also helps in solving systems of linear equations and finding the inverse of a matrix.

## 3. How do you express matrix product Ax as a linear combination of column vectors in A?

To express matrix product Ax as a linear combination of column vectors in A, we use the properties of matrix multiplication and perform the dot product between the row vector of A and the column vector of x. This is repeated for all the column vectors in A, and the resulting products are added together to form the final matrix product Ax.

## 4. Why is it important to understand how to express a matrix product as a linear combination of column vectors?

Understanding how to express a matrix product as a linear combination of column vectors is important because it helps in solving various mathematical problems involving matrices. It also provides insight into the relationship between the two matrices and can be used to simplify complex calculations.

## 5. Can matrix product Ax be expressed as a linear combination of row vectors instead of column vectors?

Yes, matrix product Ax can be expressed as a linear combination of row vectors as well. The process is similar, but instead of using the column vectors of matrix A, we use its row vectors. However, it is more common to express it as a linear combination of column vectors because of the convention in matrix notation.

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