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

[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..

 

[spamiam]

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?

 

[mitch_1211]

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?

 

[spamiam]

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.

 

[mitch_1211]

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)

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

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.

Does that sound about right?

 

[Mark44]

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.