# Find 3x3 matrix for XA=B

Homework Statement

Find the 3x3 matrix X such that XA=B, where

Code:
      1 1  1
A=   -1 0 -2
1 0 -1

1 3  0
B=   -2 1 -2
1 4 -1

The attempt at a solution

I understand how to do AX=B just fine, but XA=B is confusing me. I tried doing [x y z] A, but not sure if that's even right or what to do after that. Yes, I know it can be solved using inverses or whatever, but the professor does not want us to solve it that way (haven't gone over inverses yet).

The solution for X such that AX=B involved finding individual columns, so does that mean XA=B will find the individual rows?

Any help is appreciated!

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## Answers and Replies

ehild
Homework Helper
Homework Statement

Find the 3x3 matrix X such that XA=B, where

Code:
      1 1  1
A=   -1 0 -2
1 0 -1

1 3  0
B=   -2 1 -2
1 4 -1

The attempt at a solution

I understand how to do AX=B just fine, but XA=B is confusing me. I tried doing [x y z] A, but not sure if that's even right or what to do after that. Yes, I know it can be solved using inverses or whatever, but the professor does not want us to solve it that way (haven't learned gone over inverses yet).

The solution for X such that AX=B involved finding individual columns, so does that mean XA=B will find the individual rows?

Any help is appreciated!

Yes, find the individual rows of X. How would you do it?

ehild

If I multiply

\begin{array}{cccc}
1 & 1 & 1 \\
-1 & 0 & -2 \\
1 & 0 & -1 \\
\end{array}

by
\begin{array}{cccc}
x & y & z
\end{array}

I get

\begin{array}{cccc}
(x-y+z) & (x) & (x-2y-z) \\
\end{array}

That's where I get stuck. I don't know what to do with it. Would I transpose it to this?:

\begin{array}{cccc}
(x-y+z) \\
(x) \\
(x-2y-z) \\
\end{array}

Transpose both sides.

Okay, I transposed both sides and I see how to get each column of X now. I think I get it :). Here's my augmented matrix:

\begin{array}{cccc}
1 & 0 & 0 | & 3 & 1 & 4 \\
0 & 1 & 3 | & 1 & 0 & 2 \\
0 & 0 & -4 | & 1 & 3 & 1 \\
\end{array}

The last row seems wrong because of the -4, but if I change it to a 1 using row operations, then the answer is wrong. The correct answer is:

\begin{array}{cccc}
3 & 1 & 1 \\
1 & 0 & 3 \\
4 & 2 & 1\\
\end{array}

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Edit: another way to solve this is to multiply both sides by ##A^{-1}## from the right.

Sorry, it seems that I almost forgot one property of matrix inverse :)

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