Find the only matrix that fits the criteria

[diogomgf]

Find the only matrix \(\displaystyle A\) that fits the criteria:



So I decomposed the side to the product of elementary matrices. The result was:



If \(\displaystyle \begin{bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}\end{bmatrix} \) then the left side after all the transformations is equal to:


, wich in the end, after solving all the system of equations leaves me with:



The book says the solution is different, what am I doing wrong?

 

[pka]

Look at this solution.

 

[Steven G]

You have an equation of the form BAC=D and you want to solve for A.

B^-1(BAC)C^-1 = B^-1(D)C^-1
(B^-1B)A(CC^-1) = B^-1DC^-1
A = B^-1DC^-1

Now is your turn to show your work.

 

[diogomgf]

You have an equation of the form BAC=D and you want to solve for A.

B^-1(BAC)C^-1 = B^-1(D)C^-1
(B^-1B)A(CC^-1) = B^-1DC^-1
A = B^-1DC^-1

Now is your turn to show your work.

@Jomo while I can see where this is going, is there anything wrong with my approach?
By the way: I will solve this and post the solution.

Also what If B and C are not invertible?

 

[diogomgf]

Look at this solution.

Thank you for the resources ! I will check it

 

[diogomgf]

@Jomo

Here is the solution based on what you said:

 

[Steven G]

@Jomo while I can see where this is going, is there anything wrong with my approach?
By the way: I will solve this and post the solution.
Also what If B and C are not invertible?

I just have no idea why you would decompose the matrices and then multiply it back. Do you see that you did that. Why not just multiply the three matrices on the left, set it equal to the right side matrices and solve for A.

Also what If B and C are not invertible? Good luck solving for A.

 

[diogomgf]

I just have no idea why you would decompose the matrices and then multiply it back. Do you see that you did that. Why not just multiply the three matrices on the left, set it equal to the right side matrices and solve for A.

Also what If B and C are not invertible? Good luck solving for A.

Guess you are right, I thought doing it that way was easier this time around... Not sure why
The way you approached it was way easier though!!

I already posted the solution by the way ?

 

[Steven G]

I already posted the solution by the way ?

Did you verify your work (I won't, sorry)? if it doesn't work out then I can look for you error but it will be a silly arithmetic mistake. It's more important that you have the technique down.

 

[diogomgf]

Did you verify your work (I won't, sorry)? if it doesn't work out then I can look for you error but it will be a silly arithmetic mistake. It's more important that you have the technique down.

Yup everything is correct

 

[pka]

I would like to give the to necessary inverses:
Please not that the LaTeX complier has written three negative integers with the negative signs over the number.
\({\left[ {\begin{array}{*{20}{c}}
{ - 1}&1&0 \\
1&{ - 2}&0 \\
0&0&{ - 1}
\end{array}} \right]^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
{ - 2}&{ - 1}&0 \\
{ - 1}&{ - 1}&0 \\
0&0&{ - 1}
\end{array}} \right]\quad \quad {\left[ {\begin{array}{*{20}{c}}
0&0&1 \\
0&2&0 \\
1&0&0
\end{array}} \right]^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
0&0&1 \\
0&{\frac{1}{2}}&0 \\
1&0&0
\end{array}} \right]\)