pls check my answer

[newuser]

Given invertible matrix A and the matrix equation X ·A + B = C:
(a) clearing the matrix X.
(b) find the matrix X when

 

[pka]

Given invertible matrix A and the matrix equation X ·A + B = C:
(a) clearing the matrix X.
(b) find the matrix X when
View attachment 2058

I don't understand where the particular matrices come from. Are they given?
In any case \(\displaystyle X=(C-B)\cdot A^{-1}\).

 

[newuser]

I don't understand where the particular matrices come from. Are they given?
In any case \(\displaystyle X=(C-B)\cdot A^{-1}\).

Yes, A,B,C are given matrices

 

[pka]

Yes, A,B,C are given matrices

As you can see if your arithmetic is correct the answer should be.
You can always check it.

 

[newuser]

As you can see if your arithmetic is correct the answer should be.
You can always check it.

It means I´m correct. Thanks

 

[HallsofIvy]

Your problem, as posted, makes no sense. You cannot multiply \(\displaystyle \begin{pmatrix}x \\ y\end{pmatrix}\begin{pmatrix}1 & -2 \\ -1 & 1\end{pmatrix}\). Did you mean \(\displaystyle \begin{pmatrix}1 & -2 \\ -1 & 1\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}\)?

 

[soroban]

Hello, newuser!

\(\displaystyle \text{Given invertible matrix }A\text{ and the matrix equation: }\:X\!\cdot\! A + B \:=\: C\)

\(\displaystyle \text{where: }\:A \:=\: \begin{pmatrix}1&\text{-}2 \\ \text{-}1 & 1 \end{pmatrix} \quad B \:=\: \begin{pmatrix}1&1 \\ \text{-}2&1 \end{pmatrix} \quad C \:=\: \begin{pmatrix}3&1 \\ 1&\text{-}1 \end{pmatrix} \)

\(\displaystyle \text{(a) clearing the matrix }X\) . What does this mean?

\(\displaystyle \text{(b) Find the matrix.}\) . The equation you gave is impossible.

Even if we made \(\displaystyle X\) a row matrix, the problem makes no sense.

We have: .\(\displaystyle (x,\:y)\begin{pmatrix}1&\text{-}2 \\ \text{-}1&1\end{pmatrix} + \begin{pmatrix}1&1\\\text{-}2&1\end{pmatrix} \;=\;\begin{pmatrix}3&1\\1&\text{-}1\end{pmatrix} \)

. . . . . . . .\(\displaystyle (x,\:y)\begin{pmatrix}1&\text{-}2\\\text{-}1&1\end{pmatrix} \;=\;\begin{pmatrix}3&1\\1&\text{-}1 \end{pmatrix} - \begin{pmatrix}1&1 \\ \text{-}2&1\end{pmatrix}\)

. . . . . . . .\(\displaystyle (x,\:y)\begin{pmatrix}1&\text{-}2 \\ \text{-}1&1\end{pmatrix} \;=\;\begin{pmatrix}2&0\\3&\text{-}2\end{pmatrix}\)

Hence: . . \(\displaystyle (x-y,\:\text{-}2x+y) \;=\;\begin{pmatrix}2&0\\3&\text{-}2\end{pmatrix}\)

We have a \(\displaystyle 1\!\times\!2\) matrix equal to a \(\displaystyle 2\!\times\!2\) matrix
. . and that is impossible!


Perhaps the \(\displaystyle X\) matrix is also a \(\displaystyle 2\!\times\!2\) ?

. . Say: .\(\displaystyle X \;=\;\begin{pmatrix}w&x \\ y&z\end{pmatrix}\)

Then we could solve for \(\displaystyle w,\,x,\,y,\text{ and }z.\)