Question

Solve B to the RHS from ABX=C

Answer 1

Imagine this equation:
\[ABX=C\]

Where:\[A=\begin{bmatrix}2& 1\\1& 2\end{bmatrix}\]\[B=\begin{bmatrix}4& 6\\2& 8\end{bmatrix}\]\[C=\begin{bmatrix}3& 2\\1&5\end{bmatrix}\]

I don't want to solve them I just want to know how to move \(B\) to the RHS
\[\begin{bmatrix}2& 1\\1& 2\end{bmatrix}\times \begin{bmatrix}4& 6\\2& 8\end{bmatrix}\times X=\begin{bmatrix}3& 2\\1&5\end{bmatrix}\]

Answer 2

How would I move \(B\) to the RHS?

Answer 3

@AJ01, please help

Answer 4

hi .it multiplication or cross product?

Answer 5

Multiplication @AJ01

Answer 6

I know how to do \(AX=B\), not \(ABX=C\)

Answer 7

take inverse for B
and inverse for C

Answer 8

?

Answer 9

pre-multiply by A inverse

Answer 10

inversion method

Answer 11

Could you please show it somehow, using \(LaTeX\) or something? I find it a little bit hard understanding what you meant. Sorry :(

Answer 12

\[A =\left[\begin{matrix}a & b\\ c &d\end{matrix}\right]\]
\[A ^{-1}=\frac{ 1 }{\left| A \right| }\left[\begin{matrix}d & -b \\ -c & a\end{matrix}\right]\]

Answer 13

Yup, I understand that @AJ01

Answer 14

Help @ganeshie8 @ikram002p

Answer 15

Find its inverse and mult both side by that

Answer 16

For \(A\)? @irishboy123