Solving a Matrice Equation: AX=B

[Petrus]

Matrice equation

Hello MHB,
I am suposed to solve this matrice equation \(\displaystyle AX=B\)
\(\displaystyle A=
\left| {\begin{array}{cc} 1 & 2 & 1\\ 1 & 3 & 2\\ 1 & 6 & 6 \end{array} } \right|\)
\(\displaystyle X=
\left| {\begin{array}{cc} a & b \\ c & d \\ e & f \end{array} } \right|\)
\(\displaystyle B=
\left| {\begin{array}{cc} 1 & 10 \\ 2 & 40 \\ 3 & 60 \end{array} } \right|\)

Well I don't need to try do something I can se there will be no solution cause I will not be able to multiplication \(\displaystyle A^{-1}*B\) cause there is not same row in A as columne in B so there will be no solution? I am correct?

Regards,
\(\displaystyle |\pi\rangle\)

 

[Fernando Revilla]

Re: Matrice equation

No problem with orders, they fit perfectly. Besides, $A$ is invertible so, $$AX=B\Leftrightarrow A^{-1}(AX)=A^{-1}B\Leftrightarrow (AA^{-1})X=A^{-1}B\Leftrightarrow IX=A^{-1}B\Leftrightarrow X=A^{-1}B$$ Now, $X=A^{-1}B=\ldots$

 

[Petrus]

Re: Matrice equation

No problem with orders, they fit perfectly. Besides, $A$ is invertible so, $$AX=B\Leftrightarrow A^{-1}(AX)=A^{-1}B\Leftrightarrow (AA^{-1})X=A^{-1}B\Leftrightarrow IX=A^{-1}B\Leftrightarrow X=A^{-1}B$$ Now, $X=A^{-1}B=\ldots$

Hello Fernando Revilla,
Thanks, I see what I did misinterpret.

Regards,
\(\displaystyle |\pi\rangle\)