I was asked this question:
There's a square matrix $B$ from order $3 \times 3$, and we're given that: $$ B \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$$ $$ B\begin{pmatrix} -1 \\ -1 \\ -2 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ 2 \end{pmatrix}$$ $$ B\begin{pmatrix} 1 \\ 1 \\ 3 \end{pmatrix}= \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$$
- Is B invertible?
- Find B.
I was thinking on finding first B and then if the determinant of $B$ is not $0$ then $B$ is indeed invertible. But I've been stuck on finding $B$ for quite some time now.
I tried different ways to approach this, First I tried something like this: $$ B \begin{pmatrix} 1 & 2 & 3 \\ -1 & -1 & -2 \\ 1 & 1 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 \\ 3 & 3 & 2 \\ 2 & 1 & -1 \end{pmatrix}$$ And I thought maybe if I get the inverse of the middle matrix and multiplied it with the second matrix it will result in B, like this: $$ \begin{pmatrix} 1 & 2 & 3 \\ -1 & -1 & -2 \\ 1 & 1 & 3 \end{pmatrix}^{-1} \begin{pmatrix} 1 & 2 & 3 \\ 3 & 3 & 2 \\ 2 & 1 & -1 \end{pmatrix}=B$$ Which was incorrect, and then I thought maybe I organized the vectors in the matrix incorrectly, so I tried to transpose it, but still was incorrect.
So I gave up on this way, and tried a different way: $$ \begin{pmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \\ -1 & -1 & -2 \\ 1 & 1 & 3 \end{pmatrix}= \begin{pmatrix} 1 & 2 & 3 \\ 3 & 3 & 2 \\ 2 & 1 & -1 \end{pmatrix}$$ But I wasn't sure how to continue from here.
Thank you for your help!
I believe you indeed organized incorrectly the vectorial information you had when converting it to a matrix equality. Your three assumptions yield (the vectors must remain vertical) $$ B \begin{pmatrix} 1 & -1 & 1 \\ 2 & -1 & 1 \\ 3 & -2 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 3 & 2 \\ 2 & 3 & 1 \\ 3 & 2 & -1 \end{pmatrix} $$ which is an equality of the form $B A = C$, which indeed fully determines the matrix $B$ you are looking for if $A$ is invertible. Multiplying both sides by $A^{-1}$, you will then have $BA A^{-1} = C A^{-1}$, so $B = C A^{-1}$ (beware that matrix multiplication does not commute so you have to be cautious on which side you are multiplying).
Checking that $\det A \neq 0$, computing $A^{-1}$, computing $C A^{-1}$, and eventually $\det B$ should be a standard exercise for you.