Q.Find the matrix $A$ such that

[!FORMULA] $\begin{pmatrix} 2 & -1 \ 1 & 0 \ -3 & 4\end{pmatrix} A^T = \begin{pmatrix} -1 & -8 & -10 \ 1 & 2 & -5 \ 9 & 22 & 15\end{pmatrix}$

Three basic operations are defined on matrices:

1. Scalar multiplication. For $A=[a_{ij}]_{m\times n}$ and a scalar $k$, $kA = [k\,a_{ij}]_{m\times n}$ — every entry is scaled by $k$. Taking $k=-1$ gives $-A = [-a_{ij}]$, the negative of $A$ (never call it "a negative matrix").

2. Addition and subtraction. These are defined only when $A$ and $B$ have the same order. Then $A+B = [a_{ij}+b_{ij}]$ and $A-B = A+(-1)B = [a_{ij}-b_{ij}]$, entrywise. If the orders differ, $A+B$ and $A-B$ are simply not defined. Addition/subtraction extends to any finite number of matrices of the same order.

3. Multiplication. $A$ is conformable for the product $AB$ only when the number of columns of $A$ equals the number of rows of $B$: if $A$ is $m\times n$ and $B$ is $n\times p$, then $AB$ exists and has order $m\times p$ — schematically $(m\times n)(n\times p) = (m\times p)$, with the inner dimensions matching and cancelling. Each entry of $AB$ is a row-by-column dot product: …