# Find $XY$ given matrix $YX$ where $X$ is a row matrix and $Y$ is a column matrix

I've been given matrix $$YX$$ as below. I want to find $$XY$$ from it. I know that $$X$$ is a row matrix and $$Y$$ is a column matrix. $$X$$ has $$2$$ entries and $$Y$$ has $$2$$ entries. However I don't know the values of each entry.

$$YX= \begin{bmatrix} -2 & -3 \\ 2 & 3 \end{bmatrix}$$

I'm not quite sure how to progress from here. I know that the result will be one number. I'd like some guidance on where to go next.

Hint: $$\begin{bmatrix}a \\ b\end{bmatrix}\times\begin{bmatrix}c & d\end{bmatrix}= \begin{bmatrix}ac & ad \\ bc & bd\end{bmatrix}$$
Then: $$\begin{bmatrix}c & d\end{bmatrix}\times\begin{bmatrix}a \\ b\end{bmatrix}= \begin{bmatrix}ac +bd\end{bmatrix}$$
• Oh I see now. Thank you for explaining this to me. From the above I believe that $XY = 1$
• @Pije, $XY = [(-2)+(3)] = [1]$. The answer is a $1\times 1$ matrix and not a scalar value. May 10, 2015 at 13:07
If $Y=\begin{bmatrix}a\\b\end{bmatrix}$ and $X=\begin{bmatrix}c&d\end{bmatrix}$, we know that $ac=-2$ and $bc=2$, i.e., $b=-a$. Likewise we find $d=\frac32 c$. So $Y=u\cdot \begin{bmatrix}1\\-1\end{bmatrix}$ and $X=v\cdot \begin{bmatrix}1&\frac32\end{bmatrix}$. Moreover, $uv=-2$ and we finde $XY=uv(1+\frac32)=1$.
$YX = \pmatrix{-1\\1}\pmatrix{2&3}=\pmatrix{-2&-3\\2&3}$ therefore $XY = \pmatrix{2&3} \pmatrix{-1\\1}=1.$