Finding a matrix U such that B=UA.
A= \begin{bmatrix} 1 & 0 & 1 \\ 2 & 3 & 1 \\ \end{bmatrix}B= \begin{bmatrix} 1 & 3 & 0 \\ 4 & 3 & 3 \\ \end{bmatrix}
How does one go about solving this problem?
Find the matrix U such that B=UA and express U as a product of elementary matrices.
Basically, I tried solving the problem using trial and error trying to go from A to B and applying those elementary row operation to the identity matrix to make U. However, this method is still not efficient enough in my head.
So, is there a faster method? (Even by trial and error, I could not find the matrix. I am not good with trial and error.)
$\endgroup$4 Answers
$\begingroup$Hint: (the first two columns of $B$) is $U$ times (the first two columns of $A$). If the first two columns of $A$ formed the identity matrix, $U$ would just be the first two columns of $B$. It isn't, but you can take some row operations...
EDIT: In row reduction of $A$, the first step might be subtracting $2$ times row $1$ from row $2$. This is accomplished by the matrix multiplication $E_1 A$, where $E_1$ is the elementary matrix $\pmatrix{1 & 0\cr -2 & 1\cr}$.
One more row operation, corresponding to an elementary matrix $E_2$, gives you
a matrix whose first two columns are the identity, i.e.
$$E_2 E_1 A = \pmatrix{1 & 0 & *\cr 0 & 1 & *\cr}$$
(where I'm not specifying what the $*$ entries are).
Similarly, there are elementary matrices $E_3$, $E_4$ such that the first two
columns of $E_3 E_4 B$ form the identity matrix. If it happens that
$E_3 E_4 B = E_1 E_2 A$ (which is something you should check!), then
$B = E_4^{-1} E_3^{-1} E_1 E_2 A$, so you can take $U = E_4^{-1} E_3^{-1} E_1 E_2$ (note that the inverse of an elementary matrix is an elementary matrix).
For the first part: Firstly observe that since $A,B$ have dimension $2\times 3$ then $U$ has dimension $2\times2$. Thus $$U=\begin{pmatrix}u_1 &u_2 \\ u_3 & u_4\end{pmatrix}$$ Multiplying U with A and setting equal to B gives you the set of equations $$\begin{cases}u_1+2u_2=1 \\0u_1+3u_2=3\\u_1+u_2=0\end{cases}$$ and similarly for $u_3,u_4$. The result is $$U=\begin{pmatrix}-1 &1 \\ 2 & 1\end{pmatrix}$$ For the second part: An elementary matrix is a matrix that differs from the identity matrix by one single elementary row operation. For this I do not see another way (to the moment) than trial and error.
$\endgroup$ $\begingroup$We want to perform row operations on $A$ in order to transform it into $B$. By inspection, notice that we can do this by:
- Adding ($2$ times row $1$) to row $2$ to get: $$ \begin{bmatrix} 1 & 0 & 1 \\ 4 & 3 & 3 \\ \end{bmatrix} $$
- Scaling (row $1$) by $3$ to get: $$ \begin{bmatrix} 3 & 0 & 3 \\ 4 & 3 & 3 \\ \end{bmatrix} $$
- Adding ($-1$ times row $2$) to row $1$ to get: $$ \begin{bmatrix} -1 & -3 & 0 \\ 4 & 3 & 3 \\ \end{bmatrix} $$
- Scaling (row $1$) by $-1$ to get: $$ \begin{bmatrix} 1 & 3 & 0 \\ 4 & 3 & 3 \\ \end{bmatrix} $$
It remains to translate each row operation to an elementary matrix and multiply them together.
$\endgroup$ $\begingroup$Denote $A^T$ the matrix transpose of $A$ and $C^{-1}$ the matrix inverse of square matrix $C$. Then we have:
$$UA=B\implies U(A.A^T)=B.A^T\implies U=B.A^T (A.A^T)^{-1}$$
$$\implies U=\begin{pmatrix}-1 &1 \\ 2 & 1\end{pmatrix}$$
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