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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.)

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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).

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  • $\begingroup$ Robert, do you mean U as a matrix or as a scalar multiple? $\endgroup$ – yolo123 Nov 20 '14 at 18:41
  • $\begingroup$ Hmm... I am kind of lost. Can you provide visual explanatin $\endgroup$ – yolo123 Nov 20 '14 at 18:44
  • $\begingroup$ Thanks Robert. That was very clear. $\endgroup$ – yolo123 Nov 21 '14 at 3:20
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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.

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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.

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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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  • $\begingroup$ Ingenious method of avoiding the 2x3 problem. However, one things that irks me is that the inverse of A does not exist since it is not square. I think you should correct that. $\endgroup$ – yolo123 Nov 21 '14 at 3:21
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    $\begingroup$ $C=A.A^T$ is a 2x2 matrix and I am only inverting $C$. $\endgroup$ – mike Nov 21 '14 at 4:02

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