-2
$\begingroup$

\begin{align*} A &= \begin{bmatrix}1&0&2\\0&3&4\\5&6&0\end{bmatrix}\\ B &= \begin{bmatrix}0&3&4\\1&0&2\\50&60&0\end{bmatrix} \end{align*}

(a) How can I find a matrix C such that CA=B? Is it by using correspondence between elementary matrices and elementary row operations?

(b) How would I find the inverse of C? Are there multiple methods?

(c) How would I multiply B by the inverse of C to get A?

Thank you in advance.

$\endgroup$
  • $\begingroup$ Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, this link might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote whether to reopen it or leave it closed. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.) $\endgroup$ – Martin Sleziak Aug 23 '17 at 12:47
1
$\begingroup$

You have asked in the question: "Is it by using correspondence between elementary matrices and elementary row operations?"

The answer is yes - you simply notice that you get $B$ from $A$ by interchanging the first two rows and multiplying the third row by ten. If you do the same operation on the identity matrix, you get $$C= \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 &10 \end{pmatrix}. $$

You can check by direct computation that you indeed have: \begin{align*} CA&=B\\ \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 &10 \end{pmatrix} \begin{pmatrix} 1 & 0 & 2 \\ 0 & 3 & 4 \\ 5 & 6 & 0 \end{pmatrix} &= \begin{pmatrix} 0 & 3 & 4 \\ 1 & 0 & 2 \\ 50 &60 & 0 \end{pmatrix} \end{align*}

To get $C^{-1}$ it suffices to notice what are the "opposite" row operations - and it is again interchange of the first two rows, together with multiplying the third row by $1/10$ (i.e., dividing it by $10$): $$C^{-1}= \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 &\frac1{10} \end{pmatrix}. $$

$\endgroup$
1
$\begingroup$

(a) Supposing $A$ is invertible, $CA=B\iff C=BA^{-1}$, so you can calculate it given these two matrices.

(b) See this informal introduction. For 2x2 and 3x3 matrices, you can find formulas for the inverse here. Methods exist to calculate the inverse of bigger matrices. Or you can calculate them numerically.

(c) The inverse matrix of a matrix is still a matrix. So once you calculated the inverse matrix, you just multiply two matrices. If you want to calculate $A$ from $B$ and $C$, you would write (suppoosing $C$ is invertible) $CA=B\iff A=C^{-1}B$.

Note: you can only calculate the inverse of matrix, if the matrix is invertible.

$\endgroup$
  • 1
    $\begingroup$ Your first line $CA=B\iff C=BA^{-1}$ is only true given that $A$ is invertible. Of course you can have $CA=B$ in situations where $A$ is noninvertible for whatever reason (even if $A$ isn't square) and so you should have included the phrase "supposing $A$ is invertible" before writing the rest of the line. $\endgroup$ – JMoravitz Aug 21 '17 at 1:21
0
$\begingroup$

Hint for (a): you get $B$ from $A$ by swapping the top two rows, and multiplying the $3^{rd}$ one by $10$.

In the 2D case, for example, you would combine the following:

  • to swap two rows: $\;\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}c&d\\a&b\end{bmatrix}$

  • to multiply a row by $\lambda\,$: $\;\begin{bmatrix}\lambda&0\\0&1\end{bmatrix}\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}\lambda a & \lambda b\\c&d\end{bmatrix}$

$\endgroup$
0
$\begingroup$

Since the column vectors of $A,B$ are linearly independent, both matrices are invertible. Hence if you want $CA =B$ then you have $C = BA^{-1}$ and by our comment, such a $C$ is unique. To find the inverse of $C$ you can use the identity $C^{-1} = (BA^{-1}) = A B^{-1}$ or use the row reduction alogrithm i.e $C$ being invertible implies that you can row reduce the LHS of augmented matrix $(C \mid I)$ to the identity. Such row operations will be given by a finite number of multiplication with elementary matrices i.e $\underbrace{E_N \cdots E_1}_E C = I$ and so $(EC \mid EI) = (I \mid E)$. Hence $E = C^{-1}$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.