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$$A \cdot \begin{bmatrix} 1 & 3 & 4\\ 3 & -1 & 5\\ -2 & 4 & -3\\ \end{bmatrix} = \begin{bmatrix} 3 & -1 & 5\\ 1 & 3 & 4\\ 4 & -8 & 6\\ \end{bmatrix} $$ Find the $3 \times 3$ matrix $A$.

According to my textbook, the question requires elementary row operations on the given matrices.

I read somewhere that for an equation of the form $AB=X$ ,we can apply elementary row operation on $A$ and $X$ only. I don't know why do these contradict. Where am I wrong?

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    $\begingroup$ I'd like to edit in proper form, but I cannot understand what you are trying to write. Is the right hand side a matrix? Of which dimensions? $\endgroup$ – Andreas Caranti Jan 11 '17 at 15:38
  • $\begingroup$ Actually the question is A×(| 1 3 4|. | 3 -1 5| | 3 -1 5| ) = (| 1 3 4 | | -2 4 -3| |4 -8 6| )->rhs and lhs have 3×3 matrices ...I am not able to write them in vertical format $\endgroup$ – Piyush Kumar Jan 11 '17 at 15:47
  • $\begingroup$ Ok, so $1 3 4$ is a column? Will edit for you, so that you can learn how to do it. $\endgroup$ – Andreas Caranti Jan 11 '17 at 15:48
  • $\begingroup$ I think he means $A(1,3,4)^T=(1,3,4)^T$ for the first one. $\endgroup$ – Arnaldo Jan 11 '17 at 15:49
  • $\begingroup$ Piyush, take a look to write properly: meta.math.stackexchange.com/questions/5020/… $\endgroup$ – Arnaldo Jan 11 '17 at 15:51
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Since

$$\det \begin{bmatrix} 1 & 3 & 4\\ 3 & -1 & 5\\ -2 & 4 & -3\end{bmatrix} = 20 \neq 0$$

we can right-multiply both sides of the linear matrix equation by elementary matrices until we obtain

$$\mathrm A = \begin{bmatrix} 3 & -1 & 5\\ 1 & 3 & 4\\ 4 & -8 & 6\end{bmatrix} \begin{bmatrix} 1 & 3 & 4\\ 3 & -1 & 5\\ -2 & 4 & -3\end{bmatrix}^{-1}$$

We would be doing elementary column operations. If you must do elementary row operations, then do transpose both sides of the linear matrix equation, then do left-multiply both sides by elementary matrices, obtain $\mathrm A^{\top}$ and then transpose to obtain $\mathrm A$.

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  • $\begingroup$ I don't get you sir completely... could you solve the question ? $\endgroup$ – Piyush Kumar Jan 11 '17 at 16:45
  • $\begingroup$ Can you do Gaussian elimination? $\endgroup$ – Rodrigo de Azevedo Jan 11 '17 at 16:46
  • $\begingroup$ I am a high school student sir ...do you mean to find the inverse of the matrix? $\endgroup$ – Piyush Kumar Jan 11 '17 at 16:48
  • $\begingroup$ You don't need to compute any inverse matrices. Start by multiplying the 2nd columns of the matrices by $3$ and then add to them the 1st columns. The goal is to reduce the matrix in the LHS to the identity matrix. $\endgroup$ – Rodrigo de Azevedo Jan 11 '17 at 16:51
  • $\begingroup$ Okay..I get you ...if I apply column operation on both the matrices to convert it in the form A×I=x then x = A .But what if i apply row operations as it is on the two matrices would I be wrong in this case? $\endgroup$ – Piyush Kumar Jan 11 '17 at 16:57
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Assuming that the matrix $B$ is nonsingular, you can solve for the matrix $A$ in $AB=C$ by right-multiplying by $B^{-1}$, but that doesn’t appear to be the point of this exercise.

Compare the rows of the two given matrices: you can obtain the one on the right from the one on the left by swapping the first two rows and multiplying the last row by $-2$. Each of these corresponds to an elementary row operation—one will be a permutation matrix, the other will be diagonal. $A$ is then the product of the two matrices that represent these row operations (in this case the order of operations doesn’t matter).

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