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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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    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? – 2017-01-11
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    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 – 2017-01-11
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    Ok, so $1 3 4$ is a **column**? Will edit for you, so that you can learn how to do it. – 2017-01-11
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    I think he means $A(1,3,4)^T=(1,3,4)^T$ for the first one. – 2017-01-11
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    Piyush, take a look to write properly: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – 2017-01-11
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    Look in the lhs , the matrix multiplied by A is a 3×3 matrix which has the 1st row as (1 3 4), 2nd row as (3 -1 5) ,3rd row as (-2 4 -3) ... The matrix on the rhs side is again a 3×3 matrix with 1st row as (3 -1 5) 2nd row as (1 3 4) and 3rd row as (4 -8 6) – 2017-01-11
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    So it's rows, not columns. Will fix asap - done! – 2017-01-11
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    Thanks a lot Andreas and Arnold! – 2017-01-11

2 Answers 2

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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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    I don't get you sir completely... could you solve the question ? – 2017-01-11
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    Can you do Gaussian elimination? – 2017-01-11
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    I am a high school student sir ...do you mean to find the inverse of the matrix? – 2017-01-11
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    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. – 2017-01-11
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    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? – 2017-01-11
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    Row operations correspond to left-multiplication by elementary matrices. You would be left-multiplying $\mathrm A$ rather than the matrix you want to reduce to the identity matrix. As I mentioned before, you'd need to transpose both sides to move the given matrix to the left factor of the product and the unknown matrix to the right factor of the product. – 2017-01-11
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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).