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I need to solve the following problem:

$\det\begin{bmatrix} 7a_1 & 7a_2 & 7a_3\\4b_1+8c_1 & 4b_2+8c_2 & 4b_3+8c_3 \\ 6c_1 & 6c_2 & 6c_3\end{bmatrix}=k\det \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{bmatrix}$

I did the following: $\det\begin{bmatrix} 7a_1 & 7a_2 & 7a_3\\4b_1+8c_1 & 4b_2+8c_2 & 4b_3+8c_3 \\ 6c_1 & 6c_2 & 6c_3\end{bmatrix}=7\cdot 6\cdot \det\begin{bmatrix} a_1 & a_2 & a_3\\4b_1+8c_1 & 4b_2+8c_2 & 4b_3+8c_3 \\ c_1 & c_2 & c_3\end{bmatrix}$

I then recalled that if matrix A is obtained from matrix B by adding a multiple of one row to another, then $\det A=\det B$. So, $\det\begin{bmatrix} a_1 & a_2 & a_3\\4b_1+8c_1 & 4b_2+8c_2 & 4b_3+8c_3 \\ c_1 & c_2 & c_3\end{bmatrix} = \det\begin{bmatrix} a_1 & a_2 & a_3\\4b_1 & 4b_2 & 4b_3 \\ c_1 & c_2 & c_3\end{bmatrix} = 4\cdot \det \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{bmatrix}$

This makes $k=6\cdot 7\cdot 4=168$. Is my logic correct here?

Thanks for your time

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  • $\begingroup$ Yes, you're correct. $\endgroup$ – poyea May 27 '18 at 10:59
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Yes, your logic and step by step simplification to get $$\det\begin{bmatrix} 7a_1 & 7a_2 & 7a_3\\4b_1+8c_1 & 4b_2+8c_2 & 4b_3+8c_3 \\ 6c_1 & 6c_2 & 6c_3\end{bmatrix}=168\det \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{bmatrix}$$

is correct.

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  • $\begingroup$ It is not a good practice to simply post that the proof is correct. $\endgroup$ – José Carlos Santos May 27 '18 at 11:06
  • $\begingroup$ @Jose Isn't that what the OP asked for? $\endgroup$ – Mohammad Riazi-Kermani May 27 '18 at 11:14
  • $\begingroup$ It is. But in MSE it is considered bad practice to answer a proof-verification question simply saying that the proof is correct. $\endgroup$ – José Carlos Santos May 27 '18 at 11:21
  • $\begingroup$ Thanks, What does MSE stand for? $\endgroup$ – Mohammad Riazi-Kermani May 27 '18 at 11:57
  • $\begingroup$ Math StackExchange $\endgroup$ – José Carlos Santos May 27 '18 at 12:07

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