# Finding an unknown multiplying the determinant of a matrix A when given a modified matrix A

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?

• Yes, you're correct. – poyea May 27 '18 at 10:59

Yes, your logic and step by step simplification to get $$\det\begin{bmatrix} 7a_1 & 7a_2 & 7a_3\\4b_1+8c_1 & 4b_2+8c_2 & 4b_3+8c_3 \\ 6c_1 & 6c_2 & 6c_3\end{bmatrix}=168\det \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{bmatrix}$$