If a $4 \times 4$ matrix $A$ with rows $v_1, v_2, v_3$ and $v_4$ has determinant $\det A = 9$, then $$\det \begin{pmatrix}v_1\\ 3v_2+4v_3\\ 9v_2+3v_3\\ v_4\end{pmatrix} = \ ?$$
What I did was multiply the initial determinant by $3$ and then $3$ again as that is what the rows were being multiplied by on their own. The addition of other rows does not seem to change the determinant according to the rules. However, my answer of $81$ was incorrect.
Any help would be highly appreciated!
Hint: If your original matrix is $A$, the new one is $\pmatrix{1 & 0 & 0 & 0\cr 0 & 3 & 4 & 0\cr 0 & 9 & 3 & 0\cr 0 & 0 & 0 & 1\cr} A$.
\begin{align}\det \begin{pmatrix} v_1 \\ 3v_2+4v_3 \\ 9v_2+3v_3 \\v_4\end{pmatrix} &= \det \begin{pmatrix} v_1 \\ 3v_2+4v_3 \\ -9v_3 \\v_4\end{pmatrix} \\ &= -9\det \begin{pmatrix} v_1 \\ 3v_2+4v_3 \\ v_3 \\v_4\end{pmatrix} \\ &= -9\det \begin{pmatrix} v_1 \\ 3v_2\\ v_3 \\v_4\end{pmatrix} \\ &=-9(3) \det(A)\end{align}
You can use the linearity by row properties of the determinant. For instance, we know that:
$ \det \begin{bmatrix} v_{1} \\ 3v_{2}+4v_3 \\ 9v_2+3v_3 \\ v_4 \end{bmatrix}$ = $\det \begin{bmatrix} v_{1} \\ 3v_{2}+4v_3 \\ -9v_3 \\ v_4 \end{bmatrix}$
where we have subtracted 3 times the second row from the third row. Then we notice that in this determinant, it is equal to the determinant of where we add 4/9 times the third row to the second row. From there, the answer should follow.
