Let $A$ = $\begin{bmatrix} v_1 \hspace{2 mm} v_2 \hspace{2 mm} v_3 \end{bmatrix}$ be a 3 x 3 matrix with column vectors $v_1, v_2, v_3$.
If $det(A) = 2$, then what is $det \begin{bmatrix} v_1+v_2+v_3 \hspace{7 mm} v_1+2v_2+3v_3 \hspace{7 mm} v_1+4v_2+9v_3\end{bmatrix}$?
I then set up the matrix for the question:
$$det \left(\begin{bmatrix} 1 \hspace{5mm} 1\hspace{5mm} 1 \\ 1\hspace{5mm} 2\hspace{5mm} 4 \\ 1\hspace{5mm} 3\hspace{5mm} 9 \end{bmatrix} \begin{bmatrix}v_1\hspace{5mm} v_2\hspace{5mm} v_3\end{bmatrix}\right)$$
Through my calculation, the determinant of the first matrix (the coefficient matrix) is 2, multiplied by the determinant of the given matrix, will give 4 as a result.
I just want to check if what I am doing is right for this question.
Best Answer
According to the properties of the determinant, if we denote by $M$ the modified matrix obtained from $A$, and we apply some elementary column operations conveniently, it results that: \begin{align*} \det(M) & = \det[v_{1} + v_{2} + v_{3} \mid v_{2} + 2v_{3} \mid 2v_{2} + 6v_{3}]\\\\ & = \det[v_{1} + v_{2} + v_{3} \mid v_{2} + 2v_{3} \mid 2v_{3}]\\\\ & = 2\det[v_{1} + v_{2} + v_{3} \mid v_{2} + 2v_{3} \mid v_{3}]\\\\ & = 2\det[v_{1} + v_{2} + v_{3} \mid v_{2} \mid v_{3}]\\\\ & = 2\det[v_{1} \mid v_{2} \mid v_{3}] = 2\times 2 = 4. \end{align*}
Hopefully this helps!