Question

Let $A=\begin{bmatrix}a & b & c \\ p & q & r \\ x & y & z\end{bmatrix}$ and suppose that det.(A)=2 then the det.(B) equals, where $B=\begin{bmatrix}4 x & 2 a & -p \\ 4 y & 2 b & -q \\ 4 z & 2 c & -r\end{bmatrix}$

• A. $\operatorname{det}( B )=-2$
• B. $\operatorname{det}( B )=-8$
• C. $\operatorname{det}(B)=-16$
• D. $\operatorname{det}( B )=8$

Answer 1

Answer: (C)

$\operatorname{det}( B )=\begin{vmatrix}4 x & 2 a & - p \\ 4 y & 2 b & - q \\ 4 z & 2 c & - r \end{vmatrix}=(4)(2)(-1)\begin{vmatrix} x & a & p \\ y & b & q \\ z & c & r \end{vmatrix}=-8\begin{vmatrix} x & y & z \\ a & b & c \\ p & q & r \end{vmatrix}=-8\begin{vmatrix} a & b & c \\ p & q & r \\ x & y & z \end{vmatrix}$
$=-8 \times 2=-16$