Answer to Question #107533 in Linear Algebra for Caylin

Question

Suppose | a b c |
| d e f | =8
| g h i |

Find value of | (g+2a) (h+2b) (i + 2c) |
| 3a 3b 3c |
| 2d 2e 2f |

I got 48 as my answer. please confirm if correct.

Answer 1

$\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix}=8$

$\begin{vmatrix} g+2a & h+2b & i+2c \\ 3a & 3b & 3c \\ 2d & 2e & 2f \end{vmatrix}= 3\times 2\times \begin{vmatrix} g+2a & h+2b & i+2c \\ a & b & c \\ d & e & f \end{vmatrix}=$

$= 6\times \begin{vmatrix} g& h& i\\ a & b & c \\ d & e & f \end{vmatrix}+6\times \begin{vmatrix}2a & 2b &2c \\ a & b & c \\ d & e & f \end{vmatrix}=-6\times \begin{vmatrix} a & b & c \\ g & h & i \\ d & e & f \end{vmatrix}+6\times 0=$

$=(-1)^2 \times 6\times \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{vmatrix}=6\times 8=48$

Answer: 48.