Answer to Question #107163 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 |
| 2a 2b 2c |
I got 48, please confirm if this is the correct answer or not.

Answer 1

aei+bfg+cdh-ceg-afh-bdi=8

The total multiplier in a row (column) can be taken as the sign of the determinant

$\begin{vmatrix} (g+2a)& (h+2b)&(i+2c)\\ 3a&3b&3c\\ 2a&2b&2c \end{vmatrix}$ $\\=3*\begin{vmatrix} (g+2a)& (h+2b)&(i+2c)\\ a&b&c\\ 2a&2b&2c \end{vmatrix}$

The determinant is zero, since the second and third rows are proportional