So multiplying a row by a constant multiple the determinant by the same constant and swapping 2 rows will multiply the determinant by a negative right and adding or subtracting does not change the matrix.

What if......

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

so swapping two rows will equal to a determinant positive 6 and multiplying by (1/3) will make determinant equal -2

But the question asks......

$$\begin{vmatrix} a&b&c \\ d&e&f \\ 2a&2b&2c \end{vmatrix}$$

I don't understand how you can change the first matrix into this one at all. At the best i know 2 of row at least was added so that equations now change in determinant bu ti dunno how g h i are eliminated.

  • 1
    $\begingroup$ That last matrix has two rows that are proportional, so the determinant must be zero. $\endgroup$
    – Chappers
    Mar 26, 2015 at 2:33

1 Answer 1


The determinant you want to calculate is equal to zero because the third line is 2 times the first line. It doesn't matter the determinant of the first matrix you wrote.

  • $\begingroup$ Ahh yeah i thought of that but got thrown off by the possibility of row reducing the original. Thanks $\endgroup$
    – Ivan
    Mar 26, 2015 at 2:37

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