Question

how to evaluate 3x3 determinant using the properties of determinants for simplification?

Answer 1

\[\left|\begin{array}{ccc} a & b & c \\ d & e & f \\ g& h & i \end{array} \right|=a\left|\begin{array}{ccc} e & f \\ h & i \end{array} \right|-b\left|\begin{array}{ccc} d & f \\ g & i \end{array} \right|+c\left|\begin{array}{ccc} d & e \\ g & h\end{array} \right|\\\qquad\qquad \quad =a(ei-fh)-b(di-fg)+c(dh-eg)\]

Answer 2

if one of the rows or columns in your matrix is all zeroes, then the determinant can be simpler

Answer 3

You can also use the method of diagonals after re-writing the first two columns, which is pretty easy to remember.

Answer 4

are you still here @skit ?

Answer 5

@skit, I cannot offer anything except a warm welcome to OpenStudy :D
You may want to read the code of conduct ( http://openstudy.com/code-of-conduct ).
Don't forget to click on the button called "Best Answer" to give a medal to the person who helped you :)

Answer 6

one property thats useful is :

determinant will not change when u add linear combination of other rows/columns to a row/column.

Answer 7

for example,
determinant will not change by doing below :-

R1 -> R1-2R2 + 3R3

or

C1 -> C1-2C2 + 3C3