Let A =
and assume that det A = 3.
Compute:
det(2c^-1) where C =
I'm getting an answer of 1/3, which is wrong. The answer is 4/9.
The troubles I'm having you can read about right HERE. I know how to do everything, but I'm obviously just making a small mistake.
One specific question I would like to ask, is how do I remove the 2 from inside the det() brackets? I haven't been shown an example of that, and can't find one.
Thanks!
ok.. let's recall some theorems..
THEOREMS
Let A = [A_1 A_2 ... A_n] be an nxn matrix, where A_i are the columns of A for i=1,..,n.
let B be the transpose of A.
1) det (A) = det (B)
2) det([A_1 ... cA_k ... A_n]) = c det([A_1 ... A_k ... A_n])
3) det(cA) = det(c[A_1 A_2 ... A_n] = (c^n) det(A)
4) det([A_1 ... (A_k)+(A_j) ... A_n]) = det([A_1 ... A_k ... A_n]) + det([A_1 ... A_j ... A_n])
5) det([A_1 ... A_k cA_j ... A_n]) = 0 if A_k = A_j
6) det([A_1 ... A_k ... A_j .... A_n]) = (-1)^p det([A_1 ... A_j ... A_k .... A_n]) where p= j-k
these theorems should help you..
so, these are the steps. you are given the matrix A.
..get the transpose of A. that transpose has same determinant as A.
..looking on the matrix C, you will see that the second column is a sum, hence use THEROEM 4.
..after this step, you should notice that one of the addends has 2 a column which is a multiple of the another column (the third column is 3 times the second), hence the determinant of this addend is 0 by THEOREM 5.
..you are left with one addend. you notice that the first column has factor 2, the second has -1 and the third has 3, use THEOREM 2.
..now, switch the first column with the second column and use THEOREM 6.
..you should notice that the final matrix looks like the transpose of A.
..you can now use THEOREM 3 and you should get 4/9.
Ü