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 November 24th, 2015, 10:35 AM #1 Member   Joined: Nov 2015 From: saudi arabia Posts: 30 Thanks: 5 Need help (determinant with a parameter) hello all . I have Question about the determinant and linearly dependent and it's : Let : a) Find det(B) in terms of k; b) For what value(s) of k are the column vectors of B linearly dependent; c) For k = 0, find det(B) and . can some one help me in this question and thank you all . November 25th, 2015, 06:50 PM #2 Member   Joined: Nov 2015 From: saudi arabia Posts: 30 Thanks: 5 Any one can help me ? November 27th, 2015, 06:23 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Are you saying you do not know how to find the determinant? There are several different ways of finding a determinant. It would be best to use a method you had been taught but since you make no attempt to do this, I have no idea what that might be. The first method I was taught, though it works only for 3 by 3 determinants, is to look at "diagonals". The main diagonal, from upper left to lower right is (1, k- 3, k- 4). The product is 1(k- 3)(k- 4)= k^2- 7k+ 12. A "sub-diagonal", from upper left to lower right, is (2, 4, 3) which has product 24. Another "sub-diagonal, from upper left to lower right, is (3, 4, 2) which also has product 24. Their sum is k^2- 7k+ 60. Now, going from upper right to lower left, the "main" diagonal is (3, k- 3, 3) and their product is 9k- 27. The two "sub-diagonals", from upper right to lower left, are (2, 2, k-4) and (4, 4, 1) with products 4k- 16 and 16. Those three add to 13k- 27. So the determinant is (k^2- 7k+ 60)- (13k- 27)= k^2- 20k+ 33. To determine when the three columns, "as vectors", are dependent look at the equation $\displaystyle a\begin{matrix}1 \\ 2 \\ 3\end{pmatrix}+ b\begin{pmatrix}2 \\ k- 3\\ 4 \end{pmatrix}+ c\begin{pmatrix} 3 \\ 4 \\ k- 4 \end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0 \end{pmatrix}$. The three vectors are "dependent" if and only if there exist a, b, c, not all 0, such that the equation is true. Looking at the three components that gives the equations a+ 2b+ 3c= 0, 2a+ (k- 3)b+ 4c= 0, 3a+ 4b+ (k-4)c= 0. An obvious solution to those equations is a= b= c= 0. What values of k are such that there are other solutions? Thanks from ahmed2009 Tags determinant, parameter, parmeter Search tags for this page

### determinate of A*B

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