Davio said:
Hi guys.
1-n 3 0
3 -2-n -1
0 -1 1-n
My working for determinant :
+ 1-n x (-2-n)(1-n) - (-1).(-1)
Please use parentheses! You mean (1- n)[(-2-n)((1-n)- (-1)(-1)]
+ 1-n x (-2-n)(1-n) -1
+ 1-n x (n^2 +2n -n -2) -1
+ 1-n x (n^2 +2n -n -3)
+ 1-n x (n^2 +n -3)
(n^2 +n -3) - (n^3 -n^2 +3n)
No, if you are keeping "-" outside that last parenthesis, it is -(n^3+ n^2- 3n)
(n^2 +n -3) - n^3 -n^2 +3n
- (n^3 +4n) -3
4n - n^3 -3
Even if you had not messed up the sign, this is just the top left corner, 1-n, times its cofactor
\left|\begin{array}{cc} -2-n & -1 \\ -1 & 1- n\end{array}\right|
If you are expanding on the first row, you should also have
-3\left|\begin{array}{cc}3 & -1 \\ 0 & 1-n\end{array}\right]
If instead you are expanding on the first column, you should have
-3\left|\begin{array}{cc}3 & 0 \\ -1 & 1-n\end{array}\right]
which is obviously the same thing.
The original question is to prove the eigen values are 1, 3 and ? (negative 4 from online calculator)
Thanks in advance :S
I'm not sure how polynonial division would help me :S
In general if you no "n= a" is a root of a cubic polynomial then you know the polynomial is equal to (x- a) times quadratic polynomial. Dividing the cubic polynomial by x-a gives the remaining quadratic polynomial which you can solve, for example, using the quadratic formula.
-3 x (3 x 1-n) - 0
-3 x (3 - 3n)
-9 + 9n
Oh, here's your missing term! You should have (n^2 +n -3)-(n^3+ n^2- 3n)- 9+ 9n=
-n^3+ 13n- 12= 0.
Yes, 1 satisfies that: -1^3+ 13(1)- 12= -1+ 13- 12= 0. Now divide -n^3+ 13n- 12 by n- 1 to get a quadratic.
