SOLUTION: |a 1 1 1| |1 a 1 1| |1 1 a 1| =(a+3)(a-1)^3 |1 1 1 a|

Algebra ->  College  -> Linear Algebra -> SOLUTION: |a 1 1 1| |1 a 1 1| |1 1 a 1| =(a+3)(a-1)^3 |1 1 1 a|      Log On


   

Question 1044577: |a 1 1 1|
|1 a 1 1|
|1 1 a 1| =(a+3)(a-1)^3
|1 1 1 a|
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
abs%28matrix%284%2C4%2Ca%2C1%2C1%2C1%2C1%2Ca%2C1%2C1%2C1%2C1%2Ca%2C1%2C1%2C1%2C1%2Ca%29%29
Perform ERO3's (which do not change the value of the determinant)
-a%2AR%5B2%5D+%2B+R%5B1%5D, -1%2AR%5B2%5D+%2B+R%5B3%5D and -1%2AR%5B2%5D+%2B+R%5B4%5D, giving

.
= -abs%28matrix%283%2C3%2C1-a%5E2%2C1-a%2C1-a%2C1-a%2Ca-1%2C0%2C1-a%2C0%2Ca-1%29%29,
by Laplace expansion with the the first column as pivot.
Factoring 1-a from all three rows give
= -%281-a%29%5E3%2Aabs%28matrix%283%2C3%2C1%2Ba%2C1%2C1%2C1%2C-1%2C0%2C1%2C0%2C-1%29%29.
Now -%281-a%29%5E3+=+%28a-1%29%5E3, and so the last determinant is equal to
%28a-1%29%5E3%2Aabs%28matrix%283%2C3%2C1%2Ba%2C1%2C1%2C1%2C-1%2C0%2C1%2C0%2C-1%29%29.
Now perform the following ERO3 1%2AR%5B2%5D+%2B+R%5B1%5D, and ERO3 1%2AR%5B3%5D+%2B+R%5B1%5D IN SUCCESSION, to get
=%28a-1%29%5E3%2Aabs%28matrix%283%2C3%2C3%2Ba%2C0%2C0%2C1%2C-1%2C0%2C1%2C0%2C-1%29%29.
=%28a%2B3%29%28a-1%29%5E3%2Aabs%28matrix%282%2C2%2C-1%2C0%2C0%2C-1%29%29
By performing Laplace expansion with the first row as pivot.
=%28a%2B3%29%28a-1%29%5E3%2A%28-1%2A-1-0%2A0%29+=+highlight%28%28a%2B3%29%28a-1%29%5E3%29.