Question

Show without expanding that this determinant is zero.

Answer 1

\[\LARGE \left|\begin{matrix}0 & x-a & x-b\\ x+a& 0&x-c \\ x+b & x+c &0\end{matrix}\right|\]

Answer 2

@dan815 @ikram002p @ganeshie8

Answer 3

sorry it says show that

Answer 4

Without Expanding, show that the equation\[\LARGE \left|\begin{matrix}0 &x-a &x-b\\x+a &0 &x-c\\x+b &x+c &0\end{matrix}\right| = 0\]

has 0 as a root

Answer 5

det = (x-b)(x-a)(x+c)+(x+b)(x-a)(x-c and that dont sound like 0 :o

so u mean that if x=0 then what ?!!

Answer 6

This is simple to do now since we just plug in x=0 to show it's a root to get a skew symmetric matrix. Since for determinants:\[\LARGE |A|=|A^T|\] and a skew symmetric matrix is \[\LARGE -A=A^T\] so now we can take the determinant of this matrix to get: \[\LARGE |-A|=|A^T|\] since A has 3 rows, when we distribute the negative sign to each column, it comes out of the determinant by being factored from each row separately: \[\LARGE (-1)^3|A|=|A^T|\] So we have: \[\LARGE -|A|=|A|\] so |A|=0

Answer 7

gr8 !

Answer 8

let there exist a number such that -|A| = |A| which is not 0
This number is known as the compreal number

Answer 9

Zed=$
where $^2=0