` " if " Delta = |{:(-x,,a,,b),(b,,-x,,a),(a,,b,,-x):}|" then a factor of " Delta " is "`

Options

A. `a+b+x`
B. `x^(2)-(a-b)x+a^(2)+b^(2)+ab`
C. `x^(2)+(a+b)x+a^(2)+b^(2)-ab`
D. `a+b-x`

Answer

C, D

Explanation

Applying `C_(1) to C_(1)+C_(2)+C_(3)` we get
`Delta =|{:(a+b-x,,a,,b),(a+b-x,,-x,,a),(a+b-x,,b,,-x):}|=(a+b-x) |{:(1,,a,,b),(1,,-x,,a),(1,,b,,-x):}|`
`=(a+b-x) |{:(1,,a,,b),(0,,-x-a,,a-b),(0,,b-a,,-x-b):}|`
`"[Applying "R_(2) to R_(2) -R_(1)" and " R_(3) to R_(3)-R_(1)"]"`
`=(a+b-x)[(x+a)(x+b)+(a-b)^(2)]`
[expanding along `C_(1)`]
`=(a+b-x)[x^(2) +(a+b)x +a^(2) +b^(2) -ab]`