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suppose D= |{:(a(1),,b(1),,c(1)),(a(2)...

suppose `D= |{:(a_(1),,b_(1),,c_(1)),(a_(2),,b_(2),,c_(2)),(a_(3),,b_(3),,c_(3)):}| ` and
`D^(r )= |{:(a_(1)+pb_(1),,b_(1)+qc_(1),,c_(1)+ra_(1)),(a_(2)+pb_(2),,b_(2)+qc_(2),,c_(2)+ra_(2)),(a_(3)+pb_(3),,b_(3)+qc_(3),,c_(3)+ra_(3)):}| `. Then

A

`Delta' = 2 Delta`

B

`Delta' = Delta`(1 - pqr)

C

`Delta' = Delta` (1 + P+q+r)

D

`Delta' = Delta` (1 + pqr)

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The correct Answer is:
D
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suppose D= |{:(a_(1),,b_(1),,c_(1)),(a_(2),,b_(2),,c_(2)),(a_(3),,b_(3),,c_(3)):}| and Dprime= |{:(a_(1)+pb_(1),,b_(1)+qc_(1),,c_(1)+ra_(1)),(a_(2)+pb_(2),,b_(2)+qc_(2),,c_(2)+ra_(2)),(a_(3)+pb_(3),,b_(3)+qc_(3),,c_(3)+ra_(3)):}| . Then

Delta_(1)=det[[a_(1)+pb_(1),b_(1)+qc_(1),c_(1)+ra_(1)a_(2)+pb_(2),b_(1)+qc_(2),c_(2)+ra_(2)a_(3)+pb_(3),b_(3)+qc_(3),c_(3)+ra_(3)]]

if quad /_=[[a_(1),b_(1),c_(1)a_(2),b_(2),c_(2)a_(3),b_(3),c_(3)]]

if Delta=det[[a_(1),b_(1),c_(1)a_(2),b_(2),c_(2)a_(3),b_(3),c_(3)]]

Show that |{:(ma_(1),b_(1),nc_(1)),(ma_(2),b_(2),nc_(2)),(ma_(3),b_(3),nc_(3)):}|=-mn|{:(c_(1),b_(1),a_(1)),(c_(2),b_(2),a_(2)),(c_(3),b_(3),a_(3)):}|

If, in D={:[(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2)),(a_(3),b_(3),c_(3))]:}, the co-factor of a_(r)" is "A_(r), then , c_(1)A_(1)+c_(2)A_(2)+c_(3)A_(3)=

The determinant |(b_(1)+c_(1),c_(1)+a_(1),a_(1)+b_(1)),(b_(2)+c_(2),c_(2)+a_(2),a_(2)+b_(2)),(b_(3)+c_(3),c_(3)+a_(3),a_(3)+b_(3))|

If |(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2)),(a_(3),b_(3),c_(3))| =5 , then the value of Delta = |(b_(2) c_(3) - b_(3) c_(2),a_(3) c_(2) - a_(2) c_(3),a_(2) b_(3) -a_(3) b_(2)),(b_(3) c_(1) - b_(1) c_(3),a_(1) c_(3) - a_(3) c_(1),a_(3) b_(1) - a_(1) b_(3)),(b_(1) c_(2) - b_(2) c_(1),a_(2) c_(1) - a_(1) c_(2),a_(1) b_(2) - a_(2) b_(1))| is

In algebra, the determinant is useful value that can be computer from the elements of a square matrix. The determinant is represented as det 'A' or |A| and its value can be evaluated by the expansion of the determinant as given below (A) Expansion of two order determinant : (B) Expansion of 3^(rd) order determinant (i) With respect to first fow : |A|=|{:(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2)),(a_(3),b_(3),c_(3)):}|=a_(1)|{:(b_(1),c_(2)),(b_(x3,c_(3)):}|+b_(1)|{:(a_(2),c_(2)),(a_(2),c_(2)):}|+c_(1)|{:(b_(x),c_(2)),(b_(2),c_(2)):}| =a_(1)(b_(2)c_(3)-b_(3)c_(2))-b_(1)(a_(2)c_(3)-a_(3)c_(2))+c_(1)(a_(2)b_(3)-b_(2)a_(3)) (ii) With respect to second column : |A|=|{:(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2)),(a_(3),b_(3),c_(3)):}|=-b_(1)|{:(a_(2),c_(1)),(a_(3),c_(3)):}|+b_(2)|{:(a_(1),c_(1)),(a_(3),c_(3)):}|-b_(3)|{:(a_(1),c_(1)),(a_(2),c_(2)):}| Similarly a determinant can be expanded with respect to any row or column The value of for which determinant |{:(2,3,-1),(-1,-2,k),(1,-4,1):}| vanishes, is

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