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Let a,b,c , in R not all are equal and...

Let `a,b,c , in R` not all are equal and `Delta_(1)= |{:(a,,b,,c),(b,,c,,a),(c,,a,,b):}|`
`Delta_(2)= |{:(a+2b,,b+3c,,c+4a),(b+2c,,c+3a,,a+4b),(c+2a,,a+3b,,b+4c):}|" then " (Delta_(2))/(Delta_(1)) = "____"`

Text Solution

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The correct Answer is:
25

`Delta = |{:(a,,b+3c,,c+4a),(b,,c+3a,,a+4b),(c,,a+3b,,b+4c):}|+2 |{:(b,,b+3c,,c+4a),(c,,c+3a,,a+4b),(a,,a+3b,,b+4c):}|`
Applying `C_(3) to C_(3) -4C_(1) " in " Delta " and " C_(2) to C_(2) -C_(1) " in " Delta''` we get
`Delta_(2)= |{:(a,,b+3c,,c),(b,,c+3a,,a),(c,,a+3b,,b):}|+2xx3 |{:(b,,c,,c+4a),(c,,a,,a+4b),(a,,b,,b+4c):}|`
applying `c_(2) to C_(2)- 3C_(3) " in " Delta " and " C_(3) to C_(3) -C_(2) " in " Delta''` we get
` = |{:(a,,b,,c),(b,,c,,a),(c,,a,,b):}|+6 |{:(b,,c,,4a),(c,,a,,4b),(a,,b,,4c):}|`
`=|{:(a,,b,,c),(b,,c,,a),(c,,a,,b):}|+24 |{:(a,,b,,c),(b,,c,,a),(c,,a,,b):}|`
`=25 |{:(a,,b,,c),(b,,c,,a),(c,,a,,b):}|=25Delta_(1)`
`rArr (Delta_(2))/(Delta_(1)) =25`
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