A) \[441\times 446\times 4510\]
B) 0
C) -1
D) 1
Correct Answer: B
Solution :
We have, \[\left| \begin{matrix} 441 & 442 & 443 \\ 445 & 446 & 447 \\ 449 & 450 & 451 \\ \end{matrix} \right|\] On expanding we get, \[=441\left[ \left( 201146-201150 \right) \right]-442\] \[\left[ 200695-200703 \right]+443\left[ 200250-200254 \right]\] \[=-1764+3536-1772\] \[=3536-3536=0\] Alternate \[\left| \begin{matrix} 441 & 442 & 443 \\ 445 & 446 & 447 \\ 449 & 450 & 451 \\ \end{matrix} \right|\] \[=\left| \begin{matrix} 441 & 1 & 1 \\ 445 & 1 & 1 \\ 449 & 1 & 1 \\ \end{matrix} \right|\begin{matrix} {{C}_{2}}\to {{C}_{2}}-{{C}_{1}} \\ {{C}_{3}}\to {{C}_{3}}-{{C}_{2}} \\ \end{matrix}\]= 0 (\[\because \] two columns are in dentical)
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