A) 0
B) 1
C) \[m(m+1)\]
D) \[m(m-1)\]
Correct Answer: B
Solution :
\[\Delta =\left| \begin{matrix} 1 & 1 & 1 \\ ^{m}{{C}_{1}} & ^{m+1}{{C}_{1}} & ^{m+2}{{C}_{1}} \\ ^{m}{{C}_{2}} & ^{m+1}{{C}_{2}} & ^{m+2}{{C}_{2}} \\ \end{matrix} \right|\] \[=\left| \begin{matrix} 1 & 1 & 1 \\ ^{m}{{C}_{1}} & m+1 & m+2 \\ \frac{m(m-1)}{2} & \frac{m(m+1)}{2} & \frac{(m+1)(m+2)}{2} \\ \end{matrix} \right|\] \[=\frac{1}{2}\left| \begin{matrix} 1 & 0 & 0 \\ m & 1 & 1 \\ m(m-1) & 2m & 2(m+1) \\ \end{matrix} \right|\] \[\left[ \begin{align} & {{C}_{3}}\to {{C}_{3}}-{{C}_{2}} \\ & {{C}_{2}}\to {{C}_{2}}-{{C}_{1}} \\ \end{align} \right]\] \[=\frac{1}{2}[2.\{2(m+1)-2m\}]\] \[=\frac{1}{2}[2m+2-2m]=1\]
You need to login to perform this action.
You will be redirected in
3 sec
