1. Tardigrade
  2. Question
  3. Mathematics
  4. The value of the determinant| beginmatrix1&1&1 mC1&m+1C1&m+2C1 mC2&m+1C2&m+2C2 endmatrix|is equal to

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. The value of the determinant$\left|\begin{matrix}1&1&1\\ ^{m}C_{1}&^{m+1}C_{1}&^{m+2}C_{1}\\ ^{m}C_{2}&^{m+1}C_{2}&^{m+2}C_{2}\end{matrix}\right|$is equal to

Determinants

Solution:

$\left|\begin{matrix}1&1&1\\ \,{}^{m}C_{1}&\,{}^{m+1}C_{1}&\,{}^{m+2}C_{1}\\ \,{}^{m}C_{2}&\,{}^{m+1}C_{2}&\,{}^{m+2}C_{2}\end{matrix}\right|$
$=\left|\begin{matrix}1&1&1\\ \,{}^{m}C_{1}&\,{}^{m+1}C_{1}&\,{}^{m+1}C_{0}+\,{}^{m+1}C_{1}\\ \,{}^{m}C_{2}&\,{}^{m+1}C_{2}&\,{}^{m+1}C_{1}+\,{}^{m+1}C_{2}\end{matrix}\right|$
$=\left|\begin{matrix}1&1&0\\ \,{}^{m}C_{1}&\,{}^{m+1}C_{1}&\,{}^{m+1}C_{0}\\ \,{}^{m}C_{2}&\,{}^{m+1}C_{2}&\,{}^{m+1}C_{1}\end{matrix}\right|$
$\left[\text{Applying} C_{3}\to C_{3}-C_{2}\right]$
$=\left|\begin{matrix}1&1&0\\ \,{}^{m}C_{1}&\,{}^{m}C_{0}+\,{}^{m}C_{1}&\,{}^{m+1}C_{0}\\ \,{}^{m}C_{2}&\,{}^{m}C_{1}+\,{}^{m}C_{2}&^{m+1}C_{1}\end{matrix}\right|$
$=\left|\begin{matrix}1&0&0\\ \,{}^{m}C_{1}&\,{}^{m}C_{0}&\,{}^{m+1}C_{0}\\ \,{}^{m}C_{2}&\,{}^{m}C_{1}&\,{}^{m+1}C_{1}\end{matrix}\right|$
$\left[\text{Applying} C_{2}\to C_{2}-C_{1}\right]$
$=\,{}^{m}C_{0} \,{}^{m+1}C_{1}-\,{}^{m+1}C_{0}\,{}^{m}C_{1}=m+1-m=1$