Mathematics - Determinant Question with Solution | TestHub

Apply column operations $C_2\to C_2-C_1$ and $C_3\to C_3-C_2$. Using the identity $\left(\genfrac{}{}{0ex}{}{k+1}{r}\right)-\left(\genfrac{}{}{0ex}{}{k}{r}\right)=\left(\genfrac{}{}{0ex}{}{k}{r-1}\right)$, we transform the determinant:$D=\left|\begin{array}{ccc}1& 1-1& 1-1\\ \left(\genfrac{}{}{0ex}{}{n}{1}\right)& \left(\genfrac{}{}{0ex}{}{n+1}{1}\right)-\left(\genfrac{}{}{0ex}{}{n}{1}\right)& \left(\genfrac{}{}{0ex}{}{n+2}{1}\right)-\left(\genfrac{}{}{0ex}{}{n+1}{1}\right)\\ \left(\genfrac{}{}{0ex}{}{n}{2}\right)& \left(\genfrac{}{}{0ex}{}{n+1}{2}\right)-\left(\genfrac{}{}{0ex}{}{n}{2}\right)& \left(\genfrac{}{}{0ex}{}{n+2}{2}\right)-\left(\genfrac{}{}{0ex}{}{n+1}{2}\right)\end{array}\right|$ This simplifies to $D=\left|\begin{array}{ccc}1& 0& 0\\ \left(\genfrac{}{}{0ex}{}{n}{1}\right)& \left(\genfrac{}{}{0ex}{}{n}{0}\right)& \left(\genfrac{}{}{0ex}{}{n+1}{0}\right)\\ \left(\genfrac{}{}{0ex}{}{n}{2}\right)& \left(\genfrac{}{}{0ex}{}{n}{1}\right)& \left(\genfrac{}{}{0ex}{}{n+1}{1}\right)\end{array}\right|=\left|\begin{array}{ccc}1& 0& 0\\ n& 1& 1\\ \frac{n(n-1)}{2}& n& n+1\end{array}\right|$. Expand along the first row: $D=1\cdot \left|\begin{array}{cc}1& 1\\ n& n+1\end{array}\right|=1\cdot ((1)(n+1)-(1)(n))=n+1-n=1$.