Question

The value of the determinant $\left|\begin{array}{ccc}1& 1& 1\\ {{}^{m}C}_1& {{}^{m+1}C}_1& {{}^{m+2}C}_1\\ {{}^{m}C}_2& {{}^{m+1}C}_2& {{}^{m+2}C}_2\end{array}\right|$ is equal to

• A. $1$
• B. $-1$
• C. $0$
• D. none of these

Answer 1

Answer: (A)

$1$

$\left|\begin{array}{ccc}1& 1& 1\\ {{}^{m}C}_1& {{}^{m+1}C}_1& {{}^{m+2}C}_1\\ {{}^{m}C}_2& {{}^{m+1}C}_2& {{}^{m+2}C}_2\end{array}\right|=\left|\begin{array}{ccc}1& 1& 1\\ {{}^{m}C}_1& {{}^{m+1}C}_1& {{}^{m+1}C}_0+{{}^{m+1}C}_1\\ {{}^{m}C}_2& {{}^{m+1}C}_2& {{}^{m+1}C}_1+{{}^{m+1}C}_2\end{array}\right|=\left|\begin{array}{ccc}1& 1& 0\\ {{}^{m}C}_1& {{}^{m+1}C}_1& {{}^{m+1}C}_0\\ {{}^{m}C}_2& {{}^{m+1}C}_2& {{}^{m+1}C}_1\end{array}\right|$

applying $C_3\to C_3-C_2$ and $C_2\to C_2-C_1$

$=\left|\begin{array}{ccc}1& 1& 0\\ {{}^{m}C}_1& {{}^{m}C}_0& {{}^{m+1}C}_0\\ {{}^{m}C}_2& {{}^{m}C}_1& {{}^{m+1}C}_1\end{array}\right|={{}^{m}C}_0{{}^{m+1}C}_1-{{}^{m+1}C}_0{{}^{m+1}C}_1=m+1-m=1$