Question

The value of the determinant $\left| \begin{matrix}
   ^{5}{{C}_{0}} & ^{5}{{C}_{3}} & 14 \\
   ^{5}{{C}_{1}} & ^{5}{{C}_{4}} & 1 \\
   ^{5}{{C}_{2}} & ^{5}{{C}_{5}} & 1 \\
\end{matrix} \right|$ is
(A) $0$
(B) $-576$
(C) $80$
(D) None of these

Answer 1

Hint: For answering this question we need to simplify the matrix by applying the combinations formulae $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and derive the determinant of the given matrix $\left| \begin{matrix}
   ^{5}{{C}_{0}} & ^{5}{{C}_{3}} & 14 \\
   ^{5}{{C}_{1}} & ^{5}{{C}_{4}} & 1 \\
   ^{5}{{C}_{2}} & ^{5}{{C}_{5}} & 1 \\
\end{matrix} \right|$ by simplifying the minors and cofactors.

Complete step by step answer:
Now we have the matrix $\left| \begin{matrix}
   ^{5}{{C}_{0}} & ^{5}{{C}_{3}} & 14 \\
   ^{5}{{C}_{1}} & ^{5}{{C}_{4}} & 1 \\
   ^{5}{{C}_{2}} & ^{5}{{C}_{5}} & 1 \\
\end{matrix} \right|$ from the question by using the combinations $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and simplifying the matrix.
For simplifying let us calculate the individual value of the combinations. The values are as follows:
$^{5}{{C}_{0}}=\dfrac{5!}{0!\left( 5-0 \right)!}=1$
 $^{5}{{C}_{1}}=\dfrac{5!}{1!\left( 5-1 \right)!}=5$
$^{5}{{C}_{2}}=\dfrac{5!}{2!\left( 5-2 \right)!}=10$
$^{5}{{C}_{3}}=\dfrac{5!}{3!\left( 5-3 \right)!}=10$
$^{5}{{C}_{4}}=\dfrac{5!}{4!\left( 5-1 \right)!}=5$
$^{5}{{C}_{5}}=\dfrac{5!}{5!\left( 5-5 \right)!}=1$
By substituting these values in the matrix we have $\left| \begin{matrix}
   ^{5}{{C}_{0}} & ^{5}{{C}_{3}} & 14 \\
   ^{5}{{C}_{1}} & ^{5}{{C}_{4}} & 1 \\
   ^{5}{{C}_{2}} & ^{5}{{C}_{5}} & 1 \\
\end{matrix} \right|$ we will get the simplified matrix $\left| \begin{matrix}
   1 & 10 & 14 \\
   5 & 5 & 1 \\
   10 & 1 & 1 \\
\end{matrix} \right|$.
The minor of ${{a}_{ij}}$ is represented by ${{M}_{ij}}$ and for example for any $3\times 3$ matrix the minor of ${{a}_{21}}$ is represented by ${{M}_{21}}$ and is given by $\left| \begin{matrix}
   {{a}_{12}} & {{a}_{13}} \\
   {{a}_{32}} & {{a}_{23}} \\
\end{matrix} \right|$.
The cofactor of ${{a}_{ij}}$ is represented by ${{C}_{ij}}$ is given by${{C}_{ij}}={{\left( -1 \right)}^{i+j}}{{M}_{ij}}$ .
The determinant of any $3\times 3$ matrix is given by
$\begin{align}
  & ={{a}_{11}}{{C}_{11}}+{{a}_{12}}{{C}_{12}}+{{a}_{13}}{{C}_{13}} \\
 & \Rightarrow {{a}_{21}}{{C}_{21}}+{{a}_{22}}{{C}_{22}}+{{a}_{23}}{{C}_{23}} \\
 & \Rightarrow {{a}_{31}}{{C}_{31}}+{{a}_{32}}{{C}_{32}}+{{a}_{33}}{{C}_{33}} \\
\end{align}$
We can use any of the above three formulas.
Let us derive the determinant of this matrix. For that we will initially derive the minors of the matrix. After that we will have it as:
$\left| \begin{matrix}
   1 & 10 & 14 \\
   5 & 5 & 1 \\
   10 & 1 & 1 \\
\end{matrix} \right|=1\left| \begin{matrix}
   5 & 1 \\
   1 & 1 \\
\end{matrix} \right|-10\left| \begin{matrix}
   5 & 1 \\
   10 & 1 \\
\end{matrix} \right|+14\left| \begin{matrix}
   5 & 5 \\
   10 & 1 \\
\end{matrix} \right|$.
After performing the further simplifications we will have
$1\left( 5-1 \right)-10\left( 5-10 \right)+14\left( 5-50 \right)$ .
By further performing the calculations it will be reduced as
 $\begin{align}
  & 1\left( 4 \right)-10\left( -5 \right)+14\left( -45 \right) \\
 & \Rightarrow 4+50-630 \\
 & \Rightarrow -576 \\
\end{align}$.
Hence we can conclude that the value of the determinant $\left| \begin{matrix}
   ^{5}{{C}_{0}} & ^{5}{{C}_{3}} & 14 \\
   ^{5}{{C}_{1}} & ^{5}{{C}_{4}} & 1 \\
   ^{5}{{C}_{2}} & ^{5}{{C}_{5}} & 1 \\
\end{matrix} \right|$ is given as $-576$.

So, the correct answer is “Option B”.

Note: While answering this type of question we should be clear with the calculations. For example if we had made a mistake while calculating the determinant and had taken it as $4+50-620$ we will end up having a complete wrong answer as $-566$ . Another common mistake is interchanging the cofactors while expanding the determinant. So, perform this carefully.