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
$\text{A) 0}$
$\text{B) -576}$
$\text{C) 80}$
$\text{D) none of these}$

Answer 1

In this question we have a determinant of $9$ terms which we will solve as a determinant by taking the modulus value of the determinant. In the terms of the determinant, we have terms which are expressed in the form of the combination. We will use the combination formula which is $^{n}{{C}_{r}}=\dfrac{n!}{(n-r)!r!}$ to simplify the terms.

Complete step by step solution:
We have the determinant given as $\left| \begin{matrix} ^{5}{{C}_{0}} & ^{5}{{C}_{3}} & 14 \\\ ^{5}{{C}_{1}} & ^{5}{{C}_{4}} & 1 \\\ ^{5}{{C}_{2}} & ^{5}{{C}_{5}} & 1 \\\ \end{matrix} \right|$.
Let’s consider the given determinant to be $\text{A}$ therefore, we can write it mathematically as:
$\Rightarrow \text{A}=\left| \begin{matrix} ^{5}{{C}_{0}} & ^{5}{{C}_{3}} & 14 \\\ ^{5}{{C}_{1}} & ^{5}{{C}_{4}} & 1 \\\ ^{5}{{C}_{2}} & ^{5}{{C}_{5}} & 1 \\\ \end{matrix} \right|$
Now consider the general form of a $3\times 3$ matrix which is $\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\\ \end{matrix} \right|$. Now from this matrix we can solve the terms in the determinant before taking the value of the determinant which is denoted as $\text{ }\\!\\!|\\!\\!\text{ A }\\!\\!|\\!\\!\text{ }$
Now consider the term ${{a}_{11}}$:
$\Rightarrow {{a}_{11}}{{=}^{5}}{{C}_{0}}$
Using the formula, we get:
$\Rightarrow {{a}_{11}}=\dfrac{5!}{(5-0)!0!}=\dfrac{5!}{5!}=1$
Now consider the term ${{a}_{12}}$:
$\Rightarrow {{a}_{12}}{{=}^{5}}{{C}_{3}}$
Using the formula, we get:
$\Rightarrow {{a}_{12}}=\dfrac{5!}{(5-3)!3!}=\dfrac{5!}{2!3!}=\dfrac{5\times 4}{2\times 1}=10$
Now consider the term ${{a}_{21}}$:
$\Rightarrow {{a}_{21}}{{=}^{5}}{{C}_{1}}$
Using the formula, we get:
$\Rightarrow {{a}_{21}}=\dfrac{5!}{(5-1)!1!}=\dfrac{5!}{4!}=5$
Now consider the term ${{a}_{22}}$:
$\Rightarrow {{a}_{22}}{{=}^{5}}{{C}_{4}}$
Using the formula, we get:
$\Rightarrow {{a}_{22}}=\dfrac{5!}{(5-4)!4!}=\dfrac{5!}{4!}=5$
Now consider the term ${{a}_{31}}$:
$\Rightarrow {{a}_{31}}{{=}^{5}}{{C}_{2}}$
Using the formula, we get:
$\Rightarrow {{a}_{22}}=\dfrac{5!}{(5-2)!2!}=\dfrac{5!}{3!2!}=\dfrac{5\times 4}{2\times 1}=10$
Now consider the term ${{a}_{32}}$:
$\Rightarrow {{a}_{32}}{{=}^{5}}{{C}_{5}}$
Using the formula, we get:
$\Rightarrow {{a}_{32}}=\dfrac{5!}{(5-5)!5!}=\dfrac{5!}{5!}=1$
Now on putting all the values in the determinant $\text{A}$, we get:
$\Rightarrow \text{A}=\left| \begin{matrix} 1 & 10 & 14 \\\ 5 & 5 & 1 \\\ 10 & 1 & 1 \\\ \end{matrix} \right|$
Now on solving the determinant, we get:
$\Rightarrow \left| \text{A} \right|=1\left| \begin{matrix} 5 & 1 \\\ 1 & 1 \\\ \end{matrix} \right|-10\left| \begin{matrix} 5 & 1 \\\ 10 & 1 \\\ \end{matrix} \right|+14\left| \begin{matrix} 5 & 5 \\\ 10 & 1 \\\ \end{matrix} \right|$
Now on simplifying the values in the sub-determinant, we get:
$\Rightarrow \left| \text{A} \right|=1(5\times 1-1\times 1)-10(5\times 1-10\times 1)+14(5\times 1-5\times 10)$
Now on multiplying the terms, we get:
$\Rightarrow \left| \text{A} \right|=1(5-1)-10(5-10)+14(5-50)$
On simplifying the brackets, we get:
$\Rightarrow \left| \text{A} \right|=1(4)-10(-5)+14(-45)$
On multiplying the terms, we get:
$\Rightarrow \left| \text{A} \right|=4+50-630$
On simplifying, we get:
$\Rightarrow \left| \text{A} \right|=-576$, which is the required solution.
Therefore, the correct option is $\text{(A)}$.

Note: In this question we have used the combination formula which tells us the combination of terms when the order of the terms does not matter. There also exists the permutation formula which is $^{n}{{P}_{r}}=\dfrac{n!}{(n-r)!}$. In case of permutation, the order of the terms matter. The $!$ sign is the factorial sign which indicates the product of an integer and all the integers below it up to the integer $1$.