Question

Solve the determinant to get the value of n

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Answer 1

Question :-

If the value of given determinant is 0, find n

$\begin{gathered}\rm \:\left | \begin{array}{ccc} ^{n+2}C_{2} &^{n+3}C_{2}&^{n+4}C_{2} \\ ^{n+3}C_{2}&^{n+4}C_{2}&^{n+5}C_{2} \\ ^{n+4}C_{2}&^{n+5}C_{2}&^{n+6}C_{2}\end{array}\right| \ \end{gathered} = 0$

$\large\underline{\sf{Solution-}}$

Given determinant is

$\begin{gathered}\rm \:\left | \begin{array}{ccc} ^{n+2}C_{2} &^{n+3}C_{2}&^{n+4}C_{2} \\ ^{n+3}C_{2}&^{n+4}C_{2}&^{n+5}C_{2} \\ ^{n+4}C_{2}&^{n+5}C_{2}&^{n+6}C_{2}\end{array}\right| \ \end{gathered} = 0$

can be further rewritten as

$\begin{gathered}\rm \:\dfrac{1}{2}\left | \begin{array}{ccc} 2 \: ^{n+2}C_{2} &2 \: ^{n+3}C_{2}&2 \: ^{n+4}C_{2} \\ 2 \: ^{n+3}C_{2}&2 \: ^{n+4}C_{2}&2 \: ^{n+5}C_{2} \\ 2 \: ^{n+4}C_{2}&2 \: ^{n+5}C_{2}&2 \: ^{n+6}C_{2}\end{array}\right| \ \end{gathered} = 0$

$\begin{gathered}\rm \:\left | \begin{array}{ccc} 2 \: ^{n+2}C_{2} &2 \: ^{n+3}C_{2}&2 \: ^{n+4}C_{2} \\ 2 \: ^{n+3}C_{2}&2 \: ^{n+4}C_{2}&2 \: ^{n+5}C_{2} \\ 2 \: ^{n+4}C_{2}&2 \: ^{n+5}C_{2}&2 \: ^{n+6}C_{2}\end{array}\right| \ \end{gathered} = 0$

We know,

$\boxed{\sf{ \: ^{n}C_{2} = \frac{n(n - 1)}{2} \: \: }}$

So, using this result, we get

$\begin{gathered}\rm \:\left | \begin{array}{ccc} (n + 2)(n + 1) &(n + 3)(n + 2)&(n + 4)(n + 3) \\ (n + 3)(n + 2)&(n + 4)(n + 3)&(n + 5)(n + 4) \\ (n + 4)(n + 3)&(n + 5)(n + 4)&(n + 6)(n + 5)\end{array}\right| \ \end{gathered} = 0$

$\: \: \: \: \: \: \: \boxed{\tt{ OP \: R_3 \: \to \: R_3 - R_2 \: }}$

$\begin{gathered}\rm \:\left | \begin{array}{ccc} (n + 2)(n + 1) &(n + 3)(n + 2)&(n + 4)(n + 3) \\ (n + 3)(n + 2)&(n + 4)(n + 3)&(n + 5)(n + 4) \\ 2(n + 3)&2(n + 4)&2(n + 5)\end{array}\right| \ \end{gathered} = 0$

$\: \: \: \: \: \: \: \boxed{\tt{ OP \: R_2 \: \to \: R_2 - R_1 \: }}$

$\begin{gathered}\rm \:\left | \begin{array}{ccc} (n + 2)(n + 1) &(n + 3)(n + 2)&(n + 4)(n + 3) \\ 2(n + 2)&2(n + 3)&2(n + 4) \\ 2(n + 3)&2(n + 4)&2(n + 5)\end{array}\right| \ \end{gathered} = 0$

$\begin{gathered}\rm \:4\left | \begin{array}{ccc} (n + 2)(n + 1) &(n + 3)(n + 2)&(n + 4)(n + 3) \\ (n + 2)&(n + 3)&(n + 4) \\ (n + 3)&(n + 4)&(n + 5)\end{array}\right| \ \end{gathered} = 0$

$\: \: \: \: \: \: \: \boxed{\tt{ OP \: C_3 \: \to \: C_3 - C_2 \: }}$

$\begin{gathered}\rm \:4\left | \begin{array}{ccc} (n + 2)(n + 1) &(n + 3)(n + 2)&2(n + 3) \\ (n + 2)&(n + 3)&1 \\ (n + 3)&(n + 4)&1\end{array}\right| \ \end{gathered} = 0$

$\: \: \: \: \: \: \: \boxed{\tt{ OP \: C_2 \: \to \: C_2 - C_1 \: }}$

$\begin{gathered}\rm \:4\left | \begin{array}{ccc} (n + 2)(n + 1) &2(n + 2)&2(n + 3) \\ n + 2&1&1 \\ n + 3&1&1\end{array}\right| \ \end{gathered} = 0$

$\: \: \: \: \: \: \: \boxed{\tt{ OP \: R_3 \: \to \: R_3 - R_2 \: }}$

$\begin{gathered}\rm \:4\left | \begin{array}{ccc} (n + 2)(n + 1) &2(n + 2)&2(n + 3) \\ n + 2&1&1 \\ 1&0&0\end{array}\right| \ \end{gathered} = 0$

On expanding along 3rd row, we have

$\rm \: 4(2n + 4 - 2n - 6) = 0$

$\rm \: 4( - 2) = 0$

$\rm \: - 8 = 0$

$\rm \: which \: is \: not \: possible$

$\rm\implies \:There \: is \: no \: value \: of \: n$

So, Option (d) is correct.

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ADDITIONAL INFORMATION

1. The determinant value remains unaltered if rows and columns are interchanged.

2. The determinant value is 0, if two rows or columns are identical.

3. The determinant value is multiplied by - 1, if successive rows or columns are interchanged.

4. The determinant value remains unaltered if rows or columns are added or subtracted.