Let -r - r - 1 n- 6 r - 12 n -2 4n - 2 r - 13 n -3 3n2 - 2n- then the value of -r - 2n - 1-r -

Given , Δ_r = #160;(r 1) amp; n! amp; 6 (r 1)^2 #160; amp; (n !)^2 amp; 4n 2 (r 1)^3 amp; (n !)^3 amp; 3n^2 2n ...

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Byju's Answer

Standard XII

Mathematics

Cofactor

Let Δr = r ...

Question

Let $Δ_{r} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} (r - 1) & n! & 6 (r - 1)^{2} & (n!)^{2} & 4 n - 2 (r - 1)^{3} & (n!)^{3} & 3 n^{2} - 2 n \end{matrix} ∣ ∣ ∣ ∣ ∣$ , then the value of $n + 1 \prod r = 2 Δ_{r} =$

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Solution

Given , $Δ_{r} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} (r - 1) & n! & 6 (r - 1)^{2} & (n!)^{2} & 4 n - 2 (r - 1)^{3} & (n!)^{3} & 3 n^{2} - 2 n \end{matrix} ∣ ∣ ∣ ∣ ∣$

$∴ n + 1 \prod r = 2 Δ_{r} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} \prod_{r = 2}^{n + 1} (r - 1) & n! & 6 \prod_{r = 2}^{n + 1} (r - 1)^{2} & (n!)^{2} & 4 n - 2 \prod_{r = 2}^{n + 1} (r - 1)^{3} & (n!)^{3} & 3 n^{2} - 2 n \end{matrix} ∣ ∣ ∣ ∣ ∣$

$= ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1.2.3...... n & (n!) & 6 1^{2} . 2^{2} . 3^{2} . . . . . n^{2} & (n!)^{2} & 4 n - 2 1^{2} . 2^{3} . 3^{3} . . . . . n^{3} & (n!)^{3} & 3 n^{2} - 2 n \end{matrix} ∣ ∣ ∣ ∣ ∣$

$= ∣ ∣ ∣ ∣ ∣ \begin{matrix} (n!) & (n!) & 6 (n!)^{2} & (n!)^{2} & 4 n - 2 (n!)^{3} & (n!)^{3} & 3 n^{2} - 2 n \end{matrix} ∣ ∣ ∣ ∣ ∣$
$= 0$ (since $C_{1}$ and $C_{2}$ are identical)

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Similar questions

Q. If

Δ_{r} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} r & 2 r - 1 & 3 r - 2 \frac{n}{2} & n - 1 & a \frac{1}{2} n (n - 1) & (n - 1)^{2} & \frac{1}{2} (n - 1) (3 n - 4) \end{matrix} ∣ ∣ ∣ ∣ ∣

, then the value of

n - 1 \sum r = 1 Δ_{r}

Q. If

Δ_{r} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 2^{n} + 1 & 2^{n + 1} & n (n + 2) 1 & 1 & - 1^{n} C_{r} & 2^{r} & 2 r + 1 \end{matrix} ∣ ∣ ∣ ∣ ∣

then the value of

n \sum r = 0 Δ_{r}

Q. Let

Δ_{r} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} r - 1 & n & 6 (r - 1)^{2} & 2 n^{2} & 4 n - 2 (r - 1)^{3} & 3 n^{3} & 3 n^{2} - 3 n \end{matrix} ∣ ∣ ∣ ∣ ∣

. Then the value of

n \sum r = 1 Δ_{r}

is:

$If D_{r} = ⎛ ⎜ ⎝ \begin{matrix} 2^{r - 1} & 2 . 3^{r - 1} & 4 . 5^{r - 1} α & β & γ 2^{n} - 1 & 3^{n} - 1 & 5^{n} - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣, then the value of \sum_{r = 1}^{n} D_{r} is$

Q. Let

f (n), g (n), h (n)

be polynomials in

n

, and let

Δ_{r} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 2 r + 1 & 6 n (n + 2) & f (n) 2^{r - 1} & {3.2}^{n + 1} - 6 & g (n) r (n - r + 1) & n (n + 1) (n + 2) & h (n) \end{matrix} ∣ ∣ ∣ ∣ ∣

Find the value of

n \sum r = 1 Δ_{r}

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Let Δr=∣∣ ∣ ∣∣(r−1)n!6(r−1)2(n!)24n−2(r−1)3(n!)33n2−2n∣∣ ∣ ∣∣, then the value of n+1∏r=2Δr=

Let $Δ_{r} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} (r - 1) & n! & 6 (r - 1)^{2} & (n!)^{2} & 4 n - 2 (r - 1)^{3} & (n!)^{3} & 3 n^{2} - 2 n \end{matrix} ∣ ∣ ∣ ∣ ∣$ , then the value of $n + 1 \prod r = 2 Δ_{r} =$