If k- R 0- then det adjk I n is equal to

The correct option is B k ^ n(n 1) We know that adj A=|A|^n 1 Here, |adj (kI_n)|=|kI_n|^n 1 nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; ...

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Theorems for Differentiability

If kϵ R 0, ...

Question

If $k ϵ R_{0}$ , then $d e t {a d j (k I_{n})}$ is equal to

k^{n + 1}

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k^{n (n - 1)}

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k^{n}

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k

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Solution

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Similar questions

P = \frac{s}{{(1 + k)}^{n}}

n

p = \frac{s}{(1 + k)^{n}}

n

S_{n} = n \sum k = 1 \frac{n}{n^{2} + k n + k^{2}}

T_{n} = n - 1 \sum k = 0 \frac{n}{n^{2} + k n + k^{2}}

n = 1, 2, 3, . .

S_{n} = \sum_{k = 1}^{n} \frac{n}{n^{2} + k n + k^{2}} a n d T_{n} = \sum_{k = 0}^{n - 1} \frac{n}{n^{2} + k n + k^{2}}

a_{n} = n \sum k = 1 \frac{1}{k (n + 1 - k)}

n \geq 2

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