If Ak- k k-1- k-1 k - - then -A1-A2- -A2021- is equal to

Given, nbsp;A_k=[ k amp;k 1; k 1 amp;k ] nbsp; |A_1|= 1 amp;0 0 amp;1 nbsp; =1^2 0^2 |A_2|= 2 amp;1 1 amp;2 nbsp; =2^2 1^2 ...

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Standard XII

Mathematics

Method of Intervals

If Ak=[ k k-...

Question

If $A_{k} = [\begin{matrix} k & k - 1 k - 1 & k \end{matrix}],$ then $| A_{1} | + | A_{2} | + \dots + | A_{2021} |$ is equal to

0

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2020

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(2021)^{2}

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(2020)^{3}

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Solution

The correct option is C $(2021)^{2}$
Given, $A_{k} = [\begin{matrix} k & k - 1 k - 1 & k \end{matrix}] | A_{1} | = ∣ ∣ ∣ \begin{matrix} 1 & 0 0 & 1 \end{matrix} ∣ ∣ ∣ = 1^{2} - 0^{2} | A_{2} | = ∣ ∣ ∣ \begin{matrix} 2 & 1 1 & 2 \end{matrix} ∣ ∣ ∣ = 2^{2} - 1^{2} | A_{3} | = ∣ ∣ ∣ \begin{matrix} 3 & 2 2 & 3 \end{matrix} ∣ ∣ ∣ = 3^{2} - 2^{2} \cdot \cdot | A_{2021} | = ∣ ∣ ∣ \begin{matrix} 2021 & 2020 2020 & 2021 \end{matrix} ∣ ∣ ∣ = (2021)^{2} - (2020)^{2}$
$\Rightarrow | A_{1} | + | A_{2} | + \dots + | A_{2021} | = (2021)^{2}$

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Similar questions

Q. If

A_{k} = [\begin{matrix} k & k - 1 k - 1 & k \end{matrix}]

then

| A_{1} | + | A_{2} | + . . + | A_{2015} | = ?

Q. Let

a_{1}, a_{2}, a_{3}, . . . ., a_{11}

be real numbers satisfying

a_{1} = 15, 27 - 2 a_{2} > 0

and

a_{k} = 2 a_{k - 1} - a_{k - 2}

for

k = 3, 4, . . . . . . ., 11

\frac{a_{1}^{2} + a_{2}^{2} + . . . + a_{11}^{2}}{11} = 90

, then the value of

\frac{a_{1} + a_{2} + . . . + a_{11}}{11}

is equal to

Q.
Let

a_{1}, a_{2}, a_{3}, \dots, a_{11}

be real numbers satisfying

a_{1} = 15, 27 - 2 a_{2} > 0

and

a_{k} = 2 a_{k - 1} - a_{k - 2}

for

k = 3, 4, \dots, 11.

\frac{a_{1}^{2} + a_{2}^{2} + \dots + a_{11}^{2}}{11} = 90

, then the value of

\frac{a_{1} + a_{2} + \dots + a_{11}}{11}

is equal to

Q. Let

a_{1}, a_{2}, . . . . . . . . . a_{11}

be real numbers satisfying

a_{1} = 15, 27 - 2 a_{2} > 0

and

a_{k} = 2 a_{k - 1} - a_{k - 2}

for k=3, 4, .......11. If

\frac{a_{1}^{2} + a_{2}^{2} + . . . . . . . . a_{11}^{2}}{11} = 90

, then the value of

\frac{a_{1} + a_{2} + . . . . . . . . . . . + a_{11}}{11}

is equal to

Let $a_{1}, a_{2}, a_{3}, . . . ., a_{11}$ be real numbers satisfying $a_{1} = 15, 27 - 2 a_{2} > 0$ and $a_{k} = 2 a_{k - 1} - a_{k - 2}$ for k = 3, 4, ….11. If $\frac{a_{1}^{2} + a_{2}^{2} + a_{3}^{2} + . . . . a_{11}^{2}}{11} = 90$ , then $\frac{a_{1} + a_{2} + a_{3} + . . . . . a_{11}}{11}$ is equal to

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Method of Intervals

Standard XII Mathematics

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If Ak=[kk−1k−1k], then |A1|+|A2|+⋯+|A2021| is equal to

The correct option is C (2021)2Given, Ak=[kk−1k−1k]|A1|=∣∣∣1001∣∣∣=12−02|A2|=∣∣∣2112∣∣∣=22−12|A3|=∣∣∣3223∣∣∣=32−22⋅⋅|A2021|=∣∣∣2021202020202021∣∣∣=(2021)2−(2020)2 ⇒|A1|+|A2|+⋯+|A2021|=(2021)2

If $A_{k} = [\begin{matrix} k & k - 1 k - 1 & k \end{matrix}],$ then $| A_{1} | + | A_{2} | + \dots + | A_{2021} |$ is equal to