Let A = <math><mrow><mrow><mo>[</mo><mrow><msub><mrow><mi>a</mi></mrow><mrow><mi>i</mi><mi>j</mi></mrow></msub></mrow><mo>]</mo></mrow></mrow></math> be a square matrix of order 3 such that aij = 2ji, for all I, j = 1, 2, 3. Then, the matrix <math><mrow><msup><mrow><mi>A</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>+</mo><msup><mrow><mi>A</mi></mrow><mrow><mn>3</mn></mrow></msup><mo>+</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>+</mo><msup><mrow><mi>A</mi></mrow><mrow><mn>1</mn><mn>0</mn></mrow></msup></mrow></math> is equal to:

A=[124121214121]A2= [124121214121][124121214121]=3[124121214121]A2 = 3A A3 = 3A2A3 = 32AA4 = 33AAn = 3n-1Anow, A2 + A3 +….+A10= 3A + 32 A +…. + 39A= 3A (1 + 3 +….+ 38=3A⋅(39−1)3−1=310−32A

Let A = <math><mrow><mrow><mo>[</mo><mrow><msub><mrow><mi>a</mi></mrow><mrow><mi>i</mi><mi>j</mi></mrow></msub></mrow><mo>]</mo></mrow></mrow></math>&nbsp;be a square matrix of order 3 such that aij = 2ji, for all I, j = 1, 2, 3. Then, the matrix&nbsp;<math><mrow><msup><mrow><mi>A</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>+</mo><msup><mrow><mi>A</mi></mrow><mrow><mn>3</mn></mrow></msup><mo>+</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>+</mo><msup><mrow><mi>A</mi></mrow><mrow><mn>1</mn><mn>0</mn></mrow></msup></mrow></math>&nbsp;is equal to:

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Maths Class 12th Maths Matrices Maths NCERT Exemplar Solutions Class 12th Chapter Three

Let A = $[a_{i j}]$ be a square matrix of order 3 such that aij = 2ji, for all I, j = 1, 2, 3. Then, the matrix $A^{2} + A^{3} + . . . + A^{10}$ is equal to:

Option 1 -

$(\frac{3^{10} - 3}{2}) A$

Option 2 -

$(\frac{3^{10} - 1}{2}) A$

Option 3 -

$(\frac{3^{10} + 1}{2}) A$

Option 4 -

$(\frac{3^{10} + 3}{2}) A$

2 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 1

Detailed Solution:

$A = [\begin{matrix} 1 & 2 & 4 \\ \frac{1}{2} & 1 & 2 \\ \frac{1}{4} & \frac{1}{2} & 1 \end{matrix}]$

$A^{2} =$ $[\begin{matrix} 1 & 2 & 4 \\ \frac{1}{2} & 1 & 2 \\ \frac{1}{4} & \frac{1}{2} & 1 \end{matrix}] [\begin{matrix} 1 & 2 & 4 \\ \frac{1}{2} & 1 & 2 \\ \frac{1}{4} & \frac{1}{2} & 1 \end{matrix}]$

$= 3 [\begin{matrix} 1 & 2 & 4 \\ \frac{1}{2} & 1 & 2 \\ \frac{1}{4} & \frac{1}{2} & 1 \end{matrix}]$

A² = 3A

A³ = 3A²

A³ = 3²A

A⁴ = 3³A

Aⁿ = 3^n-1A

now, A² + A³ +….+A¹⁰

= 3A + 3² A +…. + 3⁹A

= 3A (1 + 3 +….+ 3⁸

$= 3 A \cdot \frac{(3^{9} - 1)}{3 - 1}$

$= \frac{3^{10} - 3}{2} A$

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Let A = $[a_{i j}]$ be a square matrix of order 3 such that aij = 2ji, for all I, j = 1, 2, 3. Then, the matrix $A^{2} + A^{3} + . . . + A^{10}$ is equal to:

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Let A = [aij] be a square matrix of order 3 such that aij = 2ji, for all I, j = 1, 2, 3. Then, the matrix A2+A3+...+A10 is equal to:

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Let A = $[a_{i j}]$ be a square matrix of order 3 such that aij = 2ji, for all I, j = 1, 2, 3. Then, the matrix $A^{2} + A^{3} + . . . + A^{10}$ is equal to: