Question

Let A = [a_ij] be a square matrix of order 3 such that a_ij = 2^{j – i}, for all i, j = 1, 2, 3. Then, the matrix A² + A³ + ... + A¹⁰ is equal to ______.

Options

• `((3^10 - 3)/2)A`

• `((3^10 - 1)/2)A`

• `((3^10 + 1)/2)A`

• `((3^10 + 3)/2)A`

Answer 1

Let A = [a_ij] be a square matrix of order 3 such that a_ij = 2^{j – i}, for all i, j = 1, 2, 3. Then, the matrix A² + A³ + ... + A¹⁰ is equal to `underlinebb(((3^10 - 3)/2)A)`.

Explanation:

Given: a_ij = 2^{j – i}

⇒ A = `[(2^(1 - 1), 2^(2 - 1), 2^(3 - 1)),(2^(1 - 2), 2^(2 - 2), 2^(3 - 2)), (2^(1 - 3), 2^(2 - 3), 2^(3 - 3))]`

⇒ A = `[(1, 2, 4),(1/2, 1, 2),(1/4, 1/2, 1)]`

Now, A² = `[(1, 2, 4),(1/2, 1, 2),(1/4, 1/2, 1)][(1, 2, 4),(1/2, 1, 2),(1/4, 1/2, 1)]`

⇒ A² = `[(1 + 1 + 1, 2 + 2 + 2, 4 + 4 + 4),(1/2 + 1/2 + 1/2, 1 + 1 + 1 ,2 + 2 + 2),(1/4 + 1/4 + 1/4, 1/2 + 1/2 + 1/2, 1 + 1 + 1)]`

⇒ A² = `[(3, 6, 12),(3/2, 3, 6),(3/4, 3/2, 3)]`

⇒ A² = `[(1, 2, 4),(1/2, 1, 2),(1/4, 1/2, 1)]`

⇒ A² = 3A

Also, A³ = A².A = 3A.A = 3A² = 3(3A) = 3²A

Similarly A⁴ = A³.A = 3²A.A = 3²(A²) = 3²(3A) = 3²A

⇒ A² + A³ + A⁴ + ...... + A⁹ + A¹⁰ = 3A + 3²A + 3³A + ...... + 3⁸A + 3⁹A

= 3A(1 + 3 + 3² + ...... + 3⁷ + 3⁸)

= `3A(3^(9-1)/(3 - 1))`

= `3A(3^(9-1)/2)`

= `((3^10 - 3)/2)A`