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Let u and v belong to the set of real numbers R^n. When the matrix 1 - uv^T is invertible, its inverse is represented as I + (1/k)uv^T. What is the value of k?

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Final answer:

The value of k is the dot product of vectors u and v (notated as v^Tu), which is a scalar. When the matrix (1 - uv^T) is inverted, the inverse is represented as I + (1/k)uv^T, and k is determined by solving the equation AA^{-1} = I.

Step-by-step explanation:

To find the value of k when the matrix 1 - uvT is invertible, and its inverse is given by I + (1/k)uvT, we can use the fact that a matrix multiplied by its inverse yields the identity matrix I. Let's denote the given matrix by A and its inverse by A-1. The relevant equations are:

A = 1 - uvT

A-1 = I + (1/k)uvT

To fulfill the condition AA-1 = I, we perform the multiplication and set it to the identity matrix:

(1 - uvT)(I + (1/k)uvT) = I

Let's carry out this matrix multiplication:

1 ∙ I + (1/k) ∙ uvT - uvT ∙ I - (1/k) ∙ uvTuvT = I

Simplifying and using the property that any matrix multiplied by the identity matrix I is itself, we get:

I + (1/k)uvT - uvT - (1/k)(vTu)uvT = I

The term uvT and -uvT cancel each other out and we are left with:

I - (1/k)(vTu)uvT = I

For the left side to equal the identity matrix, the matrix uvT must vanish. This implies:

(vTu)uvT = k ∙ I

Since vTu is a scalar, and the product with uvT results in a matrix proportional to the original uvT, we can infer that k equals vTu. Therefore, the value of k, in this case, is the dot product of the vectors v and u.

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