​$$\text{Let } A \text{ be an } n \times n \text{ matrix of rank 1. Let } \alpha = \det(I + A),\\ \text{ where } I \text{ is the identity matrix and let } \beta = \text{trace}(A).\\ \text{ Which of the following is true?}$$​​

Correct option is B

Let A is a n x n matrix with rank 1. this means A is a singular matrix with at least one eigenvalue as 0.

now we know that  $$A.M(\lambda)\geq G.M(\lambda) \ \{ Where,\ \lambda \ \text{is an eigen value of A} \}$$

A.M = Algebraic multiplicity .

G.M = Geometric multiplicity . 

So,

​$$A.M(0)\geq G.M(0)\\[10pt]\geq n- \rho(A-0.I)\\[10pt]\geq n-1\\[10pt]\implies \ atleast\ (n-1)\ eigen-values\ of\ A\ are \ 0.\\[10pt]\text{Now, let }\ \lambda_1,\lambda_2, \cdots \lambda_n \ \text{are eigenvalues of A .}\\[10pt]\text{As, atleast n-1 eigen values will be 0 let,}\\[10pt]\lambda_2=\lambda_3=\cdots=\lambda_n=0.\\[10pt]Trace(A)=\ \text{Sum of all eigen values.}\\[10pt]\implies \beta = \lambda_1+\lambda_2+\lambda_3+\cdots+\lambda_n\\[10pt]\implies \beta=\lambda_1\\[10pt]\text{Eigenvalues of } I + A \text{ will be:} \\[10pt](1 + \beta), 1, 1, \dots, 1 \quad (\text{n-1 times}). \\[10pt]\det(I + A) = \text{Product of all eigenvalues.} \\[10pt]\Rightarrow \det(I + A) = (1 + \beta) \cdot 1 \cdot 1 \cdot \dots \cdot 1 \\[15pt]\Rightarrow \alpha = 1 + \beta. \\\alpha - \beta = 1. \quad \text{(Answer)}$$ 

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