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Consider 3*3 matrix with every element being equal to 1. Its only non-zero eigenvalue is _________
    Correct answer is '3'. Can you explain this answer?
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    Consider 3*3 matrix with every element being equal to 1. Its only non-...
    Explanation:

    A matrix is said to have an eigenvalue if there is a scalar λ such that the determinant of the matrix minus λ times the identity matrix is zero. In other words, if A is a matrix and λ is an eigenvalue of A, then:

    det(A - λI) = 0

    where I is the identity matrix.

    For a 3x3 matrix with every element being equal to 1, we can write:

    A = [1 1 1; 1 1 1; 1 1 1]

    To find the eigenvalues of A, we need to solve the equation:

    det(A - λI) = 0

    Expanding the determinant, we get:

    (1-λ) [(1-λ)(1-λ) - 1] - (1-λ)[(1-λ)(1-λ) - 1] + (1-λ) [(1-λ)(1-λ) - 1] = 0

    Simplifying the above equation, we get:

    (1-λ)^2 (1-λ-3) = 0

    The roots of the above equation are λ = 1 and λ = 3.

    But we know that A has only one non-zero eigenvalue. Therefore, the only possible eigenvalue of A is λ = 3.

    Hence, the correct answer is 3.

    Summary:

    - A matrix has an eigenvalue if there is a scalar λ such that det(A - λI) = 0.
    - For a 3x3 matrix with every element being equal to 1, the only possible eigenvalue is 3.
    - This is because the roots of the characteristic equation are λ = 1 and λ = 3, but A has only one non-zero eigenvalue.
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