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Let A be a 3 × 3 matrix with eigen values 1, –1, 0. Then the determinant of I + A100 is :
    Correct answer is '4'. Can you explain this answer?
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    Let Abe a 3 × 3 matrix with eigen values 1, –1, 0. Then th...
    Eigen values of 
    ⇒ Eigen values of 
    Eigen values of = 1, 1, 0
    Eigen values of 
    ⇒  Eigen values of 
    Hence,  |A100 + I| = 2 × 2 × 1 = 4
    The correct answer is: 4
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    Let Abe a 3 × 3 matrix with eigen values 1, –1, 0. Then th...

    Explanation:

    Eigenvalues of the matrix A:
    - Given that the matrix A is a 3x3 matrix with eigenvalues 1, -1, and 0.

    Determinant of I + A^100:
    - Since the eigenvalues of A are 1, -1, and 0, the eigenvalues of A^100 will be 1^100, (-1)^100, and 0^100 which are 1, 1, and 0 respectively.
    - Now, the matrix I + A^100 will have eigenvalues 1+1, 1+1, and 1+0 which are 2, 2, and 1.
    - The determinant of a matrix is the product of its eigenvalues. Therefore, the determinant of I + A^100 will be 2*2*1 = 4.

    Therefore, the determinant of I + A^100 is 4.
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    Let Abe a 3 × 3 matrix with eigen values 1, –1, 0. Then the determinant of I + A100is :Correct answer is '4'. Can you explain this answer?
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    Let Abe a 3 × 3 matrix with eigen values 1, –1, 0. Then the determinant of I + A100is :Correct answer is '4'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about Let Abe a 3 × 3 matrix with eigen values 1, –1, 0. Then the determinant of I + A100is :Correct answer is '4'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let Abe a 3 × 3 matrix with eigen values 1, –1, 0. Then the determinant of I + A100is :Correct answer is '4'. Can you explain this answer?.
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