Let A be a 3 * 3 matrix with eigenvalues, 1, - 1, 0. What will be the determinant of I + A100? (Answer up to the nearest integer)Correct answer is '4'. Can you explain this answer?

Question

Answer

Understanding the Problem
To find the determinant of I + A^100, where A is a 3x3 matrix with eigenvalues 1, -1, and 0, we can leverage properties of eigenvalues.
Eigenvalues of A
- The eigenvalues of matrix A are given as 1, -1, and 0.
- When A is raised to a power, the eigenvalues of A^n become the eigenvalues of A raised to the power n.
Calculating Eigenvalues of A^100
- For A^100:
- The eigenvalue corresponding to 1 becomes 1^100 = 1.
- The eigenvalue corresponding to -1 becomes (-1)^100 = 1.
- The eigenvalue corresponding to 0 remains 0.
Thus, the eigenvalues of A^100 are 1, 1, and 0.
Eigenvalues of I + A^100
- The identity matrix I has eigenvalues of 1 for each dimension.
- Therefore, the eigenvalues of I + A^100 are calculated as follows:
- For the eigenvalue 1 from A^100: 1 + 1 = 2.
- For the eigenvalue -1 from A^100: 1 + 1 = 2.
- For the eigenvalue 0 from A^100: 1 + 0 = 1.
The eigenvalues of I + A^100 are thus 2, 2, and 1.
Determinant Calculation
- The determinant of a matrix is the product of its eigenvalues.
- Hence, the determinant of I + A^100 is calculated as:
2 * 2 * 1 = 4.
Conclusion
The determinant of I + A^100 is 4, confirming the given answer.

Answer

Eigenvalues of A are 1, - 1, 0.

Eigenvalues of A¹⁰⁰ are 1¹⁰⁰, (- 1)¹⁰⁰, (0)¹⁰⁰, i.e. 1, 1, 0.

Eigenvalues of I + A¹⁰⁰ are 1 + 1, 1 + 1, 1 + 0, i.e. 2, 2, 1.

or |I + A¹⁰⁰| = 2 * 2 * 1 = 4

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