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Let A be a 3 x 3 matrix with trace(A) = 3 and det(A) = 2. If 1 is an eigenvalue of A, then the eigenvalues of the matrix A2 — 21 are
- a)1 , 2 ( i — 1), —2 ( i + 1)
- b)— 1,2(i — 1), 2(i + 1)
- c)1 ,2 (i + 1) ,-2(i + 1)
- d)- 1,2(i - 1), —2(i + 1)
Correct answer is option 'D'. Can you explain this answer?
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Let A be a 3 x 3 matrix with trace(A) = 3 and det(A) = 2. If 1 is an e...
Let other two eigenvalues are λ1and λ2.
sum of eigenvalues = trace of matrix.
now, product of eigenvalues = detA
So, eigenvalues of A are 1,1 + i, 1 — i
eigenvalues of A2 = l2, ( l + i )2, ( l — i )2
= 1 , 2i, —2i
now, (A2 - 2I)X = A2X - 2IX
eigenvalue of (A2 — 2I) is λ2 — 2 where A are eigenvalues of A.
So, eigenvalues are — 1,2i — 2, — 2i — 2
sum of eigenvalues = trace of matrix.
now, product of eigenvalues = detA
So, eigenvalues of A are 1,1 + i, 1 — i
eigenvalues of A2 = l2, ( l + i )2, ( l — i )2
= 1 , 2i, —2i
now, (A2 - 2I)X = A2X - 2IX
eigenvalue of (A2 — 2I) is λ2 — 2 where A are eigenvalues of A.
So, eigenvalues are — 1,2i — 2, — 2i — 2
Most Upvoted Answer
Let A be a 3 x 3 matrix with trace(A) = 3 and det(A) = 2. If 1 is an e...
Let λ be an eigenvalue of A, then we have A*v = λ*v for some nonzero vector v.
Taking the determinant of both sides, we have det(A)*det(v) = det(λ*v).
Since det(v) is nonzero (because v is nonzero), we can divide both sides by det(v) to get det(A) = det(λ*v).
Using the property that det(cA) = c^n * det(A) for an n x n matrix A and a scalar c, we have det(A) = λ^n * det(v).
Since det(A) = 2 and det(v) is nonzero, we have 2 = λ^n * det(v).
Since λ = 1 is an eigenvalue, we have 2 = 1^n * det(v).
Therefore, 2 = det(v).
This means that the determinant of the eigenvector v is 2.
Now, let's consider the matrix A^2.
We have A^2*v = A*(A*v) = A*(λ*v) = λ*(A*v) = λ^2*v.
Therefore, λ^2 is an eigenvalue of A^2.
Since λ = 1 is an eigenvalue of A, we have λ^2 = 1^2 = 1 as an eigenvalue of A^2.
In addition to λ = 1, there may be other eigenvalues of A^2, but we cannot determine them with the given information.
Taking the determinant of both sides, we have det(A)*det(v) = det(λ*v).
Since det(v) is nonzero (because v is nonzero), we can divide both sides by det(v) to get det(A) = det(λ*v).
Using the property that det(cA) = c^n * det(A) for an n x n matrix A and a scalar c, we have det(A) = λ^n * det(v).
Since det(A) = 2 and det(v) is nonzero, we have 2 = λ^n * det(v).
Since λ = 1 is an eigenvalue, we have 2 = 1^n * det(v).
Therefore, 2 = det(v).
This means that the determinant of the eigenvector v is 2.
Now, let's consider the matrix A^2.
We have A^2*v = A*(A*v) = A*(λ*v) = λ*(A*v) = λ^2*v.
Therefore, λ^2 is an eigenvalue of A^2.
Since λ = 1 is an eigenvalue of A, we have λ^2 = 1^2 = 1 as an eigenvalue of A^2.
In addition to λ = 1, there may be other eigenvalues of A^2, but we cannot determine them with the given information.
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Let A be a 3 x 3 matrix with trace(A) = 3 and det(A) = 2. If 1 is an eigenvalue of A, then the eigenvalues of the matrix A2 — 21 area)1 , 2 ( i — 1), —2 ( i + 1)b)— 1,2(i — 1), 2(i + 1)c)1 ,2 (i + 1) ,-2(i + 1)d)- 1,2(i - 1), —2(i + 1)Correct answer is option 'D'. Can you explain this answer?
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Let A be a 3 x 3 matrix with trace(A) = 3 and det(A) = 2. If 1 is an eigenvalue of A, then the eigenvalues of the matrix A2 — 21 area)1 , 2 ( i — 1), —2 ( i + 1)b)— 1,2(i — 1), 2(i + 1)c)1 ,2 (i + 1) ,-2(i + 1)d)- 1,2(i - 1), —2(i + 1)Correct answer is option 'D'. Can you explain this answer? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about Let A be a 3 x 3 matrix with trace(A) = 3 and det(A) = 2. If 1 is an eigenvalue of A, then the eigenvalues of the matrix A2 — 21 area)1 , 2 ( i — 1), —2 ( i + 1)b)— 1,2(i — 1), 2(i + 1)c)1 ,2 (i + 1) ,-2(i + 1)d)- 1,2(i - 1), —2(i + 1)Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let A be a 3 x 3 matrix with trace(A) = 3 and det(A) = 2. If 1 is an eigenvalue of A, then the eigenvalues of the matrix A2 — 21 area)1 , 2 ( i — 1), —2 ( i + 1)b)— 1,2(i — 1), 2(i + 1)c)1 ,2 (i + 1) ,-2(i + 1)d)- 1,2(i - 1), —2(i + 1)Correct answer is option 'D'. Can you explain this answer?.
Let A be a 3 x 3 matrix with trace(A) = 3 and det(A) = 2. If 1 is an eigenvalue of A, then the eigenvalues of the matrix A2 — 21 area)1 , 2 ( i — 1), —2 ( i + 1)b)— 1,2(i — 1), 2(i + 1)c)1 ,2 (i + 1) ,-2(i + 1)d)- 1,2(i - 1), —2(i + 1)Correct answer is option 'D'. Can you explain this answer? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about Let A be a 3 x 3 matrix with trace(A) = 3 and det(A) = 2. If 1 is an eigenvalue of A, then the eigenvalues of the matrix A2 — 21 area)1 , 2 ( i — 1), —2 ( i + 1)b)— 1,2(i — 1), 2(i + 1)c)1 ,2 (i + 1) ,-2(i + 1)d)- 1,2(i - 1), —2(i + 1)Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let A be a 3 x 3 matrix with trace(A) = 3 and det(A) = 2. If 1 is an eigenvalue of A, then the eigenvalues of the matrix A2 — 21 area)1 , 2 ( i — 1), —2 ( i + 1)b)— 1,2(i — 1), 2(i + 1)c)1 ,2 (i + 1) ,-2(i + 1)d)- 1,2(i - 1), —2(i + 1)Correct answer is option 'D'. Can you explain this answer?.
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