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Let the eigenvalues of a 2×2 matrix A be 1, -2 with eigenvectors x1 and x2 repsectively. Then the eigenvalues and eigenvectors of the matrix A2-3A+4I would respectively, be
- a)2, 14; x1, x2
- b)2, 14; x1+x2: x1-x2
- c)2, 0; x1, x2
- d)2, 0; x1+x2: x1-x2
Correct answer is option 'A'. Can you explain this answer?
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Let the eigenvalues of a 2×2matrix A be 1, -2 with eigenvectors ...
X2 matrix A be λ1 and λ2, and let v1 and v2 be the corresponding eigenvectors. Then, we can write the matrix A as:
A = PDP^-1
where P is a matrix whose columns are the eigenvectors v1 and v2, and D is a diagonal matrix with the eigenvalues λ1 and λ2 on the diagonal.
In general, the eigenvalues of a 2x2 matrix can be found by solving the characteristic equation:
det(A - λI) = 0
where A is the given matrix, λ is the eigenvalue, and I is the identity matrix of the same size as A. The solutions to this equation are the eigenvalues of A. Once the eigenvalues are found, we can find the corresponding eigenvectors by solving the equation:
(A - λI)v = 0
where v is the eigenvector corresponding to the eigenvalue λ.
A = PDP^-1
where P is a matrix whose columns are the eigenvectors v1 and v2, and D is a diagonal matrix with the eigenvalues λ1 and λ2 on the diagonal.
In general, the eigenvalues of a 2x2 matrix can be found by solving the characteristic equation:
det(A - λI) = 0
where A is the given matrix, λ is the eigenvalue, and I is the identity matrix of the same size as A. The solutions to this equation are the eigenvalues of A. Once the eigenvalues are found, we can find the corresponding eigenvectors by solving the equation:
(A - λI)v = 0
where v is the eigenvector corresponding to the eigenvalue λ.
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Let the eigenvalues of a 2×2matrix A be 1, -2 with eigenvectors x1 and x2 repsectively. Then the eigenvalues and eigenvectors of the matrix A2-3A+4I would respectively, bea)2, 14; x1, x2b)2, 14; x1+x2: x1-x2c)2, 0; x1, x2d)2, 0; x1+x2: x1-x2Correct answer is option 'A'. Can you explain this answer?
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Let the eigenvalues of a 2×2matrix A be 1, -2 with eigenvectors x1 and x2 repsectively. Then the eigenvalues and eigenvectors of the matrix A2-3A+4I would respectively, bea)2, 14; x1, x2b)2, 14; x1+x2: x1-x2c)2, 0; x1, x2d)2, 0; x1+x2: x1-x2Correct answer is option 'A'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Let the eigenvalues of a 2×2matrix A be 1, -2 with eigenvectors x1 and x2 repsectively. Then the eigenvalues and eigenvectors of the matrix A2-3A+4I would respectively, bea)2, 14; x1, x2b)2, 14; x1+x2: x1-x2c)2, 0; x1, x2d)2, 0; x1+x2: x1-x2Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let the eigenvalues of a 2×2matrix A be 1, -2 with eigenvectors x1 and x2 repsectively. Then the eigenvalues and eigenvectors of the matrix A2-3A+4I would respectively, bea)2, 14; x1, x2b)2, 14; x1+x2: x1-x2c)2, 0; x1, x2d)2, 0; x1+x2: x1-x2Correct answer is option 'A'. Can you explain this answer?.
Let the eigenvalues of a 2×2matrix A be 1, -2 with eigenvectors x1 and x2 repsectively. Then the eigenvalues and eigenvectors of the matrix A2-3A+4I would respectively, bea)2, 14; x1, x2b)2, 14; x1+x2: x1-x2c)2, 0; x1, x2d)2, 0; x1+x2: x1-x2Correct answer is option 'A'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Let the eigenvalues of a 2×2matrix A be 1, -2 with eigenvectors x1 and x2 repsectively. Then the eigenvalues and eigenvectors of the matrix A2-3A+4I would respectively, bea)2, 14; x1, x2b)2, 14; x1+x2: x1-x2c)2, 0; x1, x2d)2, 0; x1+x2: x1-x2Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let the eigenvalues of a 2×2matrix A be 1, -2 with eigenvectors x1 and x2 repsectively. Then the eigenvalues and eigenvectors of the matrix A2-3A+4I would respectively, bea)2, 14; x1, x2b)2, 14; x1+x2: x1-x2c)2, 0; x1, x2d)2, 0; x1+x2: x1-x2Correct answer is option 'A'. Can you explain this answer?.
Solutions for Let the eigenvalues of a 2×2matrix A be 1, -2 with eigenvectors x1 and x2 repsectively. Then the eigenvalues and eigenvectors of the matrix A2-3A+4I would respectively, bea)2, 14; x1, x2b)2, 14; x1+x2: x1-x2c)2, 0; x1, x2d)2, 0; x1+x2: x1-x2Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
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Let the eigenvalues of a 2×2matrix A be 1, -2 with eigenvectors x1 and x2 repsectively. Then the eigenvalues and eigenvectors of the matrix A2-3A+4I would respectively, bea)2, 14; x1, x2b)2, 14; x1+x2: x1-x2c)2, 0; x1, x2d)2, 0; x1+x2: x1-x2Correct answer is option 'A'. Can you explain this answer?, a detailed solution for Let the eigenvalues of a 2×2matrix A be 1, -2 with eigenvectors x1 and x2 repsectively. Then the eigenvalues and eigenvectors of the matrix A2-3A+4I would respectively, bea)2, 14; x1, x2b)2, 14; x1+x2: x1-x2c)2, 0; x1, x2d)2, 0; x1+x2: x1-x2Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of Let the eigenvalues of a 2×2matrix A be 1, -2 with eigenvectors x1 and x2 repsectively. Then the eigenvalues and eigenvectors of the matrix A2-3A+4I would respectively, bea)2, 14; x1, x2b)2, 14; x1+x2: x1-x2c)2, 0; x1, x2d)2, 0; x1+x2: x1-x2Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Let the eigenvalues of a 2×2matrix A be 1, -2 with eigenvectors x1 and x2 repsectively. Then the eigenvalues and eigenvectors of the matrix A2-3A+4I would respectively, bea)2, 14; x1, x2b)2, 14; x1+x2: x1-x2c)2, 0; x1, x2d)2, 0; x1+x2: x1-x2Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice GATE tests.
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