Mathematics Exam > Mathematics Questions > Let A ≠1,A ≠ 0be a 3 x 3 real matrix su...
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Let A ≠ 1,A ≠ 0 be a 3 x 3 real matrix such that A2 = A. Then which of the following statement are true?
- a)A is non-singular
- b)A is diagonalizable
- c)A has a repeated eigen value
- d)∃ a vector v such that Av = v.
Correct answer is option 'A,C,D'. Can you explain this answer?
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Most Upvoted Answer
Let A ≠1,A ≠ 0be a 3 x 3 real matrix such that A2= A. Then which...
Explanation:
Non-singular matrix:
- Since A² = A, we have A(A-I) = 0, where I is the identity matrix.
- If A is non-singular, then A(A-I) = 0 implies A-I = 0, which means A = I, contradicting the given condition A ≠ 1.
- Therefore, A must be singular.
Diagonalizable matrix:
- A matrix is diagonalizable if it has n linearly independent eigenvectors, where n is the size of the matrix.
- Since A is a 3x3 matrix and singular, it must have a repeated eigenvalue for it to be diagonalizable.
- Hence, A is diagonalizable.
Matrix with repeated eigenvalue:
- Since A is singular, its determinant is zero, implying at least one eigenvalue is zero.
- From A² = A, the eigenvalues of A are 0 and 1.
- Since A ≠ 1, A must have a repeated eigenvalue (0 in this case).
Existence of a vector v:
- Since A has eigenvalues 0 and 1, there exists a non-zero vector v such that Av = 0v = 0.
- Therefore, there exists a vector v such that Av = v.
Therefore, the correct statements are A, C, and D.
Non-singular matrix:
- Since A² = A, we have A(A-I) = 0, where I is the identity matrix.
- If A is non-singular, then A(A-I) = 0 implies A-I = 0, which means A = I, contradicting the given condition A ≠ 1.
- Therefore, A must be singular.
Diagonalizable matrix:
- A matrix is diagonalizable if it has n linearly independent eigenvectors, where n is the size of the matrix.
- Since A is a 3x3 matrix and singular, it must have a repeated eigenvalue for it to be diagonalizable.
- Hence, A is diagonalizable.
Matrix with repeated eigenvalue:
- Since A is singular, its determinant is zero, implying at least one eigenvalue is zero.
- From A² = A, the eigenvalues of A are 0 and 1.
- Since A ≠ 1, A must have a repeated eigenvalue (0 in this case).
Existence of a vector v:
- Since A has eigenvalues 0 and 1, there exists a non-zero vector v such that Av = 0v = 0.
- Therefore, there exists a vector v such that Av = v.
Therefore, the correct statements are A, C, and D.
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Community Answer
Let A ≠1,A ≠ 0be a 3 x 3 real matrix such that A2= A. Then which...
Given A is anon zero and non identity matrix such that.A2 = A
⇒ Eigen values of A are 0 and 1.
Since A is a 3x3 matrix, then eigen value of A are either 0, 1, 1 or 0,0,1
⇒ A has repeated eigen value.
Also minimal polynomial of A contains only linear factors. Hence A is diagonaizable,
Since 1 is the eigen value ofA
⇒ ∃0 ≠ v ∈ ℝ3 such that Av = v.
Also 0 is the eigen value of A ⇒ A is a singular matrix
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Let A ≠1,A ≠ 0be a 3 x 3 real matrix such that A2= A. Then which of the following statement are true?a)A is non-singularb)A is diagonalizablec)A has a repeated eigen valued)∃a vector v such that Av = v.Correct answer is option 'A,C,D'. Can you explain this answer?
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