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Let A = (aij) be a 10 × 10 matrix such that aij = 1 for i ≠ j and aij = α + 1, where α > 0. Let λ and μ be the largest and the smallest eigenvalues of A, respectively. If λ + μ = 24, then α equals ______.
Correct answer is '7'. Can you explain this answer?
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Let A = (aij) be a 10 × 10 matrix such that aij= 1 for i ≠ j ...
A = (aij) be a 10 × 10 matrix
aij = 1 for i ≠ j
aij = α + 1
Let α + 1 = a
Now, the matrix can be formed as
We know that, eigen values are the roots of |A - λI| = 0
|A – λI| = 0
⇒ C1 → C1 + C2 + C3 + … + C10
R2 → R2 - R1
R3 → R3 - R1
⋮
R10 → R10 - R1
C2 → C2 - C1
C3 → C3 - C1
⋮
C10 → C10 - C1
Where I10 is the identify matrix of order 10.
We know that |I| = 1.
⇒ λ = a - 1, λ = a + 9
⇒ λ = α + 1 - 1, λ = α + 1 + 9
⇒ λ = α, λ = α + 10.
The possible Eigen values are: α, α + 10
Given that, smallest Eigen value = λ = α
Largest Eigen value = μ = ∝ + 10.
Given that, λ + μ = 24
⇒ α + α + 10 = 24
⇒ α = 7
Most Upvoted Answer
Let A = (aij) be a 10 × 10 matrix such that aij= 1 for i ≠ j ...
10 matrix with aij = 1 if i+j is even and aij = 0 if i+j is odd.
To prove that A is a symmetric matrix, we need to show that for all i and j, aij = aji.
Let's consider two cases:
Case 1: i+j is even.
In this case, aij = 1 and aji = 1 because i+j and j+i are both even. Therefore, aij = aji.
Case 2: i+j is odd.
In this case, aij = 0 and aji = 0 because i+j and j+i are both odd. Therefore, aij = aji.
Since aij = aji for all i and j, we conclude that A is a symmetric matrix.
To prove that A is a symmetric matrix, we need to show that for all i and j, aij = aji.
Let's consider two cases:
Case 1: i+j is even.
In this case, aij = 1 and aji = 1 because i+j and j+i are both even. Therefore, aij = aji.
Case 2: i+j is odd.
In this case, aij = 0 and aji = 0 because i+j and j+i are both odd. Therefore, aij = aji.
Since aij = aji for all i and j, we conclude that A is a symmetric matrix.
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Let A = (aij) be a 10 × 10 matrix such that aij= 1 for i ≠ j and aij= α + 1, where α > 0. Let λ and μ be the largest and the smallest eigenvalues of A, respectively. If λ + μ = 24, then α equals ______.Correct answer is '7'. Can you explain this answer?
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Let A = (aij) be a 10 × 10 matrix such that aij= 1 for i ≠ j and aij= α + 1, where α > 0. Let λ and μ be the largest and the smallest eigenvalues of A, respectively. If λ + μ = 24, then α equals ______.Correct answer is '7'. Can you explain this answer? for Civil Engineering (CE) 2025 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about Let A = (aij) be a 10 × 10 matrix such that aij= 1 for i ≠ j and aij= α + 1, where α > 0. Let λ and μ be the largest and the smallest eigenvalues of A, respectively. If λ + μ = 24, then α equals ______.Correct answer is '7'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let A = (aij) be a 10 × 10 matrix such that aij= 1 for i ≠ j and aij= α + 1, where α > 0. Let λ and μ be the largest and the smallest eigenvalues of A, respectively. If λ + μ = 24, then α equals ______.Correct answer is '7'. Can you explain this answer?.
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