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In an M × N matrix such that all non-zero entries are covered in a rows and b column. Then the maximum number of non-zero entries, such that no two are on the same row or column, is 
  • a)
    ≤ a + b    
  • b)
    ≤ max (a, b)  
  • c)
    ≤ min[ M –a, N–b]    
  • d)
    ≤ min  {a, b} 
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
In an M × N matrix such that all non-zero entries are covered in ...
Suppose a < b, for example let a = 3, b= 5, then we can put non-zero entries only in 3 rows and 5 columns. So suppose we put non-zero entries in any 3 rows in 3 different columns. Now we can’t put any other non-zero entry anywhere in matrix, because if we put it in some other row, then we will have 4 rows containing non-zeros, if we put it in one of those 3 rows, then we will have more than one non-zero entry in one row, which is not allowed.
So we can fill only “a” non-zero entries if a < b, similarly if b < a, we can put only “b” non-zero entries. So answer is ≤min(a,b), because whatever is less between a and b, we can put atmost that many non-zero entries.
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Most Upvoted Answer
In an M × N matrix such that all non-zero entries are covered in ...
Suppose a < b, for example let a = 3, b= 5, then we can put non-zero entries only in 3 rows and 5 columns. So suppose we put non-zero entries in any 3 rows in 3 different columns. Now we can’t put any other non-zero entry anywhere in matrix, because if we put it in some other row, then we will have 4 rows containing non-zeros, if we put it in one of those 3 rows, then we will have more than one non-zero entry in one row, which is not allowed.
So we can fill only “a” non-zero entries if a < b, similarly if b < a, we can put only “b” non-zero entries. So answer is ≤min(a,b), because whatever is less between a and b, we can put atmost that many non-zero entries.
So option (D) is correct.
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Community Answer
In an M × N matrix such that all non-zero entries are covered in ...
Suppose a < b, for example let a = 3, b= 5, then we can put non-zero entries only in 3 rows and 5 columns. So suppose we put non-zero entries in any 3 rows in 3 different columns. Now we can’t put any other non-zero entry anywhere in matrix, because if we put it in some other row, then we will have 4 rows containing non-zeros, if we put it in one of those 3 rows, then we will have more than one non-zero entry in one row, which is not allowed.
So we can fill only “a” non-zero entries if a < b, similarly if b < a, we can put only “b” non-zero entries. So answer is ≤min(a,b), because whatever is less between a and b, we can put atmost that many non-zero entries.
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In an M × N matrix such that all non-zero entries are covered in a rows and b column. Then the maximum number of non-zero entries, such that no two are on the same row or column, isa)≤ a + b b)≤ max (a, b) c)≤ min[ M –a, N–b] d)≤ min {a, b}Correct answer is option 'D'. Can you explain this answer?
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In an M × N matrix such that all non-zero entries are covered in a rows and b column. Then the maximum number of non-zero entries, such that no two are on the same row or column, isa)≤ a + b b)≤ max (a, b) c)≤ min[ M –a, N–b] d)≤ min {a, b}Correct answer is option 'D'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about In an M × N matrix such that all non-zero entries are covered in a rows and b column. Then the maximum number of non-zero entries, such that no two are on the same row or column, isa)≤ a + b b)≤ max (a, b) c)≤ min[ M –a, N–b] d)≤ min {a, b}Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In an M × N matrix such that all non-zero entries are covered in a rows and b column. Then the maximum number of non-zero entries, such that no two are on the same row or column, isa)≤ a + b b)≤ max (a, b) c)≤ min[ M –a, N–b] d)≤ min {a, b}Correct answer is option 'D'. Can you explain this answer?.
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