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Lat A be a 3×3 matrix with ineger entries such that det (A) =1 . What is the maximum possible number of entries of A that are even.?
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Lat A be a 3×3 matrix with ineger entries such that det (A) =1 . What ...
Introduction:
In this problem, we are given a 3x3 matrix A with integer entries and we need to find the maximum possible number of entries of A that are even, given that det(A) = 1.
Solution:
To find the maximum number of even entries in A, we need to consider the following cases:
Case 1: No even entries in A
If there are no even entries in A, then all the entries must be odd. In this case, det(A) will also be odd, which contradicts the given condition det(A) = 1. Therefore, this case is not possible.
Case 2: One even entry in A
If there is only one even entry in A, then there are three possible even entries to choose from. Let's assume that the first entry is even. Then, the other two entries must be odd. Since det(A) = 1, we have:
det(A) = a11(a22a33 - a23a32) - a12(a21a33 - a23a31) + a13(a21a32 - a22a31) = 1
Substituting the values of a11, a22, a33, a23, a32, and a13, we get:
a12(a21a33 - a23a31) - a13(a21a32 - a22a31) = 0
Since a12 and a13 are odd, either a21a33 = a23a31 or a21a32 = a22a31. In both cases, we can see that at least one of the remaining entries must be even. Therefore, this case is not possible either.
Case 3: Two even entries in A
If there are two even entries in A, then there are three possible pairs of even entries to choose from. Let's assume that a11 and a12 are even. Then, the remaining entry must be odd. Since det(A) = 1, we have:
det(A) = a11(a22a33 - a23a32) - a12(a21a33 - a23a31) + a13(a21a32 - a22a31) = 1
Substituting the values of a11, a12, a22, a33, a23, a32, and a13, we get:
2a13(a21a32 - a22a31) = 1 - 2a11a22a33 + 2a11a23a32 + 2a12a21a33 - 2a12a23a31
Since the left-hand side is even, the right-hand side must also be even. Therefore, either a11a22a33 is odd and one of the remaining entries is even, or a11a23a32 and a12a21a33 are odd and the remaining entry is even. In both cases, we can have at most two even entries in A.
Case 4: Three even entries in A
If there are three even entries in A, then all the entries must be even. In this case, det(A) will be even, which contradicts the given condition det(A) = 1. Therefore, this case is not possible
In this problem, we are given a 3x3 matrix A with integer entries and we need to find the maximum possible number of entries of A that are even, given that det(A) = 1.
Solution:
To find the maximum number of even entries in A, we need to consider the following cases:
Case 1: No even entries in A
If there are no even entries in A, then all the entries must be odd. In this case, det(A) will also be odd, which contradicts the given condition det(A) = 1. Therefore, this case is not possible.
Case 2: One even entry in A
If there is only one even entry in A, then there are three possible even entries to choose from. Let's assume that the first entry is even. Then, the other two entries must be odd. Since det(A) = 1, we have:
det(A) = a11(a22a33 - a23a32) - a12(a21a33 - a23a31) + a13(a21a32 - a22a31) = 1
Substituting the values of a11, a22, a33, a23, a32, and a13, we get:
a12(a21a33 - a23a31) - a13(a21a32 - a22a31) = 0
Since a12 and a13 are odd, either a21a33 = a23a31 or a21a32 = a22a31. In both cases, we can see that at least one of the remaining entries must be even. Therefore, this case is not possible either.
Case 3: Two even entries in A
If there are two even entries in A, then there are three possible pairs of even entries to choose from. Let's assume that a11 and a12 are even. Then, the remaining entry must be odd. Since det(A) = 1, we have:
det(A) = a11(a22a33 - a23a32) - a12(a21a33 - a23a31) + a13(a21a32 - a22a31) = 1
Substituting the values of a11, a12, a22, a33, a23, a32, and a13, we get:
2a13(a21a32 - a22a31) = 1 - 2a11a22a33 + 2a11a23a32 + 2a12a21a33 - 2a12a23a31
Since the left-hand side is even, the right-hand side must also be even. Therefore, either a11a22a33 is odd and one of the remaining entries is even, or a11a23a32 and a12a21a33 are odd and the remaining entry is even. In both cases, we can have at most two even entries in A.
Case 4: Three even entries in A
If there are three even entries in A, then all the entries must be even. In this case, det(A) will be even, which contradicts the given condition det(A) = 1. Therefore, this case is not possible
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Lat A be a 3×3 matrix with ineger entries such that det (A) =1 . What is the maximum possible number of entries of A that are even.?
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Lat A be a 3×3 matrix with ineger entries such that det (A) =1 . What is the maximum possible number of entries of A that are even.? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about Lat A be a 3×3 matrix with ineger entries such that det (A) =1 . What is the maximum possible number of entries of A that are even.? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Lat A be a 3×3 matrix with ineger entries such that det (A) =1 . What is the maximum possible number of entries of A that are even.?.
Lat A be a 3×3 matrix with ineger entries such that det (A) =1 . What is the maximum possible number of entries of A that are even.? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about Lat A be a 3×3 matrix with ineger entries such that det (A) =1 . What is the maximum possible number of entries of A that are even.? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Lat A be a 3×3 matrix with ineger entries such that det (A) =1 . What is the maximum possible number of entries of A that are even.?.
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