
Consider the set A of all determinants of order \[3\] with entries \[0\] or \[1\] only. Let B be the subset of A consisting of all determinants with value \[1\] .Let C be the subset of A consisting of all determinants with value \[ - 1\] .Then
\[\left( 1 \right)\] C is empty
\[\left( 2 \right)\] B has as many elements as C
\[\left( 3 \right)\] \[A = B \cup C\]
\[\left( 4 \right)\] B has twice as many elements as C
Answer
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Hint: Determinant of order three is the determinant of a \[3 \times 3\] matrix. By using the concept that on interchanging the rows and the columns the value of the determinant changes in terms of sign but not in magnitude we can answer the above question. And in the end we get that elements in set B have equal number of elements as in set C.
Complete step-by-step solution:
It is given that A is the determinant of order \[3\] with entries \[0\] or \[1\] only. Also it is given that \[B \subseteq A\] consists of all determinants with value one. We know that by interchanging the rows and the columns, sign changes but the magnitude will be constant. So, for every element in the determinant of B there is an element in determinant of C. That is if any determinants have value \[1\] so we can change the sign of \[1\] elements and by this we can get the value of determinant as \[ - 1\] by changing the sign of \[1\] entry of that determinant. So now we have set A as
\[A = \left\{ {{d_1},{d_2},{d_3},{d_4},...........,{d_n}} \right\}\]
where \[{d_1},{d_2},{d_3},{d_4},...........,{d_n}\] are the determinants of order \[3\]
As it is given in the question that B is the subset of A consisting of all determinants with value \[1\] .Therefore, \[B = 1\]
Also it is given that C is the subset of A consisting of all determinants with value \[ - 1\] .Therefore, \[C = - 1\]
If we change the sign of one entry in set B then it will be \[B = - 1\] . So the determinant value of B is \[ - 1\] .
Therefore we can say that both B and C subsets have equal number of elements.
Hence the correct option is \[\left( 2 \right)\] B has as many elements as C .
Note: Note that if any two rows or columns of a determinant are interchanged then the value of determinant is multiplied by \[ - 1\] . Note that the determinant is a real number, it is not a matrix. That is, a determinant is a single value that represents a square matrix.
Complete step-by-step solution:
It is given that A is the determinant of order \[3\] with entries \[0\] or \[1\] only. Also it is given that \[B \subseteq A\] consists of all determinants with value one. We know that by interchanging the rows and the columns, sign changes but the magnitude will be constant. So, for every element in the determinant of B there is an element in determinant of C. That is if any determinants have value \[1\] so we can change the sign of \[1\] elements and by this we can get the value of determinant as \[ - 1\] by changing the sign of \[1\] entry of that determinant. So now we have set A as
\[A = \left\{ {{d_1},{d_2},{d_3},{d_4},...........,{d_n}} \right\}\]
where \[{d_1},{d_2},{d_3},{d_4},...........,{d_n}\] are the determinants of order \[3\]
As it is given in the question that B is the subset of A consisting of all determinants with value \[1\] .Therefore, \[B = 1\]
Also it is given that C is the subset of A consisting of all determinants with value \[ - 1\] .Therefore, \[C = - 1\]
If we change the sign of one entry in set B then it will be \[B = - 1\] . So the determinant value of B is \[ - 1\] .
Therefore we can say that both B and C subsets have equal number of elements.
Hence the correct option is \[\left( 2 \right)\] B has as many elements as C .
Note: Note that if any two rows or columns of a determinant are interchanged then the value of determinant is multiplied by \[ - 1\] . Note that the determinant is a real number, it is not a matrix. That is, a determinant is a single value that represents a square matrix.
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