Unit matrix is a diagonal matrix in which all the diagonal elements are unity. Unit matrix of order \[n\] is denoted by \[{I_n}\] (or \[I\]) i.e. \[A = {\left[ {{a_{ij}}} \right]_n}\] is a unit matrix when \[{a_{ij}} = 0\] for \[i \ne j\] and \[{a_{ij}} = 1\]
A) True
B) False
Answer
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Hint:
Here, we are required to find whether the given statement is true or false. We will make the required matrices and by observation, we will compare whether the result is the same as the given statement or not. This will help us to find that the given statement is true or false.
Complete step by step solution:
It is given that a unit matrix is a diagonal matrix in which all the diagonal elements are unity and all the other elements are zero.
Also, a unit matrix of order \[n\] is denoted by \[{I_n}\] (or \[I\])
\[{I_n} = {\left[ \begin{array}{l}1{\rm{ }}0{\rm{ }}0{\rm{ }}....{\rm{ }}0\\0{\rm{ }}1{\rm{ }}....{\rm{ }}....{\rm{ }}0\\.{\rm{ }}.{\rm{ }}.{\rm{ }}{\rm{. }}.\\.{\rm{ }}{\rm{. }}.{\rm{ }}.{\rm{ }}.\\0{\rm{ }}0{\rm{ }}....{\rm{ }}....{\rm{ }}1\end{array} \right]_{n \times n}}\]
Now, \[A = {\left[ {{a_{ij}}} \right]_n}\] can be written as:
\[ \Rightarrow A = {\left[ \begin{array}{l}{{\rm{a}}_{11}}{\rm{ }}{{\rm{a}}_{12}}{\rm{ }}....{\rm{ }}{{\rm{a}}_{1n}}\\{{\rm{a}}_{21}}{\rm{ }}{{\rm{a}}_{22}}{\rm{ }}....{\rm{ }}{{\rm{a}}_{2n}}\\.{\rm{ }}.{\rm{ }}{\rm{. }}.\\.{\rm{ }}{\rm{. }}.{\rm{ }}.\\{{\rm{a}}_{n1}}{\rm{ }}{{\rm{a}}_{n2}}{\rm{ }}....{\rm{ }}{{\rm{a}}_{nn}}\end{array} \right]_{n \times n}}\]
Now, \[A = {\left[ {{a_{ij}}} \right]_n}\] becomes a unit matrix \[{I_n}\] when the elements \[{a_{11}},{a_{22}},{a_{33}}.....{a_{nn}} = 1\]
And all the other elements are equal to zero.
This proves that:
\[A = {\left[ {{a_{ij}}} \right]_n}\] is a unit matrix when \[{a_{ij}} = 0\] for \[i \ne j\]
and \[{a_{ij}} = 1\] for \[i = j\]
Hence, the given statement, i.e. Unit matrix is a diagonal matrix in which all the diagonal elements are unity. Unit matrix of order \[n\] is denoted by \[{I_n}\] (or \[I\]) i.e. \[A = {\left[ {{a_{ij}}} \right]_n}\] is a unit matrix when \[{a_{ij}} = 0\] for \[i \ne j\] and \[{a_{ij}} = 1\] is true.
Therefore, option A is the correct answer.
Note:
A diagonal matrix is a matrix in which all the elements except the main diagonal are zero. Now, in the case of a unit matrix, the numbers present in the diagonal should be 1. However, this need not be true in the case of just a diagonal matrix as it can have any number in its diagonal. Hence, to solve this question, we should know what exactly a unit matrix is. Hence, after drawing the matrices we will be able to answer whether the given statement is true or not.
Here, we are required to find whether the given statement is true or false. We will make the required matrices and by observation, we will compare whether the result is the same as the given statement or not. This will help us to find that the given statement is true or false.
Complete step by step solution:
It is given that a unit matrix is a diagonal matrix in which all the diagonal elements are unity and all the other elements are zero.
Also, a unit matrix of order \[n\] is denoted by \[{I_n}\] (or \[I\])
\[{I_n} = {\left[ \begin{array}{l}1{\rm{ }}0{\rm{ }}0{\rm{ }}....{\rm{ }}0\\0{\rm{ }}1{\rm{ }}....{\rm{ }}....{\rm{ }}0\\.{\rm{ }}.{\rm{ }}.{\rm{ }}{\rm{. }}.\\.{\rm{ }}{\rm{. }}.{\rm{ }}.{\rm{ }}.\\0{\rm{ }}0{\rm{ }}....{\rm{ }}....{\rm{ }}1\end{array} \right]_{n \times n}}\]
Now, \[A = {\left[ {{a_{ij}}} \right]_n}\] can be written as:
\[ \Rightarrow A = {\left[ \begin{array}{l}{{\rm{a}}_{11}}{\rm{ }}{{\rm{a}}_{12}}{\rm{ }}....{\rm{ }}{{\rm{a}}_{1n}}\\{{\rm{a}}_{21}}{\rm{ }}{{\rm{a}}_{22}}{\rm{ }}....{\rm{ }}{{\rm{a}}_{2n}}\\.{\rm{ }}.{\rm{ }}{\rm{. }}.\\.{\rm{ }}{\rm{. }}.{\rm{ }}.\\{{\rm{a}}_{n1}}{\rm{ }}{{\rm{a}}_{n2}}{\rm{ }}....{\rm{ }}{{\rm{a}}_{nn}}\end{array} \right]_{n \times n}}\]
Now, \[A = {\left[ {{a_{ij}}} \right]_n}\] becomes a unit matrix \[{I_n}\] when the elements \[{a_{11}},{a_{22}},{a_{33}}.....{a_{nn}} = 1\]
And all the other elements are equal to zero.
This proves that:
\[A = {\left[ {{a_{ij}}} \right]_n}\] is a unit matrix when \[{a_{ij}} = 0\] for \[i \ne j\]
and \[{a_{ij}} = 1\] for \[i = j\]
Hence, the given statement, i.e. Unit matrix is a diagonal matrix in which all the diagonal elements are unity. Unit matrix of order \[n\] is denoted by \[{I_n}\] (or \[I\]) i.e. \[A = {\left[ {{a_{ij}}} \right]_n}\] is a unit matrix when \[{a_{ij}} = 0\] for \[i \ne j\] and \[{a_{ij}} = 1\] is true.
Therefore, option A is the correct answer.
Note:
A diagonal matrix is a matrix in which all the elements except the main diagonal are zero. Now, in the case of a unit matrix, the numbers present in the diagonal should be 1. However, this need not be true in the case of just a diagonal matrix as it can have any number in its diagonal. Hence, to solve this question, we should know what exactly a unit matrix is. Hence, after drawing the matrices we will be able to answer whether the given statement is true or not.
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