KCET · Maths · Mathematical Reasoning
In \(\quad\) Ine \(\quad(Z, *), \quad\) if \(a^{*} b=a+b-n, \forall a, b \in Z, \quad\) where \(n\) is a fixed integer, then the inverse of \((-n)\) is
- A \(n\)
- B \(-n\)
- C \(-3 n\)
- D \(3 n\)
Answer & Solution
Correct Answer
(D) \(3 n\)
Step-by-step Solution
Detailed explanation
Given,
In a group \(\left(Z,{ }^{*}\right)\)
\(a^{*} b=a+b-n, \forall a_{1}, b \in z...(i)\)
where, \(n\) is a fixed number
\(\begin{array}{llr}\therefore & a^{*} e=a & \\ \Rightarrow & a+e-n=a \quad \text { [from Eq. (i)] } \\ \Rightarrow & e=n & \text { (identity) }\end{array}\)
For finding inverse,
\(\alpha^{*}(-n)=n\)
\(\Rightarrow \quad \alpha-n-(-n)=n \quad\) [from Eq. (i)]
\(\Rightarrow \quad \alpha-n-n=n\)
\(\Rightarrow \quad \alpha=3 n\)
So, inverse of \((-n)\) is \(3 n\).
In a group \(\left(Z,{ }^{*}\right)\)
\(a^{*} b=a+b-n, \forall a_{1}, b \in z...(i)\)
where, \(n\) is a fixed number
\(\begin{array}{llr}\therefore & a^{*} e=a & \\ \Rightarrow & a+e-n=a \quad \text { [from Eq. (i)] } \\ \Rightarrow & e=n & \text { (identity) }\end{array}\)
For finding inverse,
\(\alpha^{*}(-n)=n\)
\(\Rightarrow \quad \alpha-n-(-n)=n \quad\) [from Eq. (i)]
\(\Rightarrow \quad \alpha-n-n=n\)
\(\Rightarrow \quad \alpha=3 n\)
So, inverse of \((-n)\) is \(3 n\).
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