JEE Advanced · Mathematics · 18. Matrices
Paragraph:
Let \(p\) be an odd prime number and \(T_p\) be the following set of \(2 \times 2\) matrices
\[
T_p=\left\{A=\left[\begin{array}{ll}
a & b \\
c & a
\end{array}\right] ; a, b, c \in\{0,1,2, \ldots, p-1\}\right\}
\]Question:
The number of \(A\) in \(T_p\) such that \(A\) is either symmetric or skew-symmetric or both, and det \((A)\) is divisible by \(p\) is
- A
\((p-1)^2\)
- B
\(2(p-1)\)
- C
\((p-1)^2+1\)
- D
\(2 p-1\)
Answer & Solution
Correct Answer
(D)
\(2 p-1\)
Step-by-step Solution
Detailed explanation
Given, \(A=\left[\begin{array}{ll}a & b \\ c & a\end{array}\right]\),
\(a, b, c \in\{0,1,2, \ldots, p-1\}\)
If \(A\) is skew-symmetric matrix, then \(a=0, b=-c\)
\(\therefore \quad|A|=-b^2\).
Thus, \(P\) divides \(|A|\) only when \(b=0\)...(i)
Again, if \(A\) is symmetric matrix, then \(b=c\) and \(|A|=a^2-b^2\).
Thus, \(p\) divides \(|A|\) if either \(p\) divides \((a-b)\) or \(p\) divides \((a+b)\). \(p\) divides \((a-b)\), only when \(a=b\) ie, \(a=b \in\{0,1,2, \ldots,(p-1)\}\)
ie, pchoices
\(p\) divides \((a+b)\).
\(\Rightarrow p\) choices, including \(a=b=0\) included in (i)
\(\therefore\) Total number of choices are \((p+p-1)=2 p-1\).
Hence, (c) is the correct option.
\(a, b, c \in\{0,1,2, \ldots, p-1\}\)
If \(A\) is skew-symmetric matrix, then \(a=0, b=-c\)
\(\therefore \quad|A|=-b^2\).
Thus, \(P\) divides \(|A|\) only when \(b=0\)...(i)
Again, if \(A\) is symmetric matrix, then \(b=c\) and \(|A|=a^2-b^2\).
Thus, \(p\) divides \(|A|\) if either \(p\) divides \((a-b)\) or \(p\) divides \((a+b)\). \(p\) divides \((a-b)\), only when \(a=b\) ie, \(a=b \in\{0,1,2, \ldots,(p-1)\}\)
ie, pchoices
\(p\) divides \((a+b)\).
\(\Rightarrow p\) choices, including \(a=b=0\) included in (i)
\(\therefore\) Total number of choices are \((p+p-1)=2 p-1\).
Hence, (c) is the correct option.
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