KCET · Maths · Matrices
Let \(M\) be \(2 \times 2\) symmetric matrix with integer entries, then \(M\) is invertible if
- A the first column of \(M\) is the transpose of second row of \(M\).
- B the second row of \(M\) is the transpose of first column of \(M\).
- C \(M\) is diagonal matrix with non-zero entries in the principal diagonal.
- D The product of entries in the principal diagonal of \(M\) is the product of entries in the other diagonal.
Answer & Solution
Correct Answer
(C) \(M\) is diagonal matrix with non-zero entries in the principal diagonal.
Step-by-step Solution
Detailed explanation
Let a symmetric matrix
\(M=\left[\begin{array}{ll}
a & c \\
c & b
\end{array}\right]\)
For matrix to be invertible, determinant must not be equal to zero.
\(\begin{aligned}
& &|M| &=a b-c^{2} \neq 0 \\
\Rightarrow & & a b & \neq c^{2}
\end{aligned}\)
Therefore, \(M\) is a diagonal matrix with non-zero entries in the main diagonal and the product of entries in the main diagonal of \(M\) is not the square of an integer.
\(M=\left[\begin{array}{ll}
a & c \\
c & b
\end{array}\right]\)
For matrix to be invertible, determinant must not be equal to zero.
\(\begin{aligned}
& &|M| &=a b-c^{2} \neq 0 \\
\Rightarrow & & a b & \neq c^{2}
\end{aligned}\)
Therefore, \(M\) is a diagonal matrix with non-zero entries in the main diagonal and the product of entries in the main diagonal of \(M\) is not the square of an integer.
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