For a \( 2 \times 2 \) matrix \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), the adjugate matrix is:
\[ \text{adj } A = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}. \]If \( \text{adj } A = A \), then:
\[ \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}. \]Equating elements:
\[ d = a, \quad -b = b, \quad -c = c, \quad a = d. \]From \( -b = b \), we get \( b = 0 \), and from \( -c = c \), we get \( c = 0 \). Thus:
\[ A = \begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix}. \]The sum of the elements is:
\[ a + b + c + d = a + 0 + 0 + a = 2a. \]Final Answer: \( \boxed{2a} \)
Determine whether each of the following relations are reflexive, symmetric, and transitive.
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
\[ \begin{bmatrix} x + y & 2 \\ 5 & xy \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}, \]
then the value of\[ \left(\frac{24}{x} + \frac{24}{y}\right) \]
is: