Let \( A = \begin{pmatrix} m & n \\ p & q \end{pmatrix} \), \( d = |A| \neq 0 \), and \( |A - d(\text{Adj}(A))| = 0 \). Then
\(1 + d^2 = (m + q)^2\)
\((1 + d)^2 = m^2 + q^2\)
\((1 + d)^2 = (m + q)^2\)
\(1 + d^2 = m^2 + q^2\)
The Correct Option is C
Solution and Explanation
The given matrix is:
\[ A = \begin{bmatrix} m & n \\ p & q \end{bmatrix}. \]
Determinant of \( A \):
\[ |A| = m \cdot q - n \cdot p = d. \]
Adjugate of \( A \):
The adjugate of \( A \), denoted as \( \text{Adj}(A) \), is:
\[ \text{Adj}(A) = \begin{bmatrix} q & -n \\ -p & m \end{bmatrix}. \]
Given Condition:
We are given the condition:
\[ |A - d \cdot \text{Adj}(A)| = 0. \]
Substituting \( A \) and \( \text{Adj}(A) \) into the equation:
\[ A - d \cdot \text{Adj}(A) = \begin{bmatrix} m & n \\ p & q \end{bmatrix} - d \begin{bmatrix} q & -n \\ -p & m \end{bmatrix}. \]
This simplifies to:
\[ A - d \cdot \text{Adj}(A) = \begin{bmatrix} m - dq & n + dn \\ p + dp & q - dm \end{bmatrix}. \]
Determinant of \( A - d \cdot \text{Adj}(A) \):
The determinant is:
\[ |A - d \cdot \text{Adj}(A)| = \begin{vmatrix} m - dq & n + dn \\ p + dp & q - dm \end{vmatrix}. \]
Expanding the determinant:
\[ |A - d \cdot \text{Adj}(A)| = (m - dq)(q - dm) - (n + dn)(p + dp). \]
Simplify the terms:
\[ |A - d \cdot \text{Adj}(A)| = m \cdot q - m \cdot dm - dq \cdot q + d^2 \cdot q \cdot m - (n \cdot p + n \cdot dp + dn \cdot p + d^2 \cdot n \cdot p). \]
Using \( d = m \cdot q - n \cdot p \), the terms simplify further to:
\[ (1 + d)^2 = (m + q)^2. \]
Conclusion:
The correct answer is:
\[ (1 + d)^2 = (m + q)^2. \]
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