Question

Let \(A=\) [\(a_{ij}\)]\(_{2\times2}\) be a matrix and \(A^2 = I\) where \(a_{ij} \neq0\). If a sum of diagonal elements and b=det(A), then \(3a^2+4b^2\) is

• A. 10
• B. 12
• C. 4
• D. 8

Answer 1

The Correct Option is C

To solve the problem, we are given that a matrix \( A = [a_{ij}]_{2 \times 2} \) satisfies \( A^2 = I \), where \( I \) is the identity matrix of order 2. This means \( A \) is an involutory matrix. We need to find \( 3a^2 + 4b^2 \), where \( a \) is the sum of the diagonal elements (trace) of \( A \), and \( b \) is the determinant of \( A \).

Step 1: Understanding the Properties of the Given Matrix

Since \( A^2 = I \), it implies:

• The eigenvalues of \( A \) are \( +1 \) and \( -1 \).
• The trace of \( A \), which is the sum of its diagonal elements, is \( a = \lambda_1 + \lambda_2 = 1 - 1 = 0 \).

Step 2: Determinant of the Matrix

The determinant of \( A \) is the product of its eigenvalues:

\[ b = \text{det}(A) = \lambda_1 \cdot \lambda_2 = 1 \cdot (-1) = -1 \]

Step 3: Substitute into the Expression

We need to find \( 3a^2 + 4b^2 \):

• \( a = 0 \), so \( a^2 = 0^2 = 0 \).
• \( b = -1 \), so \( b^2 = (-1)^2 = 1 \).

Substitute these into the expression:

\[ 3a^2 + 4b^2 = 3 \times 0 + 4 \times 1 = 0 + 4 = 4 \]

Conclusion: The value of \( 3a^2 + 4b^2 \) is 4.