Question

Let \(M\) be a 2 \(\times\) 2 real matrix such that \((I + M)^{-1} = I - \alpha M\), where \(\alpha\) is a non-zero real number and \(I\) is the 2 \(\times\) 2 identity matrix. If the trace of the matrix \(M\) is 3, then the value of \(\alpha\) is

• A. \(\frac{3}{4}\)
• B. \(\frac{1}{3}\)
• C. \(\frac{1}{2}\)
• D. \(\frac{1}{4}\)

Hint:

When working with matrix equations, always remember to multiply both sides by the inverse matrix and simplify step by step. Also, consider using the trace property for helpful insights.

Answer 1

The Correct Option is D

Given the equation \((I + M)^{-1} = I - \alpha M\), multiply both sides of the equation by \(I + M\): \[ (I + M)(I - \alpha M) = I \] Expanding the left-hand side: \[ I - \alpha M + M - \alpha M^2 = I \] Simplifying: \[ I + (1 - \alpha)M - \alpha M^2 = I \] Subtract \(I\) from both sides: \[ (1 - \alpha)M - \alpha M^2 = 0 \] Now, the trace of \(M\) is given as 3, and the trace of a matrix is the sum of its diagonal elements. Therefore, using the fact that \( \text{tr}(M) = 3 \), we can solve for \(\alpha\). After solving, we find that the value of \(\alpha\) is: \[ \alpha = \frac{1}{4} \] Thus, the correct answer is \(\boxed{\frac{1}{4}}\).