Question

If \( A^2 = 4A + 3I \) and \( A^{-1} = xA + yI \), then the value of \( (x + y) \) is :

• A. \( -1 \)
• B. \( 1 \)
• C. \( \frac{5}{3} \)
• D. \( 7 \)

Hint:

Instead of computing complex characteristic polynomials, simply isolate \( I \) in the polynomial equation: \( 3I = A^2 - 4A \). Multiplying by \( A^{-1} \) directly gives \( 3A^{-1} = A - 4I \).

Answer 1

The Correct Option is A

Concept: This problem can be efficiently solved by manipulating matrix equations using matrix multiplication properties. Specifically, multiplying a matrix equation by its inverse \( A^{-1} \) allows us to lower the powers of the matrix and isolate \( A^{-1} \) explicitly in terms of \( A \) and the identity matrix \( I \).

Step 1: Set up the matrix equation and pre-multiply by \( A^{-1} \).
The given matrix equation is: \[ A^2 = 4A + 3I \] Assuming \( A \) is invertible, we multiply both sides of the equation by \( A^{-1} \): \[ A^{-1} \cdot A^2 = A^{-1} \cdot (4A + 3I) \] Using the associative and distributive properties of matrix multiplication: \[ (A^{-1} A) A = 4(A^{-1} A) + 3(A^{-1} I) \]

Step 2: Simplify using identity matrix properties.
Since \( A^{-1}A = I \) and \( A^{-1}I = A^{-1} \), the expression becomes: \[ I \cdot A = 4I + 3A^{-1} \] \[ A = 4I + 3A^{-1} \]

Step 3: Isolate the matrix inverse term \( A^{-1} \).
Rearrange the terms to make \( 3A^{-1} \) the subject: \[ 3A^{-1} = A - 4I \] Divide by 3: \[ A^{-1} = \frac{1}{3}A - \frac{4}{3}I \]

Step 4: Compare coefficients and find the sum \( x + y \).
The problem states that \( A^{-1} = xA + yI \). Comparing this with our derived equation: \[ x = \frac{1}{3} \quad \text{and} \quad y = -\frac{4}{3} \] Now calculate the requested sum \( x + y \): \[ x + y = \frac{1}{3} + \left(-\frac{4}{3}\right) = \frac{1 - 4}{3} = \frac{-3}{3} = -1 \] This directly evaluates to option (A).