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:

If a matrix equation is of the form \(aA^2 + bA + cI = 0\), the inverse \(A^{-1}\) can always be found as \(-\frac{1}{c}(aA + bI)\). This is a direct application of the Cayley-Hamilton theorem.

Answer 1

The Correct Option is A

Concept:
• Matrix equations involving powers of \(A\) and the identity matrix \(I\) can often be used to derive the expression for the inverse matrix \(A^{-1}\).
• By pre-multiplying or post-multiplying the entire equation by \(A^{-1}\), we can reduce the degree of the terms and isolate \(A^{-1}\).
• Matrix multiplication is distributive: \(A^{-1}(B + C) = A^{-1}B + A^{-1}C\).

Step 1: Manipulate the given matrix equation
We are given: \[ A^2 = 4A + 3I \] Rearrange the equation to bring the terms containing \(A\) to one side and the identity matrix to the other: \[ A^2 - 4A = 3I \]

Step 2: Multiply the equation by \(A^{-1}\)
Pre-multiply both sides of the equation by \(A^{-1}\): \[ A^{-1}(A^2 - 4A) = A^{-1}(3I) \] \[ A^{-1} \cdot A \cdot A - 4(A^{-1} \cdot A) = 3(A^{-1} \cdot I) \]

Step 3: Simplify using matrix identities
Recall that \(A^{-1} \cdot A = I\) and \(A^{-1} \cdot I = A^{-1}\). Substituting these into the equation: \[ I \cdot A - 4I = 3A^{-1} \] \[ A - 4I = 3A^{-1} \]

Step 4: Isolate \(A^{-1}\) and compare coefficients
Divide the equation by \(3\): \[ A^{-1} = \frac{1}{3}A - \frac{4}{3}I \] The given form is \(A^{-1} = xA + yI\). By comparing the two expressions, we get: \[ x = \frac{1}{3} \] \[ y = -\frac{4}{3} \]

Step 5: Calculate the final value
Find the sum of \(x\) and \(y\): \[ x + y = \frac{1}{3} + \left(-\frac{4}{3}\right) = \frac{1 - 4}{3} = \frac{-3}{3} = -1 \]