Question

If $A = \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix}$ and $A^{-1} = \alpha I + \beta A$, $\beta \in R$ where I is the identity matrix of order 2, then $4(\alpha + \beta) =$}

• A. $\frac{10}{3}$
• B. $\frac{2}{3}$
• C. $\frac{8}{3}$
• D. $\frac{1}{3}$

Hint:

Cayley-Hamilton is the fastest way to express $A^{-1}$ in terms of $A$ and $I$.

Answer 1

The Correct Option is C

Step 1: Concept 
Use the Cayley-Hamilton Theorem: A matrix satisfies its characteristic equation $|A - \lambda I| = 0$. 

Step 2: Meaning 
Characteristic equation: $(1-\lambda)(4-\lambda) - (-2) = 0 \implies \lambda^2 - 5\lambda + 6 = 0$. Thus, $A^2 - 5A + 6I = 0$. 

Step 3: Analysis 
Multiply by $A^{-1}$: $A - 5I + 6A^{-1} = 0 \implies 6A^{-1} = 5I - A$. $A^{-1} = \frac{5}{6}I - \frac{1}{6}A$. Comparing gives $\alpha = 5/6, \beta = -1/6$. 

Step 4: Conclusion 
$4(\alpha + \beta) = 4(5/6 - 1/6) = 4(4/6) = 16/6 = 8/3$.