Question

Matrix A is non-singular matrix and $(A - 3I)(A - 5I) = 0$, then $\frac{15}{8} A^{-1} = \dots\dots$

• A. $I - 8A$
• B. $2I - \frac{1}{15} A$
• C. $I - \frac{1}{8} A$
• D. $8I - 15 A$

Hint:

This technique is a direct application of the Cayley-Hamilton theorem. To quickly find an inverse from a polynomial equation, simply factor out $A$ from all higher-degree terms and push the Identity matrix to the other side!

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Expand the given matrix equation and manipulate it to isolate the term containing $A^{-1}$ by multiplying by $A^{-1}$ throughout.

Step 2: Formula Application:
$(A - 3I)(A - 5I) = A^2 - 5A - 3A + 15I = A^2 - 8A + 15I = 0$.

Step 3: Explanation:
Multiply by $A^{-1}$ on both sides: $A^{-1}(A^2 - 8A + 15I) = A^{-1}(0)$ $A - 8I + 15A^{-1} = 0$ $15A^{-1} = 8I - A$ Divide by 8: $\frac{15}{8} A^{-1} = I - \frac{1}{8} A$.

Step 4: Final Answer:
The result is $I - \frac{1}{8} A$.