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 Question:
We are given an annihilating polynomial equation for a non-singular matrix $A$. We need to manipulate this equation to express a multiple of its inverse, $A^{-1}$, in terms of $A$ and the identity matrix $I$.

Step 2: Detailed Explanation:
Start with the given matrix equation:
$(A - 3I)(A - 5I) = 0$
Expand the expression using standard matrix algebra (noting that $A$ and $I$ naturally commute):
$A^2 - 5AI - 3IA + 15I^2 = 0$
$A^2 - 8A + 15I = 0$
Since $A$ is non-singular, its inverse $A^{-1}$ exists. Multiply the entire equation by $A^{-1}$ from the left (or right):
$A^{-1} (A^2 - 8A + 15I) = A^{-1} (0)$
$A - 8I + 15A^{-1} = 0$
Now, rearrange the equation to isolate the term containing $A^{-1}$:
$15A^{-1} = 8I - A$
The question asks for the specific value of $\frac{15}{8} A^{-1}$. Divide the entire equation by 8:
$\frac{15}{8} A^{-1} = \frac{8I - A}{8}$
$\frac{15}{8} A^{-1} = I - \frac{1}{8} A$

Step 3: Final Answer:
The expression is equal to $I - \frac{1}{8} A$, matching option (c).