Question

Let $A$ be a non-singular square matrix of order 3. If $A^2 - A = 20I$, where $I$ is the unit matrix of order 3, then $A^{-1} =$}

• A. $20A$
• B. $\frac{1}{20}(A - I)$
• C. $20(A - I)$
• D. $\frac{1}{20}A$
• E. $\frac{1}{20}A^2$

Hint:

When asked for $A^{-1}$ given a polynomial equation in $A$, factor $A$ out and use the definition $A \cdot (\text{something}) = I$ to identify the inverse immediately.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For a non-singular square matrix $A$, the inverse $A^{-1}$ is the matrix such that $AA^{-1} = A^{-1}A = I$. We can manipulate the given matrix equation to isolate terms that allow us to extract $A^{-1}$.
Key Formula or Approach:
The given equation is $A^2 - A = 20I$. We can factor out $A$ from the left-hand side.

Step 2: Detailed Explanation:
1. Start with the given equation:
\[ A^2 - A = 20I \]
2. Factor $A$ from the terms on the left:
\[ A(A - I) = 20I \]
3. Since $A$ is non-singular, its inverse exists. Pre-multiply both sides by $A^{-1}$:
\[ A^{-1} [A(A - I)] = A^{-1} [20I] \]
4. Using the associative property of matrix multiplication:
\[ (A^{-1}A) (A - I) = 20 A^{-1} \]
\[ I (A - I) = 20 A^{-1} \]
\[ A - I = 20 A^{-1} \]
5. Solve for $A^{-1}$ by dividing both sides by 20:
\[ A^{-1} = \frac{1}{20} (A - I) \]

Step 3: Final Answer:
The inverse matrix $A^{-1}$ is $\frac{1}{20}(A - I)$.