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:
We have a polynomial equation in terms of a matrix \(A\).
We need to find the inverse \(A^{-1}\). We can do this by post-multiplying or pre-multiplying the equation by \(A^{-1}\).
Step 2: Key Formula or Approach:
Use the properties: \(A \cdot A^{-1} = I\) and \(I \cdot A^{-1} = A^{-1}\).
Step 3: Detailed Explanation:
Given equation:
\[ A^2 - A = 20I \] Multiply both sides by \(A^{-1}\):
\[ (A^2 - A)A^{-1} = 20I \cdot A^{-1} \] Distribute \(A^{-1}\) on the left side:
\[ A \cdot A \cdot A^{-1} - A \cdot A^{-1} = 20A^{-1} \] Apply the property \(A \cdot A^{-1} = I\):
\[ A \cdot I - I = 20A^{-1} \] Since \(A \cdot I = A\):
\[ A - I = 20A^{-1} \] Divide both sides by 20 to isolate \(A^{-1}\):
\[ A^{-1} = \frac{1}{20}(A - I) \] Step 4: Final Answer:
The inverse is \(\frac{1}{20}(A - I)\).