Question

If the matrix \( A = \begin{pmatrix} 0 & 2 \\ K & -1 \end{pmatrix} \) satisfies \( A(A^3 + 3I) = 2I \), then the value of \( K \) is :

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

Hint:

Using Cayley-Hamilton is much faster than computing $A^4$ manually. Express $A^4$ as a linear combination of $A$ and $I$ using the characteristic equation.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Every square matrix satisfies its characteristic equation \( |A - \lambda I| = 0 \) (Cayley-Hamilton Theorem).
Step 2: Detailed Explanation:
The given equation is \( A^4 + 3A = 2I \).
Characteristic equation of \( A \):

\[ \begin{vmatrix} -\lambda & 2 \\ K & -1-\lambda \end{vmatrix} = 0 \implies \lambda(1+\lambda) - 2K = 0 \implies \lambda^2 + \lambda - 2K = 0 \]

So, \( A^2 + A - 2KI = 0 \implies A^2 = 2KI - A \).
Now, \( A^4 = (A^2)^2 = (2KI - A)^2 = 4K^2 I + A^2 - 4KI A \).
Substitute \( A^2 = 2KI - A \):

\[ A^4 = 4K^2 I + (2KI - A) - 4KA = (4K^2 + 2K)I - A(1 + 4K) \]

Substitute \( A^4 \) into \( A^4 + 3A = 2I \):

\[ (4K^2 + 2K)I - A(1 + 4K) + 3A = 2I \]

\[ (4K^2 + 2K - 2)I + A(2 - 4K) = 0 \]

This must be true for all \( A \), so coefficients must be zero.
\( 2 - 4K = 0 \implies K = \frac{1}{2} \).
Check: \( 4\left(\frac{1}{2}\right)^2 + 2\left(\frac{1}{2}\right) - 2 = 1 + 1 - 2 = 0 \). (Verified)
Step 3: Final Answer:
The value of \( K \) is \( \frac{1}{2} \).