Question

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

• A. 1/2
• B. 1
• C. -1
• D. -1/2

Answer 1

Answer: (A)

K = 1/2

Answer 2

We are given:

$$A = \begin{bmatrix} 0 & 2 \\ K & -1 \end{bmatrix} \quad \text{and} \quad A\left(A^3 + 3I\right)=2I.$$

This implies:

$$A^4 + 3A = 2I.$$

Step 1. Compute $A^2$:

$$A^2 = A \cdot A = \begin{bmatrix} 0 & 2 \\ K & -1 \end{bmatrix}\begin{bmatrix} 0 & 2 \\ K & -1 \end{bmatrix} = \begin{bmatrix} 2K & -2 \\ -K & 2K+1 \end{bmatrix}.$$

Step 2. Compute $A^3 = A^2 \cdot A$:

$$A^3 = \begin{bmatrix} 2K & -2 \\ -K & 2K+1 \end{bmatrix}\begin{bmatrix} 0 & 2 \\ K & -1 \end{bmatrix} = \begin{bmatrix} -2K & 4K+2 \\ K(2K+1) & -4K-1 \end{bmatrix}.$$

Step 3. Compute $A^4 = A^3 \cdot A$:

$$A^4 = \begin{bmatrix} -2K & 4K+2 \\ K(2K+1) & -4K-1 \end{bmatrix}\begin{bmatrix} 0 & 2 \\ K & -1 \end{bmatrix} = \begin{bmatrix} 4K^2+2K & -8K-2 \\ -4K^2-K & 4K^2+6K+1 \end{bmatrix}.$$

Step 4. Set up the equation $A^4 + 3A = 2I$:

$$A^4 + 3A = \begin{bmatrix} 4K^2+2K & -8K-2 \\ -4K^2-K & 4K^2+6K+1 \end{bmatrix} + 3\begin{bmatrix} 0 & 2 \\ K & -1 \end{bmatrix}$$ $$= \begin{bmatrix} 4K^2+2K & -8K-2+6 \\ -4K^2-K+3K & 4K^2+6K+1-3 \end{bmatrix} = \begin{bmatrix} 4K^2+2K & -8K+4 \\ -4K^2+2K & 4K^2+6K-2 \end{bmatrix}.$$

For the given equation to hold, we equate with $2I = \begin{bmatrix}2 & 0 \\ 0 & 2\end{bmatrix}$:

1. $4K^2+2K = 2$
2. $-8K+4 = 0$
3. $-4K^2+2K = 0$
4. $4K^2+6K-2 = 2$

Step 5. Solve Equation (2):

$$-8K+4=0 \quad\Rightarrow\quad K=\frac{1}{2}.$$

Substitute $K=\frac{1}{2}$ into the other equations to verify:

• Equation (1): $4\left(\frac{1}{4}\right) + 2\left(\frac{1}{2}\right)= 1+1=2.$
• Equation (3): $-4\left(\frac{1}{4}\right)+2\left(\frac{1}{2}\right)= -1+1=0.$
• Equation (4): $4\left(\frac{1}{4}\right)+6\left(\frac{1}{2}\right)-2= 1+3-2=2.$

All conditions hold with $K=\frac{1}{2}$.