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

We have the matrix

$$A=\begin{bmatrix}0&2\\K&-1\end{bmatrix}$$

and the condition

$$A\bigl(A^{3}+3I\bigr)=2I.$$

First we must compute successive powers of $$A$$. The square of a 2 × 2 matrix is obtained by ordinary matrix multiplication. So

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

Carrying out the multiplication element-wise:

$$ \begin{aligned} A^{2}_{11}&=0\cdot0+2\cdot K=2K,\\ A^{2}_{12}&=0\cdot2+2\cdot(-1)=-2,\\ A^{2}_{21}&=K\cdot0+(-1)\cdot K=-K,\\ A^{2}_{22}&=K\cdot2+(-1)\cdot(-1)=2K+1. \end{aligned} $$

Hence

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

Next, $$A^{3}=A^{2}\cdot A$$, so

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

Again multiplying entry by entry,

$$ \begin{aligned} A^{3}_{11}&=2K\cdot0+(-2)\cdot K=-2K,\\[4pt] A^{3}_{12}&=2K\cdot2+(-2)\cdot(-1)=4K+2,\\[4pt] A^{3}_{21}&=-K\cdot0+(2K+1)\cdot K=K(2K+1),\\[4pt] A^{3}_{22}&=-K\cdot2+(2K+1)\cdot(-1)=-2K-(2K+1)=-4K-1. \end{aligned} $$

Thus

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

Now we form $$A^{3}+3I$$. Because the identity matrix is $$I=\begin{bmatrix}1&0\\0&1\end{bmatrix},$$ we have

$$3I=\begin{bmatrix}3&0\\0&3\end{bmatrix},$$

so that

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

We must now multiply $$A$$ by this result:

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

Computing each entry one by one,

$$ \begin{aligned} \bigl(A(A^{3}+3I)\bigr)_{11}&= 0\bigl(-2K+3\bigr)+2\cdot K(2K+1)=2K(2K+1),\\[6pt] \bigl(A(A^{3}+3I)\bigr)_{12}&= 0\bigl(4K+2\bigr)+2\bigl(-4K+2\bigr)=-8K+4,\\[6pt] \bigl(A(A^{3}+3I)\bigr)_{21}&= K\bigl(-2K+3\bigr)+(-1)\cdot K(2K+1)=K(-2K+3-2K-1)=-4K^{2}+2K,\\[6pt] \bigl(A(A^{3}+3I)\bigr)_{22}&= K\bigl(4K+2\bigr)+(-1)\bigl(-4K+2\bigr)=4K^{2}+2K+4K-2=4K^{2}+6K-2. \end{aligned} $$

Therefore

$$A(A^{3}+3I)= \begin{bmatrix} 2K(2K+1)&-8K+4\\[4pt] -4K^{2}+2K&4K^{2}+6K-2 \end{bmatrix}.$$

The given condition states that this matrix equals $$2I=\begin{bmatrix}2&0\\0&2\end{bmatrix}$$. We equate corresponding entries:

$$ \begin{aligned} 2K(2K+1)&=2, \quad\text{(1)}\\[4pt] -8K+4&=0, \quad\text{(2)}\\[4pt] -4K^{2}+2K&=0, \quad\text{(3)}\\[4pt] 4K^{2}+6K-2&=2. \quad\text{(4)} \end{aligned} $$

Equation (2) is simplest: $$-8K+4=0 \;\Rightarrow\; 8K=4 \;\Rightarrow\; K=\dfrac12.$$

We must check that this value also satisfies the remaining three equations.

Substituting $$K=\dfrac12$$ into (1):

$$2\left(\dfrac12\right)\left(2\cdot\dfrac12+1\right) =1\cdot2=2,$$

which matches the right-hand side of (1).

For (3):

$$-4\left(\dfrac12\right)^{2}+2\left(\dfrac12\right) =-4\left(\dfrac14\right)+1 =-1+1=0,$$

so (3) is also satisfied.

Finally, (4):

$$4\left(\dfrac12\right)^{2}+6\left(\dfrac12\right)-2 =4\left(\dfrac14\right)+3-2 =1+3-2=2,$$

which again matches the required value. Thus all four equations are consistent with $$K=\dfrac12$$, and no other value obtained from any single equation satisfies the complete set simultaneously.

Therefore $$K=\dfrac12$$ is the unique solution.

Hence, the correct answer is Option A.