Key Concept: The Cayley-Hamilton Theorem

The Cayley-Hamilton Theorem states that every square matrix satisfies its own characteristic equation. For a $2 \times 2$ matrix $A$, if its characteristic equation is $\lambda^2 - \text{tr}(A)\lambda + \det(A) = 0$, then $A^2 - \text{tr}(A)A + \det(A)I = 0$, where $\text{tr}(A)$ is the trace of $A$ (sum of diagonal elements) and $\det(A)$ is the determinant of $A$. This theorem is crucial for simplifying matrix polynomial expressions.

---

Step-by-Step Solution

1. Analyze the Given Information and Simplify the Matrix Equation

We are given the matrix A = \left( {\matrix{ 0 & 2 \cr K & { - 1} \cr } } \right) and the matrix equation $A({A^3} + 3I) = 2I$.

First, let's simplify the given matrix equation: $A({A^3} + 3I) = 2I$ Distribute $A$ on the left side: $A \cdot A^3 + A \cdot (3I) = 2I$ $A^4 + 3(AI) = 2I$ Since $AI = A$ (identity matrix multiplication property), we have: $A^4 + 3A = 2I$ Now, isolate $A^4$: $A^4 = 2I - 3A \quad (*)$ This equation provides a direct relationship between $A^4$ and $A$, which we will use later to find $K$.

2. Find the Characteristic Equation of Matrix A

To apply the Cayley-Hamilton Theorem, we first need to find the characteristic equation of matrix $A$. The characteristic equation is given by $\det(A - \lambda I) = 0$, where $\lambda$ is an eigenvalue and $I$ is the identity matrix.

For A = \left( {\matrix{ 0 & 2 \cr K & { - 1} \cr } } \right), we have: A - \lambda I = \left( {\matrix{ 0 & 2 \cr K & { - 1} \cr } } \right) - \lambda \left( {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right) = \left( {\matrix{ {0 - \lambda} & 2 \cr K & {-1 - \lambda} \cr } } \right) Now, calculate the determinant: $\det(A - \lambda I) = (0 - \lambda)(-1 - \lambda) - (2)(K) = 0$ $(-\lambda)(-(1 + \lambda)) - 2K = 0$ $\lambda(1 + \lambda) - 2K = 0$ $\lambda + \lambda^2 - 2K = 0$ Rearranging in standard quadratic form: $\lambda^2 + \lambda - 2K = 0$ This is the characteristic equation of matrix $A$.

3. Apply the Cayley-Hamilton Theorem

According to the Cayley-Hamilton Theorem, matrix $A$ satisfies its own characteristic equation. So, substitute $A$ for $\lambda$ and $I$ for the constant term: $A^2 + A - 2KI = 0$ This equation is a fundamental property of matrix $A$. We can express $A^2$ in terms of $A$ and $I$: $A^2 = 2KI - A \quad (**)$ This expression for $A^2$ will be useful for simplifying higher powers of $A$.

4. Express $A^4$ in terms of $A$ and $I$ using the Cayley-Hamilton Relation

We have an expression for $A^2$. To get $A^4$, we can square $A^2$: $A^4 = (A^2)^2$ Substitute the expression for $A^2$ from $(**)$: $A^4 = (2KI - A)^2$ Expand the square, remembering that matrix multiplication is not commutative in general, but $I$ commutes with any matrix ($IA = AI = A$), and scalar multiplication works as expected: $(X - Y)^2 = X^2 - XY - YX + Y^2$ Here $X = 2KI$ and $Y = A$. $(2KI - A)^2 = (2KI)(2KI) - (2KI)A - A(2KI) + A^2$ $= 4K^2I^2 - 2KIA - 2KIA + A^2$ Since $I^2 = I$ and $IA = A$: $= 4K^2I - 2KA - 2KA + A^2$ $A^4 = 4K^2I - 4KA + A^2$ Now, substitute $A^2 = 2KI - A$ (from $(**)$) back into this equation to reduce the power of $A$ further: $A^4 = 4K^2I - 4KA + (2KI - A)$ Group terms with $I$ and $A$: $A^4 = (4K^2 + 2K)I - (4K + 1)A \quad (***)$ Now we have an expression for $A^4$ solely in terms of $A$, $I$, and $K$.

5. Equate the Two Expressions for $A^4$ and Solve for K

We have two expressions for $A^4$:

1. From the given equation ($*$): $A^4 = 2I - 3A$
2. From the Cayley-Hamilton Theorem ($***$): $A^4 = (4K^2 + 2K)I - (4K + 1)A$

Equating these two expressions: $2I - 3A = (4K^2 + 2K)I - (4K + 1)A$ Rearrange the terms to group $I$ and $A$ matrices: $2I - (4K^2 + 2K)I = 3A - (4K + 1)A$ $[2 - (4K^2 + 2K)]I = [3 - (4K + 1)]A$ $(2 - 4K^2 - 2K)I = (3 - 4K - 1)A$ $(2 - 2K - 4K^2)I = (2 - 4K)A$ Factor out common terms: $-2(2K^2 + K - 1)I = 2(1 - 2K)A$ Divide both sides by 2: $-(2K^2 + K - 1)I = (1 - 2K)A$ Factor the quadratic expression $2K^2 + K - 1$: We look for two numbers that multiply to $2 \times (-1) = -2$ and add to $1$. These are $2$ and $-1$. $2K^2 + 2K - K - 1 = 2K(K+1) - 1(K+1) = (2K-1)(K+1)$ Substitute this back: $-(2K-1)(K+1)I = (1 - 2K)A$ Notice that $(1 - 2K) = -(2K - 1)$. So we can write: $-(2K-1)(K+1)I = -(2K - 1)A$ Move all terms to one side: $(2K-1)A - (2K-1)(K+1)I = 0$ Factor out $(2K-1)$: $(2K-1)[A - (K+1)I] = 0$ For this equation to hold, one of the factors must be zero. Case 1: $2K - 1 = 0$ This implies $2K = 1$, so $K = \frac{1}{2}$.

Case 2: $A - (K+1)I = 0$ This would mean $A = (K+1)I$. If A = \left( {\matrix{ 0 & 2 \cr K & { - 1} \cr } } \right) were equal to (K+1)I = \left( {\matrix{ K+1 & 0 \cr 0 & {K+1} \cr } } \right), then comparing the elements: $0 = K+1 \Rightarrow K = -1$ $2 = 0$ (This is a contradiction!) This means that $A$ cannot be a scalar multiple of the identity matrix. Therefore, the second factor $A - (K+1)I$ cannot be the zero matrix.

Since $A - (K+1)I \neq 0$, the only possibility is that the first factor is zero. $2K - 1 = 0 \Rightarrow K = \frac{1}{2}$.

6. Verify with Options

The calculated value $K = \frac{1}{2}$ matches option (A).

---

Tips and Common Mistakes:

• Matrix Algebra: Always remember that matrix multiplication is generally not commutative ($AB \neq BA$). However, the identity matrix $I$ commutes with any matrix ($AI = IA = A$). Also, scalar multiplication works just like with numbers, e.g., $(cM)^2 = c^2M^2$.
• Cayley-Hamilton Theorem: This theorem is a powerful tool for simplifying matrix polynomials. It allows you to reduce any power of a matrix $A$ to a linear combination of lower powers of $A$ (down to $A^{n-1}$ for an $n \times n$ matrix) and the identity matrix $I$.
• Factoring: When you reach an equation like $(2K-1)[A - (K+1)I] = 0$, it's crucial to analyze both factors. Don't immediately assume the scalar factor is the only one that can be zero. However, in matrix equations, if $M \cdot N = \mathbf{0}$ (zero matrix), it doesn't necessarily mean $M = \mathbf{0}$ or $N = \mathbf{0}$. But if one factor is a scalar, say $c \cdot M = \mathbf{0}$ where $M$ is a non-zero matrix, then the scalar $c$ must be zero. In our case, $A - (K+1)I$ is a matrix, and if it's not the zero matrix, then the scalar factor $(2K-1)$ must be zero.
• Careful Substitution: Be meticulous when substituting matrix expressions, especially when squaring or multiplying. A small error can propagate through the entire solution.

---

Summary and Key Takeaway:

This problem demonstrates an effective application of the Cayley-Hamilton Theorem. By first simplifying the given matrix equation and then using the characteristic equation to find a polynomial relation for $A$ (via Cayley-Hamilton), we can express higher powers of $A$ in terms of $A$ and $I$. Equating these expressions leads to a simpler equation involving $A$, $I$, and the unknown $K$. By comparing coefficients of $A$ and $I$ (or by careful factorization), we can solve for $K$. The ability to reduce