This solution leverages the powerful Cayley-Hamilton Theorem for matrices to determine the coefficients of the inverse matrix and then uses the given conditions to find the unknown parameter $\lambda$ and finally calculate the required expression.

1. Key Concepts and Formulas

   • Cayley-Hamilton Theorem: Every square matrix satisfies its own characteristic equation. For a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the characteristic equation is given by $\det(A - xI) = 0$, which simplifies to $x^2 - (\text{Tr}(A))x + \det(A) = 0$. By the Cayley-Hamilton Theorem, $A^2 - (\text{Tr}(A))A + \det(A)I = 0$.
   • Inverse of a Matrix using Cayley-Hamilton Theorem: If $\det(A) \neq 0$, we can multiply the characteristic equation by $A^{-1}$ to get $A - (\text{Tr}(A))I + \det(A)A^{-1} = 0$. Rearranging this, we get $A^{-1} = \frac{1}{\det(A)}(\text{Tr}(A)I - A) = -\frac{1}{\det(A)}A + \frac{\text{Tr}(A)}{\det(A)}I$.
   • Trace and Determinant of a $2 \times 2$ Matrix: For $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, $\text{Tr}(A) = a+d$ and $\det(A) = ad-bc$.

2. Step-by-Step Solution

   Step 1: Calculate the Trace and Determinant of Matrix A. We are given the matrix $A=\left[\begin{array}{cc}1 & 5 \\ \lambda & 10\end{array}\right]$.

   • The trace of A, denoted as $\text{Tr}(A)$, is the sum of its diagonal elements: $\text{Tr}(A) = 1 + 10 = 11$
   • The determinant of A, denoted as $\det(A)$, is calculated as $(ad - bc)$: $\det(A) = (1)(10) - (5)(\lambda) = 10 - 5\lambda$ For $A^{-1}$ to exist, $\det(A)$ must not be zero, so $10 - 5\lambda \neq 0$, which implies $\lambda \neq 2$.

   Step 2: Express $A^{-1}$ using the Cayley-Hamilton Theorem. According to the Cayley-Hamilton Theorem, the matrix $A$ satisfies its characteristic equation: $A^2 - (\text{Tr}(A))A + \det(A)I = 0$ Substitute the calculated trace and determinant: $A^2 - 11A + (10 - 5\lambda)I = 0$ To find $A^{-1}$, we multiply the entire equation by $A^{-1}$ (assuming $\det(A) \neq 0$): $A^{-1}(A^2 - 11A + (10 - 5\lambda)I) = A^{-1}(0)$ $A - 11I + (10 - 5\lambda)A^{-1} = 0$ Rearrange the equation to isolate $A^{-1}$: $(10 - 5\lambda)A^{-1} = -A + 11I$ $A^{-1} = \frac{-1}{10 - 5\lambda}A + \frac{11}{10 - 5\lambda}I$

   Step 3: Compare with the given expression for $A^{-1}$. We are given that $A^{-1}=\alpha A+\beta I$. Comparing this with our derived expression for $A^{-1}$: $\alpha = \frac{-1}{10 - 5\lambda}$ $\beta = \frac{11}{10 - 5\lambda}$

   Step 4: Use the condition $\alpha+\beta=-2$ to find $\lambda$. Substitute the expressions for $\alpha$ and $\beta$ into the given condition $\alpha+\beta=-2$: $\frac{-1}{10 - 5\lambda} + \frac{11}{10 - 5\lambda} = -2$ $\frac{10}{10 - 5\lambda} = -2$ Multiply both sides by $(10 - 5\lambda)$: $10 = -2(10 - 5\lambda)$ $10 = -20 + 10\lambda$ Add 20 to both sides: $30 = 10\lambda$ Divide by 10: $\lambda = 3$ This value of $\lambda=3$ is consistent with the condition $\lambda \neq 2$.

   Step 5: Calculate the values of $\alpha$ and $\beta$. Substitute $\lambda=3$ back into the expressions for $\alpha$ and $\beta$: $\alpha = \frac{-1}{10 - 5(3)} = \frac{-1}{10 - 15} = \frac{-1}{-5} = \frac{1}{5}$ $\beta = \frac{11}{10 - 5(3)} = \frac{11}{10 - 15} = \frac{11}{-5} = -\frac{11}{5}$ We can verify that $\alpha+\beta = \frac{1}{5} - \frac{11}{5} = -\frac{10}{5} = -2$, which matches the given condition.

   Step 6: Calculate the final expression $4 \alpha^{2}+\beta^{2}+\lambda^{2}$. Now substitute the calculated values of $\alpha$, $\beta$, and $\lambda$ into the expression $4 \alpha^{2}+\beta^{2}+\lambda^{2}$: $4 \alpha^{2}+\beta^{2}+\lambda^{2} = 4\left(\frac{1}{5}\right)^2 + \left(-\frac{11}{5}\right)^2 + (3)^2$ $= 4\left(\frac{1}{25}\right) + \frac{121}{25} + 9$ $= \frac{4}{25} + \frac{121}{25} + 9$ $= \frac{4 + 121}{25} + 9$ $= \frac{125}{25} + 9$ $= 5 + 9$ $= 14$

3. Common Mistakes & Tips

   • Incorrect Application of Cayley-Hamilton: Ensure the characteristic equation is correctly formed as $x^2 - (\text{Tr}(A))x + \det(A) = 0$. A common mistake is getting the sign of the trace term wrong.
   • Algebraic Errors in Solving for $\lambda$: Be careful with distributing negative signs and combining fractions when solving for $\lambda$ from the $\alpha+\beta$ condition.
   • Sign Errors in Squaring: Remember that squaring a negative number results in a positive number, e.g., $(-11/5)^2 = 121/25$.
   • Verification: Always double-check your values of $\alpha$ and $\beta$ using the $\alpha+\beta=-2$ condition before proceeding to the final calculation.

4. Summary

   The problem required us to find the value of an expression involving $\alpha$, $\beta$, and $\lambda$, given a matrix $A$ and a relationship for its inverse. We first calculated the trace and determinant of matrix $A$. Then, using the Cayley-Hamilton Theorem, we derived an expression for $A^{-1}$ in the form $\alpha A + \beta I$. By comparing coefficients, we found $\alpha$ and $\beta$ in terms of $\lambda$. The given condition $\alpha+\beta=-2$ allowed us to solve for $\lambda=3$. Finally, substituting the values of $\alpha=\frac{1}{5}$, $\beta=-\frac{11}{5}$, and $\lambda=3$ into the target expression $4 \alpha^{2}+\beta^{2}+\lambda^{2}$, we found the value to be 14.

5. Final Answer

The final answer is $\boxed{14}$, which corresponds to option (D).