If $$A = \begin{bmatrix} 1 & 5 \\ \lambda & 10 \end{bmatrix}$$, $$A^{-1} = \alpha A + \beta I$$ and $$\alpha + \beta = -2$$, then $$4\alpha^2 + \beta^2 + \lambda^2$$ is equal to:

We are given $$A = \begin{bmatrix} 1 & 5 \\ \lambda & 10 \end{bmatrix}$$, $$A^{-1} = \alpha A + \beta I$$, and $$\alpha + \beta = -2$$. We need to find $$4\alpha^2 + \beta^2 + \lambda^2$$.

Compute $$A^{-1}$$.

$$\det(A) = 1 \times 10 - 5 \times \lambda = 10 - 5\lambda$$

$$A^{-1} = \frac{1}{10 - 5\lambda}\begin{bmatrix} 10 & -5 \\ -\lambda & 1 \end{bmatrix}$$

Compute $$\alpha A + \beta I$$.

$$\alpha A + \beta I = \alpha\begin{bmatrix} 1 & 5 \\ \lambda & 10 \end{bmatrix} + \beta\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} \alpha + \beta & 5\alpha \\ \lambda\alpha & 10\alpha + \beta \end{bmatrix}$$

Equate and solve.

Comparing the (1,1) entries:

$$\alpha + \beta = \frac{10}{10 - 5\lambda}$$

Since $$\alpha + \beta = -2$$:

$$-2 = \frac{10}{10 - 5\lambda} \implies -2(10 - 5\lambda) = 10 \implies -20 + 10\lambda = 10 \implies \lambda = 3$$

So $$\det(A) = 10 - 15 = -5$$.

Comparing the (1,2) entries:

$$5\alpha = \frac{-5}{-5} = 1 \implies \alpha = \frac{1}{5}$$

From $$\alpha + \beta = -2$$:

$$\beta = -2 - \frac{1}{5} = -\frac{11}{5}$$

Verify with other entries.

Entry (2,1): $$\lambda\alpha = 3 \times \frac{1}{5} = \frac{3}{5}$$ and $$\frac{-\lambda}{\det(A)} = \frac{-3}{-5} = \frac{3}{5}$$. ✔

Entry (2,2): $$10\alpha + \beta = \frac{10}{5} - \frac{11}{5} = -\frac{1}{5}$$ and $$\frac{1}{\det(A)} = \frac{1}{-5} = -\frac{1}{5}$$. ✔

Compute the required expression.

$$4\alpha^2 + \beta^2 + \lambda^2 = 4 \times \frac{1}{25} + \frac{121}{25} + 9 = \frac{4 + 121}{25} + 9 = \frac{125}{25} + 9 = 5 + 9 = 14$$

The answer is Option C: 14.