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:
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:
- 10
- 12
- 14
- 19
The Correct Option is C
Solution and Explanation
Given:
\[ |A - xI| = 0 \quad \Rightarrow \quad \left| \begin{array}{cc} 1 - x & 5 \\ \lambda & 10 - x \end{array} \right| = 0 \quad \Rightarrow \quad x^2 - 11x + 10 - 5\lambda = 0 \]
From this, we get the equation:
\[ (10 - 5\lambda) A^{-1} = -A + 11I \]
From the equation above, we derive:
\[ \alpha = \frac{-1}{10 - 5\lambda} \quad \text{and} \quad \beta = \frac{+11}{10 - 5\lambda} \]
Now solving for \(\alpha + \beta = -2\):
\[ \frac{10}{10 - 5\lambda} = -2 \quad \Rightarrow \quad 10 - 5\lambda = -5 \quad \Rightarrow \quad \lambda = 3 \]
Therefore:
\[ \alpha = \frac{1}{5} \quad \text{and} \quad \beta = \frac{-11}{5} \]
Finally, we calculate:
\[ 4\alpha^2 + \beta^2 + \lambda^2 = \frac{4}{25} + \frac{121}{25} + 3^2 = 14 \]
The final answer is 14.
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