If A = begin{bmatrix} 1 & 5 lambda & 10 end{ | vedclass

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(D) By the Cayley-Hamilton theorem,every square matrix satisfies its characteristic equation $|A - xI| = 0$.$|A - xI| = \begin{vmatrix} 1 - x & 5 \ \

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$14$

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(D) By the Cayley-Hamilton theorem,every square matrix satisfies its characteristic equation $|A - xI| = 0$.$|A - xI| = \begin{vmatrix} 1 - x & 5 \ \lambda & 10 - x \end{vmatrix} = (1 - x)(10 - x) - 5\lambda = x^2 - 11x + 10 - 5\lambda = 0$.Thus,$A^2 - 11A + (10 - 5\lambda)I = 0$.Multiplying by $A^{-1}$,we get $A - 11I + (10 - 5\lambda)A^{-1} = 0$.$(10 - 5\lambda)A^{-1} = -A + 11I$.$A^{-1} = \frac{-1}{10 - 5\lambda}A + \frac{11}{10 - 5\lambda}I$.Comparing this with $A^{-1} = \alpha A + \beta I$,we have $\alpha = \frac{-1}{10 - 5\lambda}$ and $\beta = \frac{11}{10 - 5\lambda}$.Given $\alpha + \beta = -2$,so $\frac{-1 + 11}{10 - 5\lambda} = -2 \Rightarrow \frac{10}{10 - 5\lambda} = -2$.$10 = -20 + 10\lambda \Rightarrow 10\lambda = 30 \Rightarrow \lambda = 3$.Then $\alpha = \frac{-1}{10 - 15} = \frac{-1}{-5} = \frac{1}{5}$ and $\beta = \frac{11}{10 - 15} = \frac{11}{-5} = -\frac{11}{5}$.Finally,$4\alpha^2 + \beta^2 + \lambda^2 = 4(\frac{1}{25}) + (\frac{121}{25}) + 3^2 = \frac{4 + 121}{25} + 9 = \frac{125}{25} + 9 = 5 + 9 = 14$.

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:

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