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in Matrices & determinants by (50 points)
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A Let \( S=\{\sqrt{n}: 1 \leqslant n \leqslant 50 \) and \( n \) is odd \( \} \). Let \( a \in S \) and \( A =\left[\begin{array}{rrr}1 & 0 & a \\ -1 & 1 & 0 \\ - a & 0 & 1\end{array}\right] \). If \( \Sigma \operatorname{det}(\operatorname{adj} A)=100 \lambda \), then \( \lambda \) is equal to :

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\(S = \{\sqrt n;1\le n \le50\} = a\in S\{\sqrt 1, \sqrt3, \sqrt 5,....,\sqrt{49}\}\)

\(A = \begin{bmatrix}-1 &0 &a\\-1&1&0\\-a&0&1\end{bmatrix}\)

\(adj\,A = \begin{bmatrix}1 &0 &-a\\1&a^2-1&-a\\a&0&-1\end{bmatrix}\)

\(det(adj\,A) = \begin{bmatrix}1 &0 &-a\\1&a^2-1&-a\\a&0&-1\end{bmatrix}\)

\(= (a^2 - 1)\begin{vmatrix}1&-a\\a&-1\end{vmatrix}\)

\(= (a^2 - 1) (-1 + a^2)\)

\(= (a^2 - 1)^2\)

\(\because a\in \{\sqrt 1, \sqrt3, \sqrt 5,....,\sqrt{49}\}\)

\(\therefore a^2\in \{1,3,5,....,49\}\)

\(a^2 - 1\in \{0,2,4,,....,48\}\)

\(\therefore \sum det (adj \, A) = 0^2 + 2^2 + 4^2 + .... +48^2\)

\(= 2^2 (1^2 + 2^2 + 3^2 + .... + 24^2)\)

\(= 4\left(\frac{24(24+1)(48 +1)}{6}\right)\)

\(= 16 \times 25 \times 49\)

\(= 19600\)

\(= 196 \times 100\)

\(\therefore \lambda = 196\)

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