Mathematics - Matrices Question with Solution | TestHub

We know that

$\left|Adj A\right|={\left|A\right|}^{n-1}$

Here $n$ is $3\times 3$ matrix

so $\left|Adj A\right|={\left|A\right|}^{3-1}={\left|A\right|}^2$

Now find $\left|A\right|=\left|\begin{array}{ccc}1& 0& a\\ -1& 1& 0\\ -a& 0& 1\end{array}\right|$

$=1+a^2$

Now $\left|Adj A\right|={\left|A\right|}^2={\left(1+a^2\right)}^2$

Now 

$\underset{a\in S}{\Sigma }det\left(adjA\right)=\Sigma {\left(1+a^2\right)}^2=100\lambda$

$\Rightarrow 2^2+4^2...+{48}^2+{50}^2=100\lambda$

$\Rightarrow 4\left(1^2+2^2...+{25}^2\right)=100\lambda$

$\Rightarrow 4\times \frac{25\times 26\times 51}{6}=100\lambda$

$\Rightarrow 221=\lambda$