Let S= n: 1 n50, n odd.

Question

Let $S=\{\sqrt n: 1\le n\le50,\ n	ext{ odd}\}$. Let $a\in S$ and $A=\begin{pmatrix}1&0&a\-1&1&0\-a&0&1\end{pmatrix}$. If $\displaystyle\sum_{a\in S}\det(\operatorname{adj}A)=100\lambda$, then $\lambda$ is equal to:

Accepted Answer

$\det(A)=1(1-0)-0+a(0+a)=1+a^2$, so for a $3\times3$ matrix, $\det(\operatorname{adj}A)=(\det A)^2=(1+a^2)^2$.

For $a=\sqrt n$ with $n$ odd, $1\le n\le49$: $\det(\operatorname{adj}A)=(1+n)^2$.

$$\sum_{n\text{ odd},1\le n\le49}(1+n)^2 = \sum_{k=1}^{25}(2k)^2 = 4\sum_{k=1}^{25}k^2=4\cdot\frac{25\cdot26\cdot51}6=4\times5525=22100.$$

So $100\lambda=22100\Rightarrow\lambda=221$.

Solutions · 1