Let A be a 3 × 3 matrix and det(A) = 2 . If n = det( adj(adj(. ......

Given,

$\left|A\right| = 2$

And $n = det\left(\mathop{\mathop{adj\left(adj\left(adj . .. \left(A\right)\right)\right)}_{⏟}}_{2024times}\right)$

Now, using the formula $\mathop{\mathop{\left|adj\left(adj\left(adj . .. \left(A\right)\right)\right)\right|}_{⏟}}_{rtimes} = {\left|A\right|}^{{\left(n - 1\right)}^{r}}$ we get,

$\mathop{\left|\mathop{adj\left(adj\left(adj . .. \left(A\right)\right)\right)}_{⏟}\right|}_{2024times} = {\left|A\right|}^{{\left(n - 1\right)}^{2024}}$

$⇒ n = {\left|A\right|}^{{2}^{2024}}$

$⇒ n = {2}^{{2}^{2024}}$ \left\{\text{as}\left|A\right| = 2\right\}$</p><p>Now, solving</p><p>${2}^{2024} = {4}^{1012} = {\left(3 + 1\right)}^{1012} = 3k + 1$</p><p>$⇒ {2}^{{2}^{2024}} = {2}^{3k + 1}$</p><p>$⇒ {2}^{{2}^{2024}} = 2 \cdot {2}^{3k}$</p><p>$⇒ {2}^{{2}^{2024}} = 2 \cdot {\left(9 - 1\right)}^{k}$</p><p>$⇒ {2}^{{2}^{2024}} = 9m - 2$</p><p>$⇒ {2}^{{2}^{2024}} = 9t - 2 + 9$</p><p>$⇒ {2}^{{2}^{2024}} = 9m + 7$</p><p>Hence, the remainder is$7$