Let A be a 33 invertible matrix.

Question

Let $A$ be a $3	imes3$ invertible matrix. If $|\operatorname{adj}(24A)|=|\operatorname{adj}(3\operatorname{adj}(2A))|$, then $|A|^2$ is equal to:

Accepted Answer

We know, $$|adj\,A| = |A{|^{n - 1}}$$

Now, $$|adj\,24A| = |adj\,3(adj\,2A)|$$

$$ \Rightarrow |24A{|^{3 - 1}} = |3\,adj\,2A{|^{3 - 1}}$$

$$ \Rightarrow |24A{|^2} = |3\,adj\,2A{|^2}$$

Also, we know, $|KA| = {K^n}|A|$

$$ \Rightarrow {\left( {{{(24)}^2}} \right)^2}|A{|^2} = {\left( {{{(3)}^3}} \right)^2}|adj\,2A{|^2}$$

$$ \Rightarrow {(24)^6}|A{|^2} = {3^6}\,.\,{\left( {|2A{|^{3 - 1}}} \right)^2}$$

$$ \Rightarrow {(24)^6}|A{|^2} = {3^6}\,.\,|2A{|^4}$$

$$ \Rightarrow {(24)^6}|A{|^2} = {3^6}\,.\,{\left( {{2^3}} \right)^4}\,.\,|A{|^4}$$

$$ \Rightarrow {3^6}\,.\,{8^6}\,.\,|A{|^2} = {3^6}\,.\,{8^4}\,.\,|A{|^4}$$

$$ \Rightarrow {8^2} = |A{|^2}$$

$$ \Rightarrow |A{|^2} = 64$$ = 26

Solutions · 1