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Question Number 100296 by bobhans last updated on 26/Jun/20
![Given matrix A = [((3 1 4)),((1 2 5)),((0 2 6)) ] find: adj(adj A) ?](Q100296.png)
$$\mathcal{G}\mathrm{iven}\:\mathrm{matrix}\:\mathrm{A}\:=\:\begin{bmatrix}{\mathrm{3}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\mathrm{4}}\\{\mathrm{1}\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\mathrm{5}}\\{\mathrm{0}\:\:\:\:\:\mathrm{2}\:\:\:\:\:\mathrm{6}}\end{bmatrix} \\ $$$$\mathrm{find}:\:\:\mathrm{adj}\left(\mathrm{adj}\:\mathrm{A}\right)\:? \\ $$
Answered by bemath last updated on 26/Jun/20
![adj(adj A) = ∣A∣^(n−1) .A A_(3×3) ⇒∣A∣ = 3(12−10)−1(6−0)+4(2−0) ∣A∣ = 6−6+8 = 8 ⇔adj(adj A) = (8)^(3−1) .A = 64 [((3 1 4)),((1 2 5)),((0 2 6)) ]](Q100300.png)
$$\mathrm{adj}\left(\mathrm{adj}\:\mathrm{A}\right)\:=\:\mid\mathrm{A}\mid^{\mathrm{n}−\mathrm{1}} \:.\mathrm{A} \\ $$$$\mathrm{A}_{\mathrm{3}×\mathrm{3}} \:\Rightarrow\mid\mathrm{A}\mid\:=\:\mathrm{3}\left(\mathrm{12}−\mathrm{10}\right)−\mathrm{1}\left(\mathrm{6}−\mathrm{0}\right)+\mathrm{4}\left(\mathrm{2}−\mathrm{0}\right) \\ $$$$\mid\mathrm{A}\mid\:=\:\mathrm{6}−\mathrm{6}+\mathrm{8}\:=\:\mathrm{8} \\ $$$$\Leftrightarrow\mathrm{adj}\left(\mathrm{adj}\:\mathrm{A}\right)\:=\:\left(\mathrm{8}\right)^{\mathrm{3}−\mathrm{1}} \:.\mathrm{A} \\ $$$$=\:\mathrm{64}\:\begin{bmatrix}{\mathrm{3}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\mathrm{4}}\\{\mathrm{1}\:\:\:\:\:\mathrm{2}\:\:\:\:\:\mathrm{5}}\\{\mathrm{0}\:\:\:\:\:\mathrm{2}\:\:\:\:\:\mathrm{6}}\end{bmatrix} \\ $$$$ \\ $$