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if A is a $3$x$3$ matrix and let A=$2$,then what will be the value of det(adj(adj(adj($A^{-1}$)))?

1.$\dfrac{1}{512}$

2.$\dfrac{1}{1024}$

3.$\dfrac{1}{128}$

4.$\dfrac{1}{256}$

amit
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2 Answers2

1

Assume you meant $det(A)=2$. Then $det(A^{-1})=\frac{1}{2}$. Work the problem outward: \begin{align*} adj(A^{-1})&=det(A^{-1})A=\frac{1}{2}A,\\ adj(\frac{1}{2}A)&=(1/2)^2 adj(A)=\frac{1}{4}det(A)A^{-1}=\frac{1}{2}A^{-1},\\ adj(\frac{1}{2}A^{-1})&=(1/2)^2adj(A^{-1})=\frac{1}{4}det(A^{-1})A=\frac{1}{8}A,\\ det(\frac{1}{8}A)&=(1/8)^3det(A)=\frac{1}{512}2=\frac{1}{256}. \end{align*} For justifications of the steps, you can wiki "adjugate matrix" and "determinant."

Kim Jong Un
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To find: $det(adj(adj(adj(A^{-1})))$

Now denote $det(A)$ as $|A|$.

Note that by adjoint rule we must have: $$A.adj(A)=adj(A).A=|A|I_n.$$ In case, if $A$ be non singular, then deduction comes out as $|adj(A)|=|A|^{n-1}$. Hence $|adj(adj(A))|=|adj(A)|^{n-1}=(|A|^{n-1})^{n-1}=|A|^{(n-1)^2}.$ So $$|adj(adj(adj(A)))|=|adj(adj(A))|^{n-1}=|A|^{(n-1)^3}$$.

Therefore, $$|adj(adj(adj(A^{-1})))|=|A^{-1}|^{(n-1)^3}=|A|^{-(n-1)^3}$$

Since $n=3$ and $|A|=2$ so we have $$|adj(adj(adj(A^{-1})))|=2^{-8}=\frac{1}{256}$$ so that option 4 is correct

KON3
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