Adjoint and determinant

Linear Algebra
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Grigorios Kostakos
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Adjoint and determinant

#1

Post by Grigorios Kostakos »

For an invertible \(n\times n\) real matrix \(A\) (\(n\geqslant2\)) prove that:

1) \(|{\rm{adj}}A|=|A|^{n-1}\,,\)

2) \({\rm{adj}}({\rm{adj}}A)=|A|^{n-2}A\,,\)

where \(|A|\) is the determinant and \({\rm{adj}}A\) is the adjoint matrix of the matrix \(A\).
Grigorios Kostakos
Papapetros Vaggelis
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Re: Adjoint and determinant

#2

Post by Papapetros Vaggelis »

Let \(\displaystyle{A}\) be an invertible \(\displaystyle{n\times n\,\,,n\geq 2}\), real matrix. We know that \(\displaystyle{\left|A\right|\neq 0}\) .

For the adjoint matrix, \(\displaystyle{adj\,A}\), of the matrix \(\displaystyle{A}\), we have that

\(\displaystyle{A\cdot adj\,A=\left|A\right|\cdot I_{n}=adj\,A\cdot A\,\,(I)}\).

1)

\(\displaystyle{\left|adj\,A\cdot A\right|=\left|\left|A\right|\cdot I_{n}\right|\Rightarrow \left|adj\,A\right|\cdot \left|A\right|=\left|A\right|^{n}\cdot 1\Rightarrow \left|adj\,A\right|=\left|A\right|^{n-1}\neq 0}\) .

2) From \(\displaystyle{(I)}\),

\(\displaystyle{\left(\left|A\right|^{-1}\cdot A\right)\cdot adj\,A=I_{n}=adj\,A\cdot\,\left(\left|A\right|^{-1}\cdot A\right)}\)

so,

the matrix \(\displaystyle{adj\,A}\) is invertible and \(\displaystyle{\left(adj\,A\right)^{-1}=\left|A\right|^{-1}\cdot A}\).

Moreover,

\(\displaystyle{adj\,A\cdot adj\,\left(adj\,A\right)=\left|adj\,A\right|\cdot I_{n}=adj\,\left(adj\,A\right)\cdot adj\,A}\) , so

\(\displaystyle{\left(adj\,A\right)^{-1}\,\left(adj\,A\cdot adj\,\left(adj\,A\right)\right)=\left(adj\,A\right)^{-1}\cdot \left|A\right|^{n-1}\cdot I_{n}\Rightarrow}\)

\(\displaystyle{\left[\left(adj\,A\right)^{-1}\cdot adj\,A\right]\cdot adj\,\left(adj\,A\right)=\left|A\right|^{n-1}\cdot \left|A\right|^{-1}\cdot A\cdot I_{n}\Rightarrow}\)

\(\displaystyle{ I_{n}\cdot adj\,\left(adj\,A\right)=\left|A\right|^{n-2}\cdot A\Rightarrow adj\,\left(adj\,A\right)=\left|A\right|^{n-2}\cdot A}\)
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Grigorios Kostakos
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Re: Adjoint and determinant

#3

Post by Grigorios Kostakos »

Slightly different solutions :

For an invertible \(n\times n\) real matrix \(A\) (\(n\geqslant2\)) we have that \(A^{-1}=|A|^{-1}{\rm{adj}}A\) and \(|A^{-1}|=|A|^{-1}\). So:

\(\begin{aligned}
1)\quad |A|^{-1}=|A^{-1}|=\bigl|{|A|^{-1}{\rm{adj}}A}\bigr|&=(|A|^{-1})^{n}|{{\rm{adj}}A}|=|A|^{-n}|{{\rm{adj}}A}|\quad\Rightarrow\\
&|{\rm{adj}}A|=|A|^{n-1}\quad(1)\,.
\end{aligned}\)

\(\begin{aligned}
2)\quad A&=(A^{-1})^{-1}\\
&=\bigl({|A|^{-1}{\rm{adj}}A}\bigr)^{-1}\\
&=(|A|^{-1})^{-1}\bigl({{\rm{adj}}A}\bigr)^{-1}\\
&=|A|\,\bigl|{{\rm{adj}}A}\bigr|^{-1}{\rm{adj}}({\rm{adj}}A)\\
&\stackrel{(1)}{=}|A|\,|A|^{-(n-1)}\,{\rm{adj}}({\rm{adj}}A)\\
&=|A|^{-(n-2)}\,{\rm{adj}}({\rm{adj}}A)\quad\Rightarrow\\
&{\rm{adj}}({\rm{adj}}A)=|A|^{n-2}A\,.
\end{aligned}\)
Grigorios Kostakos
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