What will be the value of $2(|AB-I_5|+|BA-I_4|) +1 ?$
I am stuck , can't even start the problem, is there some identify which I am missing?
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@ajotatxe edited – Piesquareisg Jan 18 '18 at 18:44
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1@vikrant Try some matrices – For the love of maths Jan 18 '18 at 18:46
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@MohammadZuhairKhan ok – Piesquareisg Jan 18 '18 at 18:48
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Try those with $1$s in their diagonal first, they are generally easier. – For the love of maths Jan 18 '18 at 18:51
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1The two $I$ need different sizes. – coffeemath Jan 18 '18 at 18:51
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What do the brackets mean? – Hans Lundmark Jan 18 '18 at 18:56
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1The value of matrix (as we calculate of determinants) – Piesquareisg Jan 18 '18 at 18:57
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This is impossible since $AB$ is $5\times 5$ and $BA$ is $4\times 4$ – Mostafa Ayaz Jan 18 '18 at 19:02
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Check out the Cauchy-Binet formula. You will find that $|AB-I_5| = - |BA-I_4|$, so that the original formula always evaluates to 1. – ChocolateAndCheese Jan 19 '18 at 00:03
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You are missing the Sylvester's determinant identity. Following Wikipedia, suppose $A$, $B$, $C$, and $D$ are matrices of dimension $n × n$, $n × m$, $m × n$, and $m × m$, respectively. When $A$ is invertible, one has $$ \det {\begin{pmatrix}A&B\\C&D\end{pmatrix}}=\det(A)\det(D-CA^{-1}B), $$ and if $D$ is invertible $$ \det {\begin{pmatrix}A&B\\C&D\end{pmatrix}}=\det(D)\det(A-BD^{-1}C). $$
Using the matrices from your question with the identities above: $$ \det {\begin{pmatrix}1_4&B\\A&1_5\end{pmatrix}}=\det(1_5-AB)=\det(1_4-BA). $$
Recall that $\det(cM)=c^n\det(M)$ for a constant $c$ and an $n\times n$ matrix $M$, then $$ \det(1_5-AB)=(-1)^5\det(AB-1_5)=\det(1_4-BA)=(-1)^4\det(BA-1_4), $$ or simply $$ \det(AB-1_5)=-\det(BA-1_4). $$
From this last identity, the result to your question is immediate: 1.
There are already many questions about the Sylvester's identity, as the following:
Joca Ramiro
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