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David White
David White
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Let $B=A+D$. With $B_1,\dots,B_n$ the columns of $B$, $d_1,\dots,d_n$ the diagonal $D$ $$ \det A=(B_1-d_1e_1)\wedge\dots\wedge (B_n-d_ne_n) $$ so that $\det A$ is an explicit polynomial in $d$, whose constant coefficient is $\det B$ and the term of highest degree is $(-1)^nd_1\dots d_n$. For instance, the coefficient of $d_1$ is $ -e_1\wedge B_2\wedge \dots\wedge B_n, $ and all the coefficients can be expressed explicitelyexplicitly.

Let $B=A+D$. With $B_1,\dots,B_n$ the columns of $B$, $d_1,\dots,d_n$ the diagonal $D$ $$ \det A=(B_1-d_1e_1)\wedge\dots\wedge (B_n-d_ne_n) $$ so that $\det A$ is an explicit polynomial in $d$, whose constant coefficient is $\det B$ and the term of highest degree is $(-1)^nd_1\dots d_n$. For instance, the coefficient of $d_1$ is $ -e_1\wedge B_2\wedge \dots\wedge B_n, $ and all the coefficients can be expressed explicitely.

Let $B=A+D$. With $B_1,\dots,B_n$ the columns of $B$, $d_1,\dots,d_n$ the diagonal $D$ $$ \det A=(B_1-d_1e_1)\wedge\dots\wedge (B_n-d_ne_n) $$ so that $\det A$ is an explicit polynomial in $d$, whose constant coefficient is $\det B$ and the term of highest degree is $(-1)^nd_1\dots d_n$. For instance, the coefficient of $d_1$ is $ -e_1\wedge B_2\wedge \dots\wedge B_n, $ and all the coefficients can be expressed explicitly.

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Bazin
Bazin
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Let $B=A+D$. With $B_1,\dots,B_n$ the columns of $B$, $d_1,\dots,d_n$ the diagonal $D$ $$ \det A=(B_1-d_1e_1)\wedge\dots\wedge (B_n-d_ne_n) $$ so that $\det A$ is an explicit polynomial in $d$, whose constant coefficient is $\det B$ and the term of highest degree is $(-1)^nd_1\dots d_n$. For instance, the coefficient of $d_1$ is $ -e_1\wedge B_2\wedge \dots\wedge B_n, $ and all the coefficients can be expressed explicitely.