Es ist

${\displaystyle {}{\begin{aligned}\det {\begin{pmatrix}0&3-4{\mathrm {i} }&2{\mathrm {i} }\\2-3{\mathrm {i} }&4+7{\mathrm {i} }&3-{\mathrm {i} }\\0&1-5{\mathrm {i} }&-4-{\mathrm {i} }\end{pmatrix}}&=-{\left(2-3{\mathrm {i} }\right)}\det {\begin{pmatrix}3-4{\mathrm {i} }&2{\mathrm {i} }\\1-5{\mathrm {i} }&-4-{\mathrm {i} }\end{pmatrix}}\\&={\left(-2+3{\mathrm {i} }\right)}{\left({\left(3-4{\mathrm {i} }\right)}{\left(-4-{\mathrm {i} }\right)}-2{\mathrm {i} }{\left(1-5{\mathrm {i} }\right)}\right)}\\&={\left(-2+3{\mathrm {i} }\right)}{\left(-16+13{\mathrm {i} }-2{\mathrm {i} }-10\right)}\\&={\left(-2+3{\mathrm {i} }\right)}{\left(-26+11{\mathrm {i} }\right)}\\&=19-100{\mathrm {i} }.\end{aligned}}}$