Matrix/C/2+5i 1-2i 3-4i 6-2i/Invertible/Solution for 54+72i 0/Exercise/Solution

a) We compute the determinant of the matrix. It is

{}{\begin{aligned}\det M&=(2+5{\mathrm {i} })(6-2{\mathrm {i} })-(3-4{\mathrm {i} })(1-2{\mathrm {i} })\\&=12+10+30{\mathrm {i} }-4{\mathrm {i} }-(3-8-4{\mathrm {i} }-6{\mathrm {i} })\\&=27+36{\mathrm {i} }.\end{aligned}}

In particular the matrix is invertible.

b) We have

{}{\begin{aligned}{\begin{pmatrix}54+72{\mathrm {i} }\\0\end{pmatrix}}&=2{\begin{pmatrix}\det M\\0\end{pmatrix}}\\\end{aligned}}

We can see directly a solution, namely

{}{\begin{pmatrix}z_{1}\\z_{2}\end{pmatrix}}=2{\begin{pmatrix}6-2{\mathrm {i} }\\-(3-4{\mathrm {i} })\end{pmatrix}}={\begin{pmatrix}12-4{\mathrm {i} }\\-6+8{\mathrm {i} }\end{pmatrix}}\,,

because we have

{}{\begin{aligned}{\begin{pmatrix}2+5{\mathrm {i} }&1-2{\mathrm {i} }\\3-4{\mathrm {i} }&6-2{\mathrm {i} }\end{pmatrix}}{\begin{pmatrix}2(6-2{\mathrm {i} })\\2(-3+4{\mathrm {i} })\end{pmatrix}}&=2{\begin{pmatrix}(2+5{\mathrm {i} })(6-2{\mathrm {i} })+(1-2{\mathrm {i} })(-3+4{\mathrm {i} })\\(3-4{\mathrm {i} })(6-2{\mathrm {i} })+(6-2{\mathrm {i} })(-3+4{\mathrm {i} })\end{pmatrix}}\\&={\begin{pmatrix}2\det M\\0\end{pmatrix}}.\end{aligned}}