Let alpha=a+bi and beta=c+di be complex scalars and let A and B be matrices with complex entries.

BenoguigoliB 2021-09-16 Answered

Let \(\displaystyle\alpha={a}+{b}{i}\ \text{ and }\ \beta={c}+{d}{i}\) be complex scalars and let A and B be matrices with complex entries.
(a) Show that
\(\displaystyle\overline{{\alpha+\beta}}=\overline{{\alpha}}+\overline{{\beta}}\ \text{ and }\ \overline{{\alpha\beta}}=\overline{{\alpha}}\overline{{\beta}}\)
(b) Show that the (i,j) entries of\( \overline{AB} \text{ and } \bar{A}\bar{B}\) are equal and hence that
\(\displaystyle\overline{{{A}{B}}}=\overline{{{A}}}\overline{{{B}}}\)

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Expert Answer

Brittany Patton
Answered 2021-09-17 Author has 18187 answers
Step 1
(a)
Let \(\displaystyle\alpha={a}+{b}{i}\ \text{ and }\ \beta={c}+{d}{i}\)
\(\displaystyle\overline{{\alpha+\beta}}=\overline{{{\left({a}+{b}{i}\right)}+{\left({c}+{d}{i}\right)}}}\)
\(\displaystyle=\overline{{{\left({a}+{c}\right)}+{i}{\left({b}+{d}\right)}}}\)
\(\displaystyle={\left({a}+{c}\right)}-{i}{\left({b}+{d}\right)}\)
\(\displaystyle={a}+{c}-{i}{b}-{i}{d}\)
\(\displaystyle={\left({a}-{i}{b}\right)}+{\left({c}-{i}{d}\right)}\)
\(\displaystyle=\overline{{\alpha}}+\overline{{\beta}}\)
Similarly,
\(\displaystyle\overline{{\alpha\beta}}=\overline{{{\left({a}+{b}{i}\right)}{\left({c}+{d}{i}\right)}}}\)
\(\displaystyle=\overline{{{a}{c}-{b}{d}+{i}{\left({b}{c}+{a}{d}\right)}}}\)
\(\displaystyle={a}{c}-{b}{d}-{i}{\left({b}{c}+{a}{d}\right)}\)
\(\displaystyle={\left({a}-{i}{b}\right)}{\left({c}-{i}{d}\right)}\)
\(\displaystyle=\overline{{\alpha}}\overline{{\beta}}\)
Thus, \(\displaystyle\overline{{\alpha+\beta}}=\overline{{\alpha}}+\overline{{\beta}}\ \text{ and }\ \overline{{\alpha\beta}}=\overline{{\alpha}}\overline{{\beta}}\)
Step 2
(b)
Let A nad B are the matrices whose entries are from complex numbers. (i,j)th entry of AB is \(\displaystyle{a}_{{{i}{1}}}{b}_{{{1}{j}}}+{a}_{{{i}{2}}}{b}_{{{2}{j}}}+\dot{{s}}+{a}_{{\in}}{b}_{{{n}{j}}}\)
From part (a),
\(\displaystyle\overline{{{a}_{{{i}{1}}}{b}_{{{1}{j}}}+{a}_{{{i}{2}}}{b}_{{{2}{j}}}+\dot{{s}}+{a}_{{\in}}{b}_{{{n}{j}}}}}=\overline{{{a}_{{{i}{1}}}{b}_{{{1}{j}}}}}+\overline{{{a}_{{{i}{2}}}{b}_{{{2}{j}}}}}+\dot{{s}}+\overline{{{a}_{{\in}}{b}_{{{n}{j}}}}}\)
\(\displaystyle=\overline{{{a}_{{{i}{1}}}}}\overline{{{b}_{{{1}{j}}}}}+\overline{{{a}_{{{i}{2}}}}}\overline{{{b}_{{{2}{j}}}}}+\dot{{s}}+\overline{{{a}_{{\in}}}}\overline{{{b}_{{{n}{j}}}}}\)
This implies that i,j) th entry of \(\displaystyle\overline{{{A}{B}}}\ \text{ and }\ \overline{{{A}}}\overline{{{B}}}\) are same.
It is known that if each of entry of two matrices are same then they are equal.
Hence, \(\displaystyle\overline{{{A}{B}}}=\overline{{{A}}}\overline{{{B}}}\)
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