Simplify the following matrices , then find the multiplication AB A=begin{bmatrix}4i & sqrt{-1} sqrt{2}e^{ipi/4} & 5sin(53.13) sqrt{-9} & sqrt{1} end{bmatrix} and B=begin{bmatrix}ln e^3& log_{x}{x^2}& sqrt{-1}ln e^i i log_{y}{y^{2i}} & 2e^{ipi} & ln e^i end{bmatrix}

Simplify the following matrices , then find the multiplication AB
$A=\left[\begin{array}{cc}4i& \sqrt{-1}\\ \sqrt{2}{e}^{i\pi /4}& 5\mathrm{sin}\left(53.13\right)\\ \sqrt{-9}& \sqrt{1}\end{array}\right]$ and $B=\left[\begin{array}{ccc}\mathrm{ln}{e}^{3}& {\mathrm{log}}_{x}{x}^{2}& \sqrt{-1}\mathrm{ln}{e}^{i}\\ i{\mathrm{log}}_{y}{y}^{2i}& 2{e}^{i\pi }& \mathrm{ln}{e}^{i}\end{array}\right]$
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Step 1
We have given two matrices,
$A=\left[\begin{array}{cc}4i& \sqrt{-1}\\ \sqrt{2}{e}^{i\pi /4}& 5\mathrm{sin}\left(53.13\right)\\ \sqrt{-9}& \sqrt{1}\end{array}\right]$ and $B=\left[\begin{array}{ccc}\mathrm{ln}{e}^{3}& {\mathrm{log}}_{x}{x}^{2}& \sqrt{-1}\mathrm{ln}{e}^{i}\\ i{\mathrm{log}}_{y}{y}^{2i}& 2{e}^{i\pi }& \mathrm{ln}{e}^{i}\end{array}\right]$
These matrices will be simplified by some formulas:
${i}^{2}=-1$
${e}^{ia\pi }=\left(-1{\right)}^{a}$
And in a right angled triagle ,
$\mathrm{sin}\left(53.13\right)=0.8$
So $5\mathrm{sin}\left(53.13\right)=4$
Logarithmic formulas
$\mathrm{ln}{e}^{a}=a$
${\mathrm{log}}_{a}{x}^{b}=b{\mathrm{log}}_{a}x$
${\mathrm{log}}_{x}x=1$
Step 2
So for the matrix A,
$A=\left[\begin{array}{cc}4i& \sqrt{-1}\\ \sqrt{2}{e}^{i\pi /4}& 5\mathrm{sin}\left(53.13\right)\\ \sqrt{-9}& \sqrt{1}\end{array}\right]$
$A=\left[\begin{array}{cc}4i& \sqrt{{i}^{2}}\\ \sqrt{2}\left(-1{\right)}^{1/4}& 4\\ \sqrt{9{i}^{2}}& 1\end{array}\right]$
$A=\left[\begin{array}{cc}4i& i\\ \sqrt{2i}& 4\\ 3i& 1\end{array}\right]$ For matrix B
$B=\left[\begin{array}{ccc}\mathrm{ln}{e}^{3}& {\mathrm{log}}_{x}{x}^{2}& \sqrt{-1}\mathrm{ln}{e}^{i}\\ i{\mathrm{log}}_{y}{y}^{2i}& 2{e}^{i\pi }& \mathrm{ln}{e}^{i}\end{array}\right]$
$B=\left[\begin{array}{ccc}3& 2& i\cdot i\\ i\cdot 2i& 2\cdot \left(-1\right)& i\end{array}\right]$
$B=\left[\begin{array}{ccc}3& 2& -1\\ 2{i}^{2}& -2& i\end{array}\right]$
$B=\left[\begin{array}{ccc}3& 2& {i}^{2}\\ -2& -2& i\end{array}\right]$
Step 3
Now we multiply both the matrices,
$AB=\left[\begin{array}{cc}4i& i\\ \sqrt{2i}& 4\\ 3i& 1\end{array}\right]\left[\begin{array}{ccc}3& 2& {i}^{2}\\ -2& -2& i\end{array}\right]$
$AB=\left[\begin{array}{ccc}12i-2i& 8i-2i& -4i+{i}^{2}\\ 3\sqrt{2i}-8& 2\sqrt{2i}-8& -\sqrt{2i}+4i\\ 9i-2& 6i-2& -3i+i\end{array}\right]$
$AB=\left[\begin{array}{ccc}10i& 6i& -4i+{i}^{2}\\ 3\sqrt{2i}-8& 2\sqrt{2i}-8& -\sqrt{2i}+4i\\ 9i-2& 6i-2& -2i\end{array}\right]$
Jeffrey Jordon