Simplify the following matrices , then find the multiplication AB A=[4i−12eiπ/45sin⁡(53.13)−91] and B=[ln⁡e3logx⁡x2−1ln⁡eiilogy⁡y2i2eiπln⁡ei]

Step 1 We have given two matrices, A=[4i−12eiπ/45sin⁡(53.13)−91] and B=[ln⁡e3logx⁡x2−1ln⁡eiilogy⁡y2i2eiπln⁡ei] These matrices will be simplified by some formulas: i2=−1 eiaπ=(−1)a And in a right angled triagle , sin⁡(53.13)=0.8 So 5sin⁡(53.13)=4 Logarithmic formulas ln⁡ea=a loga⁡xb=bloga⁡x logx⁡x=1 Step 2 So for the matrix A, A=[4i−12eiπ/45sin⁡(53.13)−91] A=[4ii22(−1)1/449i21] A=[4ii2i43i1] For matrix B B=[ln⁡e3logx⁡x2−1ln⁡eiilogy⁡y2i2eiπln⁡ei] B=[32i⋅ii⋅2i2⋅(−1)i] B=[32−12i2−2i] B=[32i2−2−2i] Step 3 Now we multiply both the matrices, AB=[4ii2i43i1][32i2−2−2i] AB=[12i−2i8i−2i−4i+i232i−822i−8−2i+4i9i−26i−2−3i+i] AB=[10i6i−4i+i232i−822i−8−2i+4i9i−26i−2−2i]

Answered: "Simplify the following matrices , then find the..."

nicekikah

Answered question

2021-01-02

Simplify the following matrices , then find the multiplication AB

A = [\begin{matrix} 4 i & \sqrt{- 1} \\ \sqrt{2} e^{i π / 4} & 5 \sin (53.13) \\ \sqrt{- 9} & \sqrt{1} \end{matrix}]

and

B = [\begin{matrix} \ln e^{3} & \log_{x} x^{2} & \sqrt{- 1} \ln e^{i} \\ i \log_{y} y^{2 i} & 2 e^{i π} & \ln e^{i} \end{matrix}]

Answer & Explanation

delilnaT

Skilled2021-01-03Added 94 answers

Step 1
We have given two matrices,

A = [\begin{matrix} 4 i & \sqrt{- 1} \\ \sqrt{2} e^{i π / 4} & 5 \sin (53.13) \\ \sqrt{- 9} & \sqrt{1} \end{matrix}]

and

B = [\begin{matrix} \ln e^{3} & \log_{x} x^{2} & \sqrt{- 1} \ln e^{i} \\ i \log_{y} y^{2 i} & 2 e^{i π} & \ln e^{i} \end{matrix}]

These matrices will be simplified by some formulas:

i^{2} = - 1

e^{i a π} = (- 1)^{a}

And in a right angled triagle ,

\sin (53.13) = 0.8

5 \sin (53.13) = 4

Logarithmic formulas

\ln e^{a} = a

\log_{a} x^{b} = b \log_{a} x

\log_{x} x = 1

Step 2
So for the matrix A,

A = [\begin{matrix} 4 i & \sqrt{- 1} \\ \sqrt{2} e^{i π / 4} & 5 \sin (53.13) \\ \sqrt{- 9} & \sqrt{1} \end{matrix}]

A = [\begin{matrix} 4 i & \sqrt{i^{2}} \\ \sqrt{2} (- 1)^{1 / 4} & 4 \\ \sqrt{9 i^{2}} & 1 \end{matrix}]

A = [\begin{matrix} 4 i & i \\ \sqrt{2 i} & 4 \\ 3 i & 1 \end{matrix}]

For matrix B

B = [\begin{matrix} \ln e^{3} & \log_{x} x^{2} & \sqrt{- 1} \ln e^{i} \\ i \log_{y} y^{2 i} & 2 e^{i π} & \ln e^{i} \end{matrix}]

B = [\begin{matrix} 3 & 2 & i \cdot i \\ i \cdot 2 i & 2 \cdot (- 1) & i \end{matrix}]

B = [\begin{matrix} 3 & 2 & - 1 \\ 2 i^{2} & - 2 & i \end{matrix}]

B = [\begin{matrix} 3 & 2 & i^{2} \\ - 2 & - 2 & i \end{matrix}]

Step 3
Now we multiply both the matrices,

A B = [\begin{matrix} 4 i & i \\ \sqrt{2 i} & 4 \\ 3 i & 1 \end{matrix}] [\begin{matrix} 3 & 2 & i^{2} \\ - 2 & - 2 & i \end{matrix}]

A B = [\begin{matrix} 12 i - 2 i & 8 i - 2 i & - 4 i + i^{2} \\ 3 \sqrt{2 i} - 8 & 2 \sqrt{2 i} - 8 & - \sqrt{2 i} + 4 i \\ 9 i - 2 & 6 i - 2 & - 3 i + i \end{matrix}]

A B = [\begin{matrix} 10 i & 6 i & - 4 i + i^{2} \\ 3 \sqrt{2 i} - 8 & 2 \sqrt{2 i} - 8 & - \sqrt{2 i} + 4 i \\ 9 i - 2 & 6 i - 2 & - 2 i \end{matrix}]

Jeffrey Jordon

Expert2022-01-23Added 2605 answers

Answer is given below (on video)

Jeffrey Jordon

Expert2022-08-23Added 2605 answers

Answer is given below (on video)

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