Multiply the Matrices

Problem

[[2*e^x,−e^(−x),e^(−3*x)],[−e^x,−e^(−x),−2*e^(−3*x)],[3*e^x,6*e^(−x),6*e^(−3*x)]]*[[2/3*e^(−x),4/3*e^(−x),1/3*e^(−x)],[0,e^x,1/3*e^x],[−1/3*e^(3*x),−15/9*e^(3*x),−1/3*e^(3*x)]]

Solution

1. Identify the dimensions of the matrices. Both are 3×3 matrices, so the resulting product will also be a 3×3 matrix.

2. Calculate the first row of the product by multiplying the first row of the left matrix by each column of the right matrix.

(R_1)*(C_1)=(2*e^x)*(2/3*e^(−x))+(−e^(−x))*(0)+(e^(−3*x))*(−1/3*e^(3*x))=4/3+0−1/3=1

(R_1)*(C_2)=(2*e^x)*(4/3*e^(−x))+(−e^(−x))*(e^x)+(e^(−3*x))*(−15/9*e^(3*x))=8/3−1−5/3=0

(R_1)*(C_3)=(2*e^x)*(1/3*e^(−x))+(−e^(−x))*(1/3*e^x)+(e^(−3*x))*(−1/3*e^(3*x))=2/3−1/3−1/3=0

3. Calculate the second row of the product.

(R_2)*(C_1)=(−e^x)*(2/3*e^(−x))+(−e^(−x))*(0)+(−2*e^(−3*x))*(−1/3*e^(3*x))=−2/3+0+2/3=0

(R_2)*(C_2)=(−e^x)*(4/3*e^(−x))+(−e^(−x))*(e^x)+(−2*e^(−3*x))*(−15/9*e^(3*x))=−4/3−1+10/3=1

(R_2)*(C_3)=(−e^x)*(1/3*e^(−x))+(−e^(−x))*(1/3*e^x)+(−2*e^(−3*x))*(−1/3*e^(3*x))=−1/3−1/3+2/3=0

4. Calculate the third row of the product.

(R_3)*(C_1)=(3*e^x)*(2/3*e^(−x))+(6*e^(−x))*(0)+(6*e^(−3*x))*(−1/3*e^(3*x))=2+0−2=0

(R_3)*(C_2)=(3*e^x)*(4/3*e^(−x))+(6*e^(−x))*(e^x)+(6*e^(−3*x))*(−15/9*e^(3*x))=4+6−10=0

(R_3)*(C_3)=(3*e^x)*(1/3*e^(−x))+(6*e^(−x))*(1/3*e^x)+(6*e^(−3*x))*(−1/3*e^(3*x))=1+2−2=1

Final Answer

[[2*e^x,−e^(−x),e^(−3*x)],[−e^x,−e^(−x),−2*e^(−3*x)],[3*e^x,6*e^(−x),6*e^(−3*x)]]*[[2/3*e^(−x),4/3*e^(−x),1/3*e^(−x)],[0,e^x,1/3*e^x],[−1/3*e^(3*x),−15/9*e^(3*x),−1/3*e^(3*x)]]=[[1,0,0],[0,1,0],[0,0,1]]

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