Multiply the Matrices

Problem

[[2*ex,−e(−x),e(−3*x)],[−ex,−e(−x),−2*e(−3*x)],[3*ex,6*e(−x),6*e(−3*x)]]*[[2/3*e(−x),4/3*e(−x),1/3*e(−x)],[0,ex,1/3*ex],[−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*ex)*(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*ex)*(4/3*e(−x))+(−e(−x))*(ex)+(e(−3*x))*(−15/9*e(3*x))=8/3−1−5/3=0

(R_1)*(C_3)=(2*ex)*(1/3*e(−x))+(−e(−x))*(1/3*ex)+(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)=(−ex)*(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)=(−ex)*(4/3*e(−x))+(−e(−x))*(ex)+(−2*e(−3*x))*(−15/9*e(3*x))=−4/3−1+10/3=1

(R_2)*(C_3)=(−ex)*(1/3*e(−x))+(−e(−x))*(1/3*ex)+(−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*ex)*(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*ex)*(4/3*e(−x))+(6*e(−x))*(ex)+(6*e(−3*x))*(−15/9*e(3*x))=4+6−10=0

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

Final Answer

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