Simplify the Matrix

Problem

[[0,−2*i,−3]]*[[2*i,0,4],[0,i,5−i],[−4,−5−i,4*i]]*[[0],[2*i],[−3]]

Solution

1. Multiply the row vector by the 3×3 matrix to find the intermediate row vector.

[[0,−2*i,−3]]*[[2*i,0,4],[0,i,5−i],[−4,−5−i,4*i]]=[[(R_1),(R_2),(R_3)]]

(R_1)=(0)*(2*i)+(−2*i)*(0)+(−3)*(−4)=12

(R_2)=(0)*(0)+(−2*i)*(i)+(−3)*(−5−i)=−2*i2+15+3*i=2+15+3*i=17+3*i

(R_3)=(0)*(4)+(−2*i)*(5−i)+(−3)*(4*i)=−10*i+2*i2−12*i=−22*i−2

2. Combine the results into the intermediate row vector.

[[12,17+3*i,−2−22*i]]

3. Multiply the intermediate row vector by the final column vector.

[[12,17+3*i,−2−22*i]]*[[0],[2*i],[−3]]=(12)*(0)+(17+3*i)*(2*i)+(−2−22*i)*(−3)

4. Simplify the resulting expression using the property i2=−1

0+(34*i+6*i2)+(6+66*i)

34*i−6+6+66*i

100*i

Final Answer

[[0,−2*i,−3]]*[[2*i,0,4],[0,i,5−i],[−4,−5−i,4*i]]*[[0],[2*i],[−3]]=100*i

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