Find the Inverse A=[[-3-3i,1-3i],[-3+i,2+4i]]

Problem

A=[[−3−3*i,1−3*i],[−3+i,2+4*i]]

Solution

1. Identify the formula for the inverse of a 2×2 matrix A=[[a,b],[c,d]] which is A(−1)=1/det(A)*[[d,−b],[−c,a]]

2. Calculate the determinant det(A)=a*d−b*c

det(A)=(−3−3*i)*(2+4*i)−(1−3*i)*(−3+i)

3. Expand the products using the property i2=−1

(−3−3*i)*(2+4*i)=−6−12*i−6*i−12*i2=−6−18*i+12=6−18*i

(1−3*i)*(−3+i)=−3+i+9*i−3*i2=−3+10*i+3=10*i

4. Subtract the products to find the final determinant value.

det(A)=(6−18*i)−(10*i)=6−28*i

5. Construct the adjugate matrix by swapping the main diagonal elements and changing the signs of the off-diagonal elements.

adj(A)=[[2+4*i,−1+3*i],[3−i,−3−3*i]]

6. Divide each element of the adjugate matrix by the determinant. To simplify, multiply the numerator and denominator by the conjugate of the determinant, 6+28*i

det(A)⋅(det(A))=6+28=36+784=820

7. Multiply each term in the adjugate by 6+28*i

(2+4*i)*(6+28*i)=12+56*i+24*i−112=−100+80*i

(−1+3*i)*(6+28*i)=−6−28*i+18*i−84=−90−10*i

(3−i)*(6+28*i)=18+84*i−6*i+28=46+78*i

(−3−3*i)*(6+28*i)=−18−84*i−18*i+84=66−102*i

8. Simplify the resulting fractions by dividing by the common denominator 820

(−100+80*i)/820=(−5+4*i)/41

(−90−10*i)/820=(−9−i)/82

(46+78*i)/820=(23+39*i)/410

(66−102*i)/820=(33−51*i)/410

Final Answer

A(−1)=[[(−5+4*i)/41,(−9−i)/82],[(23+39*i)/410,(33−51*i)/410]]

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