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

Problem

[[2,2],[−1+3*i,−1−3*i]]

Solution

1. Identify the matrix A and its elements.

A=[[a,b],[c,d]]=[[2,2],[−1+3*i,−1−3*i]]

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

det(A)=(2)*(−1−3*i)−(2)*(−1+3*i)

3. Simplify the determinant expression.

det(A)=−2−6*i−(−2+6*i)

det(A)=−2−6*i+2−6*i

det(A)=−12*i

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

A(−1)=1/(−12*i)*[[−1−3*i,−2],[−(−1+3*i),2]]

5. Simplify the scalar 1/(−12*i) by multiplying the numerator and denominator by i

1/(−12*i)⋅i/i=i/(−12*i2)=i/12

6. Distribute the scalar i/12 into each element of the matrix.

A(−1)=[[(i*(−1−3*i))/12,(−2*i)/12],[(i*(1−3*i))/12,(2*i)/12]]

7. Perform the final arithmetic for each entry.

(i*(−1−3*i))/12=(−i−3*i2)/12=(3−i)/12

(i*(1−3*i))/12=(i−3*i2)/12=(3+i)/12

Final Answer

[[2,2],[−1+3*i,−1−3*i]](−1)=[[(3−i)/12,−i/6],[(3+i)/12,i/6]]

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