Find the Inverse [[1,-2],[3,-4]]^3

Problem

([[1,−2],[3,−4]]3)(−1)

Solution

1. Identify the property of matrix inverses that allows the exponent and the inverse to be swapped.

(An)(−1)=(A(−1))n

2. Calculate the determinant of the matrix A=[[1,−2],[3,−4]]

det(A)=(1)*(−4)−(−2)*(3)

det(A)=−4+6=2

3. Find the inverse of the matrix A using the formula A(−1)=1/det(A)*[[d,−b],[−c,a]]

A(−1)=1/2*[[−4,2],[−3,1]]

A(−1)=[[−2,1],[−1.5,0.5]]

4. Compute the square of the inverse matrix A(−2)

A(−2)=[[−2,1],[−1.5,0.5]]*[[−2,1],[−1.5,0.5]]

A(−2)=[[(−2)*(−2)+(1)*(−1.5),(−2)*(1)+(1)*(0.5)],[(−1.5)*(−2)+(0.5)*(−1.5),(−1.5)*(1)+(0.5)*(0.5)]]

A(−2)=[[2.5,−1.5],[2.25,−1.25]]

5. Compute the cube of the inverse matrix A(−3) by multiplying A(−2) by A(−1)

A(−3)=[[2.5,−1.5],[2.25,−1.25]]*[[−2,1],[−1.5,0.5]]

A(−3)=[[(2.5)*(−2)+(−1.5)*(−1.5),(2.5)*(1)+(−1.5)*(0.5)],[(2.25)*(−2)+(−1.25)*(−1.5),(2.25)*(1)+(−1.25)*(0.5)]]

A(−3)=[[−5+2.25,2.5−0.75],[−4.5+1.875,2.25−0.625]]

A(−3)=[[−2.75,1.75],[−2.625,1.625]]

Final Answer

([[1,−2],[3,−4]]3)(−1)=[[−2.75,1.75],[−2.625,1.625]]

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