Given that
A=⎣⎡124120−130⎦⎤ and B=⎣⎡124120−130⎦⎤
Then
AT=⎣⎡11−1223400⎦⎤ and BT=⎣⎡11−1223400⎦⎤
det(A)=∣∣124120−130∣∣
det(A)=(1)(0−0)−(1)(0−12)+(−1)(0−8)
det(A)=0+12+8
det(A)=20
And
det(B)=∣∣124120−130∣∣
det(B)=(1)(0−0)−(1)(0−12)+(−1)(0−8)
det(B)=0+12+8
det(B)=20
For the matrix A=⎣⎡124120−130⎦⎤
A−1=det(A)adjointA
The adjoint of A is calculated by the cofactor method
A−1=(20)[012−80445−50]
A−1=⎣⎡200201220−8200204204205203200⎦⎤
A−1=⎣⎡00.6−0.400.20.20.25−0.250⎦⎤
Similarly
for the matrix B=⎣⎡124120−130⎦⎤
B−1=det(B)adjointB
The adjoint of B is calculated by the cofactor method
B−1=(20)[012−80445−50]
B−1=⎣⎡200201220−8200204204205203200⎦⎤
B−1=⎣⎡00.6−0.400.20.20.25−0.250⎦⎤
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