Suppose A is a 3x3 matrix such that det(A) =16.Find the value of det[(4(A^-1)^T]
Taking into account that det(M−1)=1det(M)\det(M^{-1})=\frac{1}{\det (M)}det(M−1)=det(M)1 and det(MT)=det(M)\det(M^T)=\det (M)det(MT)=det(M) for any matrix MMM, we conclude that det((A−1)T)=det(A−1)=1detA=116.\det ((A^{-1})^T)=\det (A^{-1})=\frac{1}{\det A}=\frac{1}{16}.det((A−1)T)=det(A−1)=detA1=161.
Since det(kM)=k3det(M)\det(kM)=k^3\det(M)det(kM)=k3det(M) for any 3×33\times 33×3 matrix MMM,
det(4(A−1)T)=43det((A−1)T)=43116=4.\det (4(A^{-1})^T)=4^3\det ((A^{-1})^T)=4^3\frac{1}{16}=4.det(4(A−1)T)=43det((A−1)T)=43161=4.
Answer: 4.4.4.
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