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Suppose that the eigenvalues of matrix $A$ are $1, 2, 4$. The determinant of $\left(A^{-1}\right)^{T}$ is _________.
asked in Linear Algebra by Boss (43.6k points)
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2 Answers

+40 votes
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Determinant of Matrix A = product of eigen values = $1 \times 2 \times 4 =8$

Determinant of Inverse Matrix of A, $\text{det}(A^{-1}) = \frac{1}{\text{det}(A)} =\frac{1}{8}$

Determinant remains same after the Transpose

So, determinant of $(A^{-1})^T$ = $\text{det}(A^{-1}) = \frac{1}{8}$ = 0.125
answered by Veteran (55.9k points)
edited by
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Inverse of a matrix comprises of transpose of cofactor matrix divided by det of that matrix, so do we not need to consider the value of transpose of cofactor matrix turned into determinant? pls explain.
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+16 votes
0.125

Eigen value of A inverse is 1,1/2,1/4. Product of those eigen values gives determinant value.

Transposing a matrix doesn't change is eigen value.
answered by Junior (691 points)
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