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Suppose that the eigenvalues of matrix $A$ are $1, 2, 4$. The determinant of $\left(A^{-1}\right)^{T}$ is _________.

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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 operation.

So, determinant of $(A^{-1})^T = \text{det}(A^{-1}) = \frac{1}{8} = 0.125$
by Veteran (57.1k points)
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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.
0

+1

We can use any of these 2 properties:

1) det(A$^{-1}$) = 1 / (det(A))

Proof:

det(AA$^{-1}$) = det(I)

i;e

det(A) det(A$^{-1}$) = 1

det(A$^{-1}$) = 1 / (det(A))

2) if $\lambda$ is the eigen value of A, then (1/$\lambda$) is the eigen value of A$^{-1}$

Proof: (source: stackoverflow)

Av=λv

Pre multiply by A$^{-1}$, we have

⟹A$^{-1}$Av=λA$^{-1}$v

⟹Iv=λA$^{-1}$v

⟹(1/λ)v=A$^{-1}$v

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.
by Junior (699 points)