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Chapter 4: Problem 7

Prove that \(\operatorname{det}\left(A^{t}\right)=\operatorname{det}(A)\) for any \(A \in \mathrm{M}_{2 \times 2}(F)\).

Short Answer

Expert verified
Given a 2x2 matrix \(A = \begin{bmatrix} a & b\\ c & d \end{bmatrix}\) with \(\operatorname{det}(A) = ad - bc\), its transpose is \(A^{t} = \begin{bmatrix} a & c\\ b & d \end{bmatrix}\) with determinant \(\operatorname{det}\left(A^{t}\right) = ad - cb\). Since \(ad - bc = ad - cb\), we have \(\operatorname{det}\left(A^{t}\right)=\operatorname{det}(A)\) for any \(A \in \mathrm{M}_{2 \times 2}(F)\).

Step by step solution

01

Find the determinant of A

Let \(A \in \mathrm{M}_{2 \times 2}(F)\) be given by: \[A = \begin{bmatrix} a & b\\ c & d \end{bmatrix}\] The determinant of A, denoted as \(\operatorname{det}(A)\), can be computed using the definition: \[\operatorname{det}(A) = ad-bc\]
02

Compute the transpose of A

The transpose of a 2x2 matrix is obtained by interchanging its rows and columns, thus: \[A^{t} = \begin{bmatrix} a & c\\ b & d \end{bmatrix}\]
03

Find the determinant of the transpose of A

To find the determinant of the transpose of A, denoted as \(\operatorname{det}\left(A^{t}\right)\), use the definition of a determinant: \[\operatorname{det}\left(A^{t}\right) = ad - cb\]
04

Establish the equality between the two determinants

We need to show that \(\operatorname{det}\left(A^{t}\right)=\operatorname{det}(A)\). By comparing the expressions for both determinants: \[\operatorname{det}\left(A^{t}\right) = ad - cb = ad - bc=\operatorname{det}(A)\] Since the two determinants are found to be equal, we have proved that \[\operatorname{det}\left(A^{t}\right)=\operatorname{det}(A)\] for any \(A \in \mathrm{M}_{2 \times 2}(F)\).

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