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Chapter 2: Problem 48

Prove Property 4 of Theorem 2.8: If \(A\) is an invertible matrix, then \(\left(A^{T}\right)^{-1}=\left(A^{-1}\right)^{T}\)

Short Answer

Expert verified
This property is verified true by showing that when \(A^{-1}A = I\) and \(AA^{-1} = I\) are transposed, both result in equations that satisfy the condition for a matrix inverse of the transposed of \(A\), proving that \(\left(A^{T}\right)^{-1}=\left(A^{-1}\right)^{T}\).

Step by step solution

01

Set Up the Equation

Starting with the definition of an inverse matrix, we have that \(AA^{-1} = A^{-1}A = I\), where \(I\) is the identity matrix.
02

Using Transpose Properties

Use the property \((AB)^T = B^TA^T\), where \(A\) and \(B\) are any matrices that can be multiplied. Then \(AA^{-1} = I\) becomes \((AA^{-1})^T = I^T\), this deconstructs into \((A^{-1})^TA^T = I\), showing one side of the proof.
03

Proving the second half

Similarly for \(A^{-1}A = I\), that becomes \((A^{-1}A)^T = I^T\), deconstructing into \(A^{T}(A^{-1})^T = I\), showing the second half of the proof. These steps confirm that \(\left( A^T \right)^{-1}= \left(A^{-1}\right)^T\).

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