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Understanding how to calculate the determinant of a matrix is crucial to assessing whether a matrix can be inverted. For a simple 2x2 matrix like the one in this exercise:\[A = \begin{pmatrix} a & -b \ b & a \end{pmatrix}\]The determinant is calculated using the formula for a 2x2 matrix:\[\text{det}(A) = a \cdot d - b \cdot c\]Here we replace with the values:\[\text{det}(A) = a \cdot a - (-b) \cdot b = a^2 + b^2\]The determinant tells us about the transformation properties of the matrix. If it's zero, the matrix collapses space into a lower dimension and becomes non-invertible. As long as the determinant is not zero, the matrix remains invertible, indicating a viable path to calculate its inverse.

The key to determining the invertibility of a matrix is checking its determinant. For any 2x2 matrix:- The matrix is invertible if the determinant is non-zero.- Therefore, a matrix like this remains invertible as long as \( a^2 + b^2 \) is not zero.Let's delve into why this condition is critical:- A matrix with a zero determinant implies that its columns (or rows) are linearly dependent. This means they do not span the full 2-dimensional space and there isn't a unique solution to the linear system.- Contrastingly, a non-zero determinant indicates that the matrix can "undo" transformations through its inverse, ensuring that every possible outcome has a unique antecedent.In this particular exercise, unless both \( a \) and \( b \) are zero, \( a^2 + b^2 \) won't be zero, ensuring that the matrix you are dealing with is indeed invertible.

Theorem 3.8 provides us with a practical and reliable method to find the inverse of a 2x2 matrix when it's invertible.For a generic 2x2 matrix:\[A = \begin{pmatrix} p & q \ r & s \end{pmatrix}\]The inverse, according to Theorem 3.8, is derived using:\[A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} s & -q \ -r & p \end{pmatrix}\]Applying this to our specific matrix:\[A = \begin{pmatrix} a & -b \ b & a \end{pmatrix}\]The determinant is already calculated as \( a^2 + b^2 \). Therefore, the inverse is:\[A^{-1} = \frac{1}{a^2 + b^2} \begin{pmatrix} a & b \ -b & a \end{pmatrix}\]This procedure is crucial because it not only ensures that a system of linear equations represented by this matrix can be solved, but it also provides insight into the properties of the matrix, such as rotation and scaling transformations.