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Matrix inversion is the process of finding a matrix that, when multiplied by a given matrix, results in the identity matrix. This 'inverse matrix' is denoted as \(A^{-1}\). If you multiply a matrix by its inverse, the result is the identity matrix, denoted as \(I\). The identity matrix is like the number 1 for matrices—it doesn't change other matrices when multiplied. For instance, if \(A^{-1}\) is the inverse of \(A\), then \(AA^{-1} = I\) and \(A^{-1}A = I\). Not every matrix has an inverse. It's possible only if the determinant of the matrix is non-zero.

The determinant is a special number that can be computed from a square matrix. It is crucial for understanding whether a matrix has an inverse. For a 2x2 matrix: \[ A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \] The determinant, denoted as \( \text{det}(A) \), is calculated as: \[ \text{det}(A) = ad - bc \]. If \( \text{det}(A) \) is zero, the matrix \(A\) does not have an inverse. If \( \text{det}(A) \) is non-zero, the matrix is invertible. In our exercise, the determinant of matrix \( A \) is: \[ \text{det}(A) = 1 \times 8 - 4 \times 3 = 8 - 12 = -4 \]. Since \( -4 \) is non-zero, matrix \( A \) has an inverse.

The adjugate (or adjoint) matrix is a key step in finding the inverse of a matrix. It involves modifying the original matrix in a specific way. For a 2x2 matrix: \[ A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \], the adjugate matrix, \( \text{adj}(A) \), is formed by swapping the elements on the main diagonal and changing the signs of the off-diagonal elements: \[ \text{adj}(A) = \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \]. In our example, for matrix \( A \): \[ A = \begin{bmatrix} 1 & 4 \ 3 & 8 \end{bmatrix} \], the adjugate matrix is: \[ \text{adj}(A) = \begin{bmatrix} 8 & -4 \ -3 & 1 \end{bmatrix} \]. This matrix will be used in the next step to find the inverse.

Matrix multiplication is an operation that takes a pair of matrices and produces another matrix. When checking if a matrix and its inverse multiply to the identity matrix, you need to perform matrix multiplication. The formula for multiplying two 2x2 matrices \[ A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \] and \[ B = \begin{bmatrix} e & f \ g & h \end{bmatrix} \] is: \[ AB = \begin{bmatrix} ae + bg & af + bh \ ce + dg & cf + dh \end{bmatrix} \]. To confirm that \( A^{-1} \) is indeed the inverse of \( A \), you must show: \[ AA^{-1} = I \] and \[ A^{-1}A = I \]. In our exercise, we verified both: \[ A \begin{bmatrix} -2 & 1 \ \frac{3}{4} & -\frac{1}{4} \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = I \] and \[ \begin{bmatrix} -2 & 1 \ \frac{3}{4} & -\frac{1}{4} \end{bmatrix} A = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = I \].