91Ó°ÊÓ

Understanding how to calculate a determinant is crucial for finding the inverse of a matrix. In this exercise, we have a 2x2 matrix:\[A = \begin{bmatrix} \frac{\sqrt{2}}{2} & -\sqrt{2} \2\sqrt{2} & \sqrt{2} \end{bmatrix}\]
For a 2x2 matrix, the formula to compute the determinant \( \Delta \) is given by:\[\Delta = ad - bc\]where \( a, b, c, \) and \( d \) are the elements of the matrix. In our matrix:

• \( a = \frac{\sqrt{2}}{2} \)
• \( b = -\sqrt{2} \)
• \( c = 2\sqrt{2} \)
• \( d = \sqrt{2} \)

Plugging these values into the determinant formula:\[\Delta = \left(\frac{\sqrt{2}}{2}\right)(\sqrt{2}) - (-\sqrt{2})(2\sqrt{2})\]It simplifies to:\[\Delta = \frac{2}{2} + 4 = 1 + 4 = 5\]Since the determinant is non-zero, the matrix is invertible, which is a necessary condition for finding an inverse matrix.

Calculating an adjugate matrix is a vital step towards finding the inverse of a matrix. The adjugate matrix, often called the adjoint, is formed by taking the transpose of the cofactor matrix. For a 2x2 matrix, the process is straightforward:
Consider the original matrix:\[A = \begin{bmatrix} \frac{\sqrt{2}}{2} & -\sqrt{2} \2\sqrt{2} & \sqrt{2} \end{bmatrix}\]
The adjugate of matrix \( A \), denoted as \( \text{adj}(A) \), can be quickly found by swapping the diagonal elements and changing the signs of the off-diagonal elements. Thus:
1. Swap the diagonal elements:

• \( \sqrt{2} \) becomes the new first element, and \( \frac{\sqrt{2}}{2} \) becomes the new bottom-right element.

2. Change the signs of the off-diagonal elements:

• \( -\sqrt{2} \) becomes \( \sqrt{2} \)
• \( 2\sqrt{2} \) becomes \( -2\sqrt{2} \)

Constructing the adjugate matrix:\[\text{adj}(A) = \begin{bmatrix} \sqrt{2} & \sqrt{2} \-2\sqrt{2} & \frac{\sqrt{2}}{2} \end{bmatrix}\]This matrix plays a key role in determining the inverse.

Verifying the inverse matrix confirms that our calculations were correct and that our inverse matrix is valid. After calculating the inverse matrix \( A^{-1} \), we check its validity by multiplying it with the original matrix \( A \). If the result is the identity matrix, our inverse is correct. The identity matrix for a 2x2 has ones on the diagonal and zeros elsewhere:\[I = \begin{bmatrix} 1 & 0 \0 & 1 \end{bmatrix}\]
For our example, we calculated the inverse as:\[A^{-1} = \frac{1}{5} \begin{bmatrix} \sqrt{2} & \sqrt{2} \-2\sqrt{2} & \frac{\sqrt{2}}{2} \end{bmatrix} = \begin{bmatrix} \frac{\sqrt{2}}{5} & \frac{\sqrt{2}}{5} \\frac{-2\sqrt{2}}{5} & \frac{\sqrt{2}}{10} \end{bmatrix}\]
Now, verify by multiplying with the original matrix:\[\begin{bmatrix} \frac{\sqrt{2}}{2} & -\sqrt{2} \2\sqrt{2} & \sqrt{2} \end{bmatrix} \cdot \begin{bmatrix} \frac{\sqrt{2}}{5} & \frac{\sqrt{2}}{5} \\frac{-2\sqrt{2}}{5} & \frac{\sqrt{2}}{10} \end{bmatrix} = \begin{bmatrix} 1 & 0 \0 & 1 \end{bmatrix}\]
If this multiplication results in the identity matrix, our work verifies that the calculated inverse is correct and that all steps to find \( A^{-1} \) were executed properly.