Determine which of the matrices in Exercises 7–12 are orthogonal.If orthogonal, find the inverse.7. 1=p2 1=p21=p2 1=p2

To determine whether the given matrix is orthogonal, we will proceed with the following steps:

1. ### Break Down the Problem

We need to determine if the matrix is orthogonal. A matrix $A$ is orthogonal if $A^T A = I$, where $A^T$ is the transpose of $A$ and $I$ is the identity matrix. If the matrix is orthogonal, we will also find its inverse, which for orthogonal matrices is $A^T$.

2. ### Relevant Concepts

Let’s denote the provided matrix as:

$A = \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix}$

The steps to check orthogonality will involve computing the transpose of the matrix and the product $A^T A$:

1. Calculate $A^T$
2. Compute $A^T A$
3. Check if $A^T A = I$

3. ### Analysis and Detail

1. Calculate the Transpose:

   $A^T = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix}$

2. Compute $A^T A$:

   $A^T A = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix}$

   Performing the multiplication:

   • First row, first column:

   $\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2} + \frac{1}{2} = 1$

   • First row, second column:

   $\frac{1}{\sqrt{2}} \cdot \left(-\frac{1}{\sqrt{2}}\right) + \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = -\frac{1}{2} + \frac{1}{2} = 0$

   • Second row, first column:

   $-\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = -\frac{1}{2} + \frac{1}{2} = 0$

   • Second row, second column:

   $-\frac{1}{\sqrt{2}} \cdot \left(-\frac{1}{\sqrt{2}}\right) + \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2} + \frac{1}{2} = 1$

   Thus:

   $A^T A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I$

4. ### Verify and Summarize

Since $A^T A = I$, the matrix is orthogonal.

Final Answer

• The given matrix is orthogonal.
• The inverse of an orthogonal matrix is its transpose: $A^{-1} = A^T = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix}$