Key Concept: Orthogonal Matrices

A square matrix $A$ is called an orthogonal matrix if its transpose is equal to its inverse, i.e., $A^T = A^{-1}$. This fundamental property implies that $AA^T = I$ and $A^T A = I$, where $I$ is the identity matrix.

A crucial characteristic of orthogonal matrices is that their rows form an orthonormal set of vectors, and similarly, their columns form an orthonormal set of vectors. This means:

1. The dot product of any row (or column) with itself is 1 (i.e., its magnitude squared is 1).
2. The dot product of any two distinct rows (or columns) is 0 (i.e., they are orthogonal).

Step-by-Step Derivation

1. Identify the given condition: We are given the matrix $A = \begin{pmatrix} 0 & 2q & r \\ p & q & -r \\ p & -q & r \end{pmatrix}$ and the condition $AA^T = I_3$. This condition directly tells us that $A$ is an orthogonal matrix.

2. Recall the properties of an orthogonal matrix: Since $A$ is an orthogonal matrix, its columns must form an orthonormal set of vectors. This means:

   • The sum of the squares of the elements in each column must be equal to 1.
   • The dot product of any two distinct columns must be 0.

3. Apply the property to the first column of A: Let's consider the first column of matrix $A$, denoted as $C_1 = \begin{pmatrix} 0 \\ p \\ p \end{pmatrix}$. According to the property of orthogonal matrices, the square of the magnitude of this column vector must be 1. Mathematically, this means $C_1 \cdot C_1 = 1$.

   So, we have: $0^2 + p^2 + p^2 = 1$

4. Solve for $|p|$: Simplify the equation: $0 + p^2 + p^2 = 1$ $2p^2 = 1$

   Now, isolate $p^2$: $p^2 = \frac{1}{2}$

   To find $|p|$, take the square root of both sides: $\sqrt{p^2} = \sqrt{\frac{1}{2}}$ $|p| = \frac{1}{\sqrt{2}}$

Explanation for Each Step:

• Given $AA^T = I_3$: This is the defining property of an orthogonal matrix. Recognizing this is the quickest way to solve the problem. If you didn't recognize it, you would have to perform the full matrix multiplication $AA^T$ and then equate it to $I_3$, which is more tedious.
• Using column properties: We chose to use the property that the sum of the squares of elements in each column is 1 because the first column conveniently contains only $p$ (and a 0), allowing us to directly form an equation for $p$. We could also use row properties (e.g., $R_2 \cdot R_2 = p^2 + q^2 + (-r)^2 = 1$), but that would involve $q$ and $r$ as well, requiring more equations to solve. The first column property is the most direct path to $|p|$.
• $0^2 + p^2 + p^2 = 1$: This equation arises from the fact that the magnitude squared of the first column vector $\begin{pmatrix} 0 \\ p \\ p \end{pmatrix}$ must be 1. The elements $0, p, p$ are the components of this vector.
• Solving for $|p|$: The algebraic steps are straightforward. Remember that $\sqrt{x^2} = |x|$, hence $\sqrt{p^2} = |p|$.

Tips and Common Mistakes:

• Recognize Orthogonal Matrices: Always be on the lookout for the $AA^T = I$ or $A^T A = I$ condition. It's a strong indicator of an orthogonal matrix, and its properties can significantly simplify calculations.
• Properties of Rows/Columns: Remember that for an orthogonal matrix, the rows (and columns) form an orthonormal basis. This means:
  • Dot product of a row/column with itself = 1 (magnitude squared is 1).
  • Dot product of distinct rows/columns = 0.
• Careful with Signs: While calculating squares like $(-r)^2 = r^2$ or $(-q)^2 = q^2$, signs cancel out. However, if you were calculating dot products of distinct rows/columns (which is not needed for this problem), be careful with signs.
• Don't Overcomplicate: While you could multiply $A$ by $A^T$ explicitly, it's often faster to use the row/column properties of orthogonal matrices, especially when only one variable is needed.

Summary and Key Takeaway:

This problem is a direct application of the properties of orthogonal matrices. When given that $AA^T = I$, we immediately know that $A$ is orthogonal. The most efficient way to find the value of a variable like $p$ is to use the property that the square of the magnitude of each column (or row) vector is 1. By focusing on the first column, which only involved $p$, we quickly arrived at the solution.

The final answer is $\boxed{{1 \over {\sqrt 2 }}}$.