Question

If the matrix A = \(\begin{bmatrix} 0 & 2y & z
x & y & -z
x & -y & z \end{bmatrix}\) satisfies the equation AA' = I, then find the values of x, y and z.

Hint:

The condition \(AA' = I\) for a square matrix A signifies that A is an orthogonal matrix. This implies that its row vectors (and column vectors) form an orthonormal set. That is, each row vector has a magnitude of 1, and the dot product of any two different row vectors is 0. This insight can sometimes provide a faster way to set up the equations.

Answer 1

Step 1: Understanding the Concept:
The given equation is \(AA' = I\), where A' is the transpose of matrix A and I is the identity matrix.
A matrix A that satisfies this condition is called an orthogonal matrix.
For an orthogonal matrix, the dot product of any row vector with itself is 1, and the dot product of any two distinct row vectors is 0.
Step 2: Key Formula or Approach:
1. Find the transpose of matrix A, denoted as A'.
2. Calculate the product AA'.
3. Equate the resulting matrix with the identity matrix \(I = \begin{bmatrix} 1 & 0 & 0
0 & 1 & 0
0 & 0 & 1 \end{bmatrix}\).
4. Solve the system of equations obtained by equating the corresponding elements.
Step 3: Detailed Explanation or Calculation:
Given matrix \(A = \begin{bmatrix} 0 & 2y & z
x & y & -z
x & -y & z \end{bmatrix}\).
The transpose of A is \(A' = \begin{bmatrix} 0 & x & x
2y & y & -y
z & -z & z \end{bmatrix}\).
Now, we compute the product \(AA'\):
\[ AA' = \begin{bmatrix} 0 & 2y & z
x & y & -z
x & -y & z \end{bmatrix} \begin{bmatrix} 0 & x & x
2y & y & -y
z & -z & z \end{bmatrix} \] \[ = \begin{bmatrix} (0)^2+(2y)^2+(z)^2 & 0(x)+2y(y)+z(-z) & 0(x)+2y(-y)+z(z)
x(0)+y(2y)+(-z)(z) & x^2+y^2+(-z)^2 & x^2+y(-y)+(-z)(z)
x(0)+(-y)(2y)+z(z) & x(x)+(-y)(y)+z(-z) & x^2+(-y)^2+z^2 \end{bmatrix} \] \[ = \begin{bmatrix} 4y^2+z^2 & 2y^2-z^2 & -2y^2+z^2
2y^2-z^2 & x^2+y^2+z^2 & x^2-y^2-z^2
-2y^2+z^2 & x^2-y^2-z^2 & x^2+y^2+z^2 \end{bmatrix} \] Equating \(AA'\) with \(I\):
\[ \begin{bmatrix} 4y^2+z^2 & 2y^2-z^2 & -2y^2+z^2
2y^2-z^2 & x^2+y^2+z^2 & x^2-y^2-z^2
-2y^2+z^2 & x^2-y^2-z^2 & x^2+y^2+z^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0
0 & 1 & 0
0 & 0 & 1 \end{bmatrix} \] From the non-diagonal elements, we get:
\(2y^2 - z^2 = 0 \implies z^2 = 2y^2\) (Equation 1)
From the diagonal elements, we get:
\(4y^2 + z^2 = 1\) (Equation 2)
\(x^2 + y^2 + z^2 = 1\) (Equation 3)
Substitute \(z^2 = 2y^2\) from Eq. 1 into Eq. 2:
\(4y^2 + 2y^2 = 1 \implies 6y^2 = 1 \implies y^2 = \frac{1}{6} \implies y = \pm \frac{1}{\sqrt{6}}\).
Now find \(z^2\) using Eq. 1:
\(z^2 = 2y^2 = 2(\frac{1}{6}) = \frac{1}{3} \implies z = \pm \frac{1}{\sqrt{3}}\).
Finally, substitute the values of \(y^2\) and \(z^2\) into Eq. 3 to find x:
\(x^2 + \frac{1}{6} + \frac{1}{3} = 1 \implies x^2 + \frac{1+2}{6} = 1 \implies x^2 + \frac{3}{6} = 1 \implies x^2 + \frac{1}{2} = 1\).
\(x^2 = 1 - \frac{1}{2} = \frac{1}{2} \implies x = \pm \frac{1}{\sqrt{2}}\).
Step 4: Final Answer
The values are \(x = \pm \frac{1}{\sqrt{2}}\), \(y = \pm \frac{1}{\sqrt{6}}\), and \(z = \pm \frac{1}{\sqrt{3}}\).