If `A=[(0,2b,c),(a,b,-c),(a,-b,c)]`, then find the value of a, b and c. Such that `A^(T)A=I`

Answer

To solve the problem, we need to find the values of \( a \), \( b \), and \( c \) such that the matrix \( A \) satisfies the condition \( A^T A = I \), where \( I \) is the identity matrix.

Given:
\[ A = \begin{pmatrix} 0 & 2b & c \\ a & b & -c \\ a & -b & c \end{pmatrix} \]

### Step 1: Calculate \( A^T \)
The transpose of matrix \( A \) is obtained by swapping rows and columns:
\[ A^T = \begin{pmatrix} 0 & a & a \\ 2b & b & -b \\ c & -c & c \end{pmatrix} \]

### Step 2: Calculate \( A^T A \)
Now, we will multiply \( A^T \) with \( A \):
\[ A^T A = \begin{pmatrix} 0 & a & a \\ 2b & b & -b \\ c & -c & c \end{pmatrix} \begin{pmatrix} 0 & 2b & c \\ a & b & -c \\ a & -b & c \end{pmatrix} \]

### Step 3: Compute the elements of \( A^T A \)
We will compute each element of the resulting matrix:

1. **First Row**:
- \( (0 \cdot 0 + a \cdot a + a \cdot a) = 2a^2 \)
- \( (0 \cdot 2b + a \cdot b + a \cdot -b) = 0 \)
- \( (0 \cdot c + a \cdot -c + a \cdot c) = 0 \)

2. **Second Row**:
- \( (2b \cdot 0 + b \cdot a + -b \cdot a) = 0 \)
- \( (2b \cdot 2b + b \cdot b + -b \cdot -b) = 6b^2 \)
- \( (2b \cdot c + b \cdot -c + -b \cdot c) = 0 \)

3. **Third Row**:
- \( (c \cdot 0 + -c \cdot a + c \cdot a) = 0 \)
- \( (c \cdot 2b + -c \cdot b + c \cdot -b) = 0 \)
- \( (c \cdot c + -c \cdot -c + c \cdot c) = 3c^2 \)

Putting it all together, we have:
\[ A^T A = \begin{pmatrix} 2a^2 & 0 & 0 \\ 0 & 6b^2 & 0 \\ 0 & 0 & 3c^2 \end{pmatrix} \]

### Step 4: Set \( A^T A = I \)
We know that \( A^T A \) must equal the identity matrix \( I \):
\[ I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

This gives us the following equations:
1. \( 2a^2 = 1 \)
2. \( 6b^2 = 1 \)
3. \( 3c^2 = 1 \)

### Step 5: Solve for \( a \), \( b \), and \( c \)
1. From \( 2a^2 = 1 \):
\[ a^2 = \frac{1}{2} \implies a = \pm \frac{1}{\sqrt{2}} \]

2. From \( 6b^2 = 1 \):
\[ b^2 = \frac{1}{6} \implies b = \pm \frac{1}{\sqrt{6}} \]

3. From \( 3c^2 = 1 \):
\[ c^2 = \frac{1}{3} \implies c = \pm \frac{1}{\sqrt{3}} \]

### Final Values
Thus, the values of \( a \), \( b \), and \( c \) are:
\[ a = \pm \frac{1}{\sqrt{2}}, \quad b = \pm \frac{1}{\sqrt{6}}, \quad c = \pm \frac{1}{\sqrt{3}} \]