For `k=1/sqrt(50)`, the value of a, b, c such that PP' = I, where `P=[(2//3,3k,a),(-1//3,-4k,b),(2//3,-5k,c)]` is

Options

A. `pm13/(5sqrt(2)),pm16/(5sqrt(2)),pm1/(3sqrt(2))`
B. `pm3/(5sqrt(2)),pm2/(5sqrt(2)),pm1/(3sqrt(2))`
C. `pm21/(5sqrt(2)),pm19/(5sqrt(2)),pm1/(3sqrt(2))`
D. None of these

Answer

To solve the problem, we need to find the values of \(a\), \(b\), and \(c\) such that the product of matrix \(P\) and its transpose \(P'\) equals the identity matrix \(I\).

Given:
\[ k = \frac{1}{\sqrt{50}} \]
\[ P = \begin{pmatrix} \frac{2}{3} & 3k & a \\ -\frac{1}{3} & -4k & b \\ \frac{2}{3} & -5k & c \end{pmatrix} \]

### Step 1: Calculate \(P'\)
The transpose of matrix \(P\) is obtained by interchanging rows and columns:
\[ P' = \begin{pmatrix} \frac{2}{3} & -\frac{1}{3} & \frac{2}{3} \\ 3k & -4k & -5k \\ a & b & c \end{pmatrix} \]

### Step 2: Calculate \(PP'\)
Now, we will multiply \(P\) and \(P'\):
\[ PP' = P \cdot P' = \begin{pmatrix} \frac{2}{3} & 3k & a \\ -\frac{1}{3} & -4k & b \\ \frac{2}{3} & -5k & c \end{pmatrix} \cdot \begin{pmatrix} \frac{2}{3} & -\frac{1}{3} & \frac{2}{3} \\ 3k & -4k & -5k \\ a & b & c \end{pmatrix} \]

### Step 3: Calculate the elements of \(PP'\)
1. **Element (1,1)**:
\[ \left(\frac{2}{3} \cdot \frac{2}{3}\right) + (3k \cdot 3k) + (a \cdot a) = \frac{4}{9} + 9k^2 + a^2 \]
Set this equal to 1:
\[ \frac{4}{9} + 9k^2 + a^2 = 1 \]

2. **Element (2,2)**:
\[ \left(-\frac{1}{3} \cdot -\frac{1}{3}\right) + (-4k \cdot -4k) + (b \cdot b) = \frac{1}{9} + 16k^2 + b^2 \]
Set this equal to 1:
\[ \frac{1}{9} + 16k^2 + b^2 = 1 \]

3. **Element (3,3)**:
\[ \left(\frac{2}{3} \cdot \frac{2}{3}\right) + (-5k \cdot -5k) + (c \cdot c) = \frac{4}{9} + 25k^2 + c^2 \]
Set this equal to 1:
\[ \frac{4}{9} + 25k^2 + c^2 = 1 \]

### Step 4: Substitute \(k^2\)
Since \(k = \frac{1}{\sqrt{50}}\), we have:
\[ k^2 = \frac{1}{50} \]
Now, substitute \(k^2\) into the equations.

1. For \(a\):
\[ \frac{4}{9} + 9 \cdot \frac{1}{50} + a^2 = 1 \]
\[ \frac{4}{9} + \frac{9}{50} + a^2 = 1 \]
\[ a^2 = 1 - \left(\frac{4}{9} + \frac{9}{50}\right) \]

Calculate \(\frac{4}{9} + \frac{9}{50}\):
\[ \text{LCM of } 9 \text{ and } 50 = 450 \]
\[ \frac{4}{9} = \frac{200}{450}, \quad \frac{9}{50} = \frac{81}{450} \]
\[ \frac{4}{9} + \frac{9}{50} = \frac{200 + 81}{450} = \frac{281}{450} \]
\[ a^2 = 1 - \frac{281}{450} = \frac{450 - 281}{450} = \frac{169}{450} \]
\[ a = \pm \frac{13}{\sqrt{450}} = \pm \frac{13}{15\sqrt{2}} \]

2. For \(b\):
\[ \frac{1}{9} + 16 \cdot \frac{1}{50} + b^2 = 1 \]
\[ b^2 = 1 - \left(\frac{1}{9} + \frac{16}{50}\right) \]
Calculate \(\frac{1}{9} + \frac{16}{50}\):
\[ \frac{1}{9} = \frac{50}{450}, \quad \frac{16}{50} = \frac{144}{450} \]
\[ b^2 = 1 - \frac{50 + 144}{450} = 1 - \frac{194}{450} = \frac{256}{450} \]
\[ b = \pm \frac{16}{\sqrt{450}} = \pm \frac{16}{15\sqrt{2}} \]

3. For \(c\):
\[ \frac{4}{9} + 25 \cdot \frac{1}{50} + c^2 = 1 \]
\[ c^2 = 1 - \left(\frac{4}{9} + \frac{25}{50}\right) \]
Calculate \(\frac{4}{9} + \frac{25}{50}\):
\[ \frac{25}{50} = \frac{225}{450} \]
\[ c^2 = 1 - \left(\frac{200 + 225}{450}\right) = 1 - \frac{425}{450} = \frac{25}{450} \]
\[ c = \pm \frac{5}{\sqrt{450}} = \pm \frac{5}{15\sqrt{2}} \]

### Final Values
Thus, the values of \(a\), \(b\), and \(c\) are:
\[ a = \pm \frac{13}{15\sqrt{2}}, \quad b = \pm \frac{16}{15\sqrt{2}}, \quad c = \pm \frac{5}{15\sqrt{2}} \]