Let `A=[(2,0,7),(0,1,0),(1,-2,1)] and B=[(-k,14k,7k),(0,1,0),(k,-4k,-2k)]`. If `AB=I`, where I is an identity matrix of order 3, then the sum of all elements of matrix B is equal to

Options

A. 2
B. `(3)/(7)`
C. `(10)/(3)`
D. 4

Answer

To solve the problem, we need to find the value of \( k \) such that the product of matrices \( A \) and \( B \) equals the identity matrix \( I \). Then, we will calculate the sum of all elements in matrix \( B \).

### Step-by-Step Solution:

1. **Define the matrices**:
\[ A = \begin{pmatrix} 2 & 0 & 7 \\ 0 & 1 & 0 \\ 1 & -2 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} -k & 14k & 7k \\ 0 & 1 & 0 \\ k & -4k & -2k \end{pmatrix} \]

2. **Multiply matrices \( A \) and \( B \)**:
We need to compute \( AB \):
\[ AB = \begin{pmatrix} 2 & 0 & 7 \\ 0 & 1 & 0 \\ 1 & -2 & 1 \end{pmatrix} \begin{pmatrix} -k & 14k & 7k \\ 0 & 1 & 0 \\ k & -4k & -2k \end{pmatrix} \]

- **First row of \( AB \)**:
- First element: \( 2(-k) + 0(0) + 7(k) = -2k + 7k = 5k \)
- Second element: \( 2(14k) + 0(1) + 7(-4k) = 28k - 28k = 0 \)
- Third element: \( 2(7k) + 0(0) + 7(-2k) = 14k - 14k = 0 \)

- **Second row of \( AB \)**:
- First element: \( 0(-k) + 1(0) + 0(k) = 0 \)
- Second element: \( 0(14k) + 1(1) + 0(-4k) = 1 \)
- Third element: \( 0(7k) + 1(0) + 0(-2k) = 0 \)

- **Third row of \( AB \)**:
- First element: \( 1(-k) + -2(0) + 1(k) = -k + k = 0 \)
- Second element: \( 1(14k) + -2(1) + 1(-4k) = 14k - 2 - 4k = 10k - 2 \)
- Third element: \( 1(7k) + -2(0) + 1(-2k) = 7k - 2k = 5k \)

Thus,
\[ AB = \begin{pmatrix} 5k & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 10k - 2 & 5k \end{pmatrix} \]

3. **Set \( AB \) equal to the identity matrix \( I \)**:
\[ I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

From the equality \( AB = I \), we get the following equations:
- \( 5k = 1 \)
- \( 10k - 2 = 0 \)
- \( 5k = 1 \)

4. **Solve for \( k \)**:
From \( 5k = 1 \):
\[ k = \frac{1}{5} \]

Check with the second equation:
\[ 10\left(\frac{1}{5}\right) - 2 = 2 - 2 = 0 \quad \text{(True)} \]

5. **Substitute \( k \) back into matrix \( B \)**:
\[ B = \begin{pmatrix} -\frac{1}{5} & 14\left(\frac{1}{5}\right) & 7\left(\frac{1}{5}\right) \\ 0 & 1 & 0 \\ \frac{1}{5} & -4\left(\frac{1}{5}\right) & -2\left(\frac{1}{5}\right) \end{pmatrix} \]
Simplifying \( B \):
\[ B = \begin{pmatrix} -\frac{1}{5} & \frac{14}{5} & \frac{7}{5} \\ 0 & 1 & 0 \\ \frac{1}{5} & -\frac{4}{5} & -\frac{2}{5} \end{pmatrix} \]

6. **Calculate the sum of all elements of matrix \( B \)**:
\[ \text{Sum} = -\frac{1}{5} + \frac{14}{5} + \frac{7}{5} + 0 + 1 + 0 + \frac{1}{5} - \frac{4}{5} - \frac{2}{5} \]
Simplifying:
\[ = \left(-\frac{1}{5} + \frac{14}{5} + \frac{7}{5} + \frac{1}{5} - \frac{4}{5} - \frac{2}{5}\right) + 1 \]
\[ = \left(\frac{15}{5}\right) + 1 = 3 + 1 = 4 \]

### Final Answer:
The sum of all elements of matrix \( B \) is \( 4 \).