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 sum of all elements of matrix B given that \( AB = I \), where \( I \) is the identity matrix of order 3. Here are the steps to find the solution:

### Step 1: Write down matrices A and B
Given:
\[ A = \begin{pmatrix} 2 & 0 & 7 \\ 0 & 1 & 0 \\ 1 & -2 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} -k & 14k & 7k \\ 0 & 1 & 0 \\ k & -4k & -2k \end{pmatrix} \]

### Step 2: Calculate the product \( AB \)
We need to compute the product \( AB \):
\[ AB = A \cdot B = \begin{pmatrix} 2 & 0 & 7 \\ 0 & 1 & 0 \\ 1 & -2 & 1 \end{pmatrix} \cdot \begin{pmatrix} -k & 14k & 7k \\ 0 & 1 & 0 \\ k & -4k & -2k \end{pmatrix} \]

Calculating each element of the resulting matrix:

- **Element (1,1)**:
\[ 2 \cdot (-k) + 0 \cdot 0 + 7 \cdot k = -2k + 7k = 5k \]

- **Element (1,2)**:
\[ 2 \cdot 14k + 0 \cdot 1 + 7 \cdot (-4k) = 28k - 28k = 0 \]

- **Element (1,3)**:
\[ 2 \cdot 7k + 0 \cdot 0 + 7 \cdot (-2k) = 14k - 14k = 0 \]

- **Element (2,1)**:
\[ 0 \cdot (-k) + 1 \cdot 0 + 0 \cdot k = 0 \]

- **Element (2,2)**:
\[ 0 \cdot 14k + 1 \cdot 1 + 0 \cdot (-4k) = 1 \]

- **Element (2,3)**:
\[ 0 \cdot 7k + 1 \cdot 0 + 0 \cdot (-2k) = 0 \]

- **Element (3,1)**:
\[ 1 \cdot (-k) + (-2) \cdot 0 + 1 \cdot k = -k + k = 0 \]

- **Element (3,2)**:
\[ 1 \cdot 14k + (-2) \cdot 1 + 1 \cdot (-4k) = 14k - 2 - 4k = 10k - 2 \]

- **Element (3,3)**:
\[ 1 \cdot 7k + (-2) \cdot 0 + 1 \cdot (-2k) = 7k - 2k = 5k \]

Thus, the product \( AB \) is:
\[ AB = \begin{pmatrix} 5k & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 10k - 2 & 5k \end{pmatrix} \]

### Step 3: Set \( AB \) equal to the identity matrix \( I \)
We know that \( AB = I \):
\[ AB = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

From this, we can set up the following equations:
1. \( 5k = 1 \)
2. \( 10k - 2 = 0 \)
3. \( 5k = 1 \)

### Step 4: Solve for \( k \)
From the first equation:
\[ 5k = 1 \implies k = \frac{1}{5} \]

### Step 5: Calculate the sum of all elements of matrix B
Now, substituting \( k = \frac{1}{5} \) into matrix B:
\[ B = \begin{pmatrix} -\frac{1}{5} & 14 \cdot \frac{1}{5} & 7 \cdot \frac{1}{5} \\ 0 & 1 & 0 \\ \frac{1}{5} & -4 \cdot \frac{1}{5} & -2 \cdot \frac{1}{5} \end{pmatrix} \]
\[ = \begin{pmatrix} -\frac{1}{5} & \frac{14}{5} & \frac{7}{5} \\ 0 & 1 & 0 \\ \frac{1}{5} & -\frac{4}{5} & -\frac{2}{5} \end{pmatrix} \]

Now, we calculate the sum of all elements of matrix B:
\[ \text{Sum} = -\frac{1}{5} + \frac{14}{5} + \frac{7}{5} + 0 + 1 + 0 + \frac{1}{5} - \frac{4}{5} - \frac{2}{5} \]
\[ = (-\frac{1}{5} + \frac{1}{5}) + (\frac{14}{5} + \frac{7}{5} - \frac{4}{5} - \frac{2}{5}) + 1 \]
\[ = 0 + \frac{15}{5} + 1 = 3 + 1 = 4 \]

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