Let `A` and `B` be two `3xx3` matrices such that `A+B = 2 B'` and `3A + 2B= I ` then

Options

A. `A-B=O`
B. `A+B=I`
C. `A-B=I`
D. `A+2B=O`

Answer

To solve the problem, we will follow a systematic approach using the given equations involving matrices \( A \) and \( B \).

### Given:
1. \( A + B = 2B' \) (where \( B' \) is the transpose of \( B \))
2. \( 3A + 2B = I \) (where \( I \) is the identity matrix)

### Step 1: Transpose the first equation
Taking the transpose of both sides of the first equation:
\[ (A + B)' = (2B')' \]
Using the property of transposes, we have:
\[ A' + B' = 2B \]
This gives us our second equation:
\[ A' + B' = 2B \quad \text{(Equation 2)} \]

### Step 2: Transpose the second equation
Now, we take the transpose of the second equation:
\[ (3A + 2B)' = I' \]
Again, using the property of transposes:
\[ 3A' + 2B' = I \quad \text{(Equation 3)} \]

### Step 3: Substitute Equation 2 into Equation 3
From Equation 2, we can express \( A' \):
\[ A' = 2B - B' \]
Substituting this into Equation 3:
\[ 3(2B - B') + 2B' = I \]
Expanding this:
\[ 6B - 3B' + 2B' = I \]
Combining like terms:
\[ 6B - B' = I \quad \text{(Equation 4)} \]

### Step 4: Express \( B' \) in terms of \( B \)
From Equation 4, we can express \( B' \):
\[ B' = 6B - I \]

### Step 5: Substitute \( B' \) back into Equation 1
Now, substitute \( B' \) into the first equation:
\[ A + B = 2(6B - I) \]
Expanding this:
\[ A + B = 12B - 2I \]
Rearranging gives:
\[ A = 12B - 2I - B \]
So,
\[ A = 11B - 2I \quad \text{(Equation 5)} \]

### Step 6: Substitute Equation 5 into Equation 3
Now substitute \( A \) from Equation 5 into the second equation \( 3A + 2B = I \):
\[ 3(11B - 2I) + 2B = I \]
Expanding this:
\[ 33B - 6I + 2B = I \]
Combining like terms:
\[ 35B - 6I = I \]
Rearranging gives:
\[ 35B = 7I \]
Thus,
\[ B = \frac{1}{5}I \]

### Step 7: Substitute \( B \) back to find \( A \)
Now substitute \( B \) back into Equation 5 to find \( A \):
\[ A = 11\left(\frac{1}{5}I\right) - 2I \]
Calculating this gives:
\[ A = \frac{11}{5}I - 2I = \frac{11}{5}I - \frac{10}{5}I = \frac{1}{5}I \]

### Step 8: Find \( A - B \)
Now we can find \( A - B \):
\[ A - B = \frac{1}{5}I - \frac{1}{5}I = 0 \]

### Conclusion
Thus, we have:
\[ A - B = 0 \quad \text{(Zero matrix)} \]

### Final Answer
The correct option is \( A - B = 0 \) matrix.

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