For two `3xx3` matrices A and B, let `A+B=2B'` and `3A+2B=I_3` where `B'` is the transpose of B and `I_3` is `3xx3` identity matrix, Then:

Options

A. `10A+5B=3I_3`
B. `5A+10B=2I_3`
C. `3A+6B=2I_3`
D. `B+2A=I_3`

Answer

To solve the problem, we will follow the steps outlined in the video transcript and provide a step-by-step solution.

### Step 1: Write down the given equations
We have two equations based on the problem statement:
1. \( A + B = 2B' \) (Equation 1)
2. \( 3A + 2B = I_3 \) (Equation 2)

### Step 2: Take the transpose of Equation 1
Taking the transpose of both sides of Equation 1:
\[ (A + B)' = (2B')' \]
Using the property of transposes, we have:
\[ A' + B' = 2B \]
This gives us Equation 3:
\[ A' + B' = 2B \quad \text{(Equation 3)} \]

### Step 3: Take the transpose of Equation 2
Now, we take the transpose of Equation 2:
\[ (3A + 2B)' = I_3' \]
Using the property of transposes, we have:
\[ 3A' + 2B' = I_3 \]
This gives us Equation 4:
\[ 3A' + 2B' = I_3 \quad \text{(Equation 4)} \]

### Step 4: Substitute Equation 3 into Equation 4
From Equation 3, we can express \( B' \):
\[ B' = \frac{1}{2}(A' + B') \]
Substituting \( B' \) into Equation 4:
\[ 3A' + 2\left(\frac{1}{2}(A + B)\right) = I_3 \]
This simplifies to:
\[ 3A' + A + B = I_3 \]

### Step 5: Rearranging the equation
Rearranging gives us:
\[ 3A' + A + B = I_3 \implies 3A' + A = I_3 - B \]

### Step 6: Substitute \( B \) from Equation 1 into the rearranged equation
From Equation 1, we have:
\[ B = 2B' - A \]
Substituting this into the rearranged equation:
\[ 3A' + A = I_3 - (2B' - A) \]
This leads to:
\[ 3A' + A = I_3 - 2B' + A \]
Thus:
\[ 3A' = I_3 - 2B' \]

### Step 7: Solve for \( A \) and \( B \)
From the equations, we can express \( A \) in terms of \( B \):
\[ A = B \]
Substituting \( A = B \) back into Equation 2:
\[ 3A + 2A = I_3 \implies 5A = I_3 \implies A = \frac{1}{5}I_3 \]
Thus, we also have:
\[ B = \frac{1}{5}I_3 \]

### Conclusion
The final result is:
\[ A = B = \frac{1}{5}I_3 \]