Let `A+2B=[{:(2,4,0),(6,-3,3),(-5,3,5):}] and 2A-B=[{:(6,-2,4),(6,1,5),(6,3,4):}]` , then tr (A) - tr (B) is equal to (where , tr (A) =n trace of matrix x A i.e. . Sum of the principle diagonal elements of matrix A)

Answer

To solve the problem, we need to find the values of matrices A and B from the given equations and then calculate the trace of both matrices to find \( \text{tr}(A) - \text{tr}(B) \).

### Step 1: Write down the equations
We have the following equations:
1. \( A + 2B = \begin{pmatrix} 2 & 4 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 5 \end{pmatrix} \)
2. \( 2A - B = \begin{pmatrix} 6 & -2 & 4 \\ 6 & 1 & 5 \\ 6 & 3 & 4 \end{pmatrix} \)

### Step 2: Multiply the first equation by 2
To eliminate \( B \), we can manipulate the equations. Multiply the first equation by 2:
\[ 2(A + 2B) = 2 \begin{pmatrix} 2 & 4 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 5 \end{pmatrix} \]
This gives:
\[ 2A + 4B = \begin{pmatrix} 4 & 8 & 0 \\ 12 & -6 & 6 \\ -10 & 6 & 10 \end{pmatrix} \]

### Step 3: Add the two equations
Now we can add the modified first equation to the second equation:
\[ (2A + 4B) + (2A - B) = \begin{pmatrix} 4 & 8 & 0 \\ 12 & -6 & 6 \\ -10 & 6 & 10 \end{pmatrix} + \begin{pmatrix} 6 & -2 & 4 \\ 6 & 1 & 5 \\ 6 & 3 & 4 \end{pmatrix} \]
This simplifies to:
\[ 4A + 3B = \begin{pmatrix} 10 & 6 & 4 \\ 18 & -5 & 11 \\ -4 & 9 & 14 \end{pmatrix} \]

### Step 4: Solve for B
Now, we can isolate \( B \) by rearranging the equation:
\[ 3B = \begin{pmatrix} 10 & 6 & 4 \\ 18 & -5 & 11 \\ -4 & 9 & 14 \end{pmatrix} - 4A \]
We need to express \( A \) in terms of \( B \) from the first equation:
\[ A = \begin{pmatrix} 2 & 4 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 5 \end{pmatrix} - 2B \]

### Step 5: Substitute A back into the equation for B
Substituting \( A \) into the equation for \( B \):
\[ 3B = \begin{pmatrix} 10 & 6 & 4 \\ 18 & -5 & 11 \\ -4 & 9 & 14 \end{pmatrix} - 4\left(\begin{pmatrix} 2 & 4 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 5 \end{pmatrix} - 2B\right) \]
This will yield a new equation in terms of \( B \).

### Step 6: Solve for A and B
After solving the equations, we find:
\[ A = \begin{pmatrix} 14/5 & -1/5 & 13/5 \\ 18/5 & -1/5 & 13/5 \\ 7/5 & 9/5 & 13/5 \end{pmatrix} \]
\[ B = \begin{pmatrix} -2/5 & 10/5 & -4/5 \\ 6/5 & -7/5 & 1/5 \\ -16/5 & 3/5 & 6/5 \end{pmatrix} \]

### Step 7: Calculate the traces
Now we calculate the traces:
\[ \text{tr}(A) = \frac{14}{5} + \frac{-1}{5} + \frac{13}{5} = \frac{26}{5} \]
\[ \text{tr}(B) = \frac{-2}{5} + \frac{10}{5} + \frac{-4}{5} = \frac{4}{5} \]

### Step 8: Find \( \text{tr}(A) - \text{tr}(B) \)
Now we can find:
\[ \text{tr}(A) - \text{tr}(B) = \frac{26}{5} - \frac{4}{5} = \frac{22}{5} \]

### Final Answer
Thus, \( \text{tr}(A) - \text{tr}(B) = \frac{22}{5} \).

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