1. Key Concepts and Formulas

To solve this problem, we'll utilize fundamental properties of matrices, including addition, scalar multiplication, and transposition. For matrices $X$ and $Y$ of appropriate dimensions, and a scalar $k$:

• Matrix Addition: $X+Y$ is defined if $X$ and $Y$ have the same dimensions.
• Scalar Multiplication: $kX$ involves multiplying every element of $X$ by $k$.
• Transpose of a Matrix ($X^T$):
  • $(X+Y)^T = X^T + Y^T$ (The transpose of a sum is the sum of the transposes).
  • $(kX)^T = kX^T$ (The transpose of a scalar multiple is the scalar multiple of the transpose).
  • $(X^T)^T = X$ (The transpose of a transpose is the original matrix).
  • The transpose of an identity matrix is itself: $I_n^T = I_n$.
• Identity Matrix ($I_n$): A square matrix with ones on the main diagonal and zeros elsewhere. It acts as the multiplicative identity for matrices.
• Solving System of Linear Matrix Equations: Similar to solving systems of linear algebraic equations, we can use substitution or elimination methods to find unknown matrices.

2. Step-by-Step Solution

We are given two equations involving two $3 \times 3$ matrices, $A$ and $B$:

1. $A + B = 2B^T \quad \ldots (1)$
2. $3A + 2B = I_3 \quad \ldots (2)$

Our goal is to find the matrices $A$ and $B$ and then check which of the given options holds true.

• Step 1: Express A in terms of B and $B^T$ from Equation (1). We start by isolating $A$ from the first equation. This allows us to substitute $A$ into the second equation, reducing the system to an equation involving only $B$ and $B^T$. $A = 2B^T - B \quad \ldots (3)$

• Step 2: Substitute the expression for A into Equation (2). Now, substitute the expression for $A$ from Equation (3) into Equation (2). This eliminates $A$ from the system, leaving us with an equation solely in terms of $B$ and $B^T$. $3(2B^T - B) + 2B = I_3$

• Step 3: Simplify the resulting equation. Distribute the scalar and combine the terms involving $B$: $6B^T - 3B + 2B = I_3$ $6B^T - B = I_3 \quad \ldots (4)$ This simplified equation is a key intermediate result.

• Step 4: Take the transpose of Equation (4). To create a system of equations for $B$ and $B^T$, we take the transpose of both sides of Equation (4). Remember the transpose properties: $(X+Y)^T = X^T+Y^T$, $(kX)^T = kX^T$, $(X^T)^T = X$, and $I_3^T = I_3$. $(6B^T - B)^T = I_3^T$ $6(B^T)^T - B^T = I_3$ $6B - B^T = I_3 \quad \ldots (5)$ Now we have a system of two linear matrix equations involving $B$ and $B^T$.

• Step 5: Solve the system of equations (4) and (5) for B and $B^T$. Our system is: $6B^T - B = I_3 \quad \ldots (4)$ $-B^T + 6B = I_3 \quad \ldots (5)$ To eliminate $B^T$, multiply Equation (5) by 6: $-6B^T + 36B = 6I_3 \quad \ldots (6)$ Now, add Equation (4) and Equation (6): $(6B^T - B) + (-6B^T + 36B) = I_3 + 6I_3$ $35B = 7I_3$ $B = \frac{7}{35}I_3$ $B = \frac{1}{5}I_3$ This determines matrix $B$.

• Step 6: Find $B^T$. Substitute the value of $B$ back into Equation (4) (or Equation (5)) to find $B^T$: $6B^T - \frac{1}{5}I_3 = I_3$ $6B^T = I_3 + \frac{1}{5}I_3$ $6B^T = \frac{6}{5}I_3$ $B^T = \frac{1}{5}I_3$ Notice that $B = B^T$, which means $B$ is a symmetric matrix. This is expected since $B$ is a scalar multiple of the identity matrix.

• Step 7: Solve for Matrix A. Now that we have $B$ and $B^T$, substitute these into Equation (3) to find $A$: $A = 2B^T - B$ $A = 2\left(\frac{1}{5}I_3\right) - \left(\frac{1}{5}I_3\right)$ $A = \frac{2}{5}I_3 - \frac{1}{5}I_3$ $A = \frac{1}{5}I_3$ So, we have found that $A = \frac{1}{5}I_3$ and $B = \frac{1}{5}I_3$.

• Step 8: Verify the options. Substitute $A = \frac{1}{5}I_3$ and $B = \frac{1}{5}I_3$ into each option:

  • (A) $5A + 10B = 2I_3$ $5\left(\frac{1}{5}I_3\right) + 10\left(\frac{1}{5}I_3\right) = I_3 + 2I_3 = 3I_3$ Since $3I_3 \neq 2I_3$, option (A) is incorrect.

  • (B) $10A + 5B = 3I_3$ $10\left(\frac{1}{5}I_3\right) + 5\left(\frac{1}{5}I_3\right) = 2I_3 + I_3 = 3I_3$ Since $3I_3 = 3I_3$, option (B) is correct.

  • (C) $B + 2A = I_3$ $\frac{1}{5}I_3 + 2\left(\frac{1}{5}I_3\right) = \frac{1}{5}I_3 + \frac{2}{5}I_3 = \frac{3}{5}I_3$ Since $\frac{3}{5}I_3 \neq I_3$, option (C) is incorrect.

  • (D) $3A + 6B = 2I_3$ $3\left(\frac{1}{5}I_3\right) + 6\left(\frac{1}{5}I_3\right) = \frac{3}{5}I_3 + \frac{6}{5}I_3 = \frac{9}{5}I_3$ Since $\frac{9}{5}I_3 \neq 2I_3$, option (D) is incorrect.

3. Common Mistakes & Tips

• Transpose Properties: Always be careful when applying transpose properties, especially to sums and scalar multiples. Remember $(X^T)^T = X$ and $I_n^T = I_n$.
• Systematic Elimination/Substitution: Treat matrix equations like algebraic equations. A systematic approach to solving simultaneous equations (like substitution or elimination) is crucial for accuracy.
• Verify Solutions: After finding the matrices $A$ and $B$, it's good practice to substitute them back into the original given equations to ensure they satisfy the conditions. This helps catch calculation errors early.

4. Summary

We were given a system of two linear matrix equations involving $3 \times 3$ matrices $A$ and $B$. By using substitution, transposition properties, and solving the resulting system, we found that $A = \frac{1}{5}I_3$ and $B = \frac{1}{5}I_3$. Substituting these values into the given options, we determined that only option (B) holds true, as $10A + 5B = 10(\frac{1}{5}I_3) + 5(\frac{1}{5}I_3) = 2I_3 + I_3 = 3I_3$, which matches the right-hand side of option (B).

5. Final Answer

The final answer is $\boxed{\text{A}}$.