If `2A+3B=[(2,-1,4),(3,2,5)]` and `A+2B=[(5,0,3),(1,6,2)]`, then `B` is

Options

A. `[(8,-1,2),(-1,10,-1)]`
B. `[(8,1,2),(-1,10,-1)]`
C. `[(8,1,-2),(-1,10,-1)]`
D. `[(8,1,2),(1,10,1)]`

Explanation

To solve the problem, we have the following two equations involving matrices \( A \) and \( B \):

1. \( 2A + 3B = \begin{pmatrix} 2 & -1 & 4 \\ 3 & 2 & 5 \end{pmatrix} \) (Equation 1)
2. \( A + 2B = \begin{pmatrix} 5 & 0 & 3 \\ 1 & 6 & 2 \end{pmatrix} \) (Equation 2)

We need to find the matrix \( B \).

### Step 1: Multiply Equation 2 by 2
We will first multiply the entire Equation 2 by 2 to make it easier to eliminate \( A \) later.

\[ 2(A + 2B) = 2 \begin{pmatrix} 5 & 0 & 3 \\ 1 & 6 & 2 \end{pmatrix} \]

This simplifies to:

\[ 2A + 4B = \begin{pmatrix} 10 & 0 & 6 \\ 2 & 12 & 4 \end{pmatrix} \quad \text{(Equation 3)} \]

### Step 2: Subtract Equation 1 from Equation 3
Now, we will subtract Equation 1 from Equation 3 to eliminate \( A \):

\[ (2A + 4B) - (2A + 3B) = \begin{pmatrix} 10 & 0 & 6 \\ 2 & 12 & 4 \end{pmatrix} - \begin{pmatrix} 2 & -1 & 4 \\ 3 & 2 & 5 \end{pmatrix} \]

This simplifies to:

\[ 4B - 3B = \begin{pmatrix} 10 - 2 & 0 - (-1) & 6 - 4 \\ 2 - 3 & 12 - 2 & 4 - 5 \end{pmatrix} \]

\[ B = \begin{pmatrix} 8 & 1 & 2 \\ -1 & 10 & -1 \end{pmatrix} \]

### Final Result
Thus, the matrix \( B \) is:

\[ B = \begin{pmatrix} 8 & 1 & 2 \\ -1 & 10 & -1 \end{pmatrix} \]