Question

If \[ 2A + B + X = 0 \] where \[ A = \begin{bmatrix} -1 & 2 \\ 3 & 4 \end{bmatrix} \quad \text{and} \quad B = \begin{bmatrix} 3 & 1 \\ 5 & -2 \end{bmatrix}, \] then \[ X = \]

• A. \( \begin{bmatrix} 1 & -13 \\ -7 & 2 \end{bmatrix} \)
• B. \( \begin{bmatrix} 1 & 7 \\ 2 & 13 \end{bmatrix} \)
• C. \( \begin{bmatrix} -1 & -7 \\ -2 & -13 \end{bmatrix} \)
• D. \( \begin{bmatrix} 1 & 13 \\ -7 & -2 \end{bmatrix} \)

Hint:

When solving for a matrix in an equation, always isolate the matrix first, and then perform the matrix operations step-by-step.

Answer 1

The Correct Option is C

From the given equation: \[ 2A + B + X = 0 \] Rearranging for \( X \): \[ X = - (2A + B) \]

Step 1: Calculate \( 2A \) \[ 2A = 2 \begin{bmatrix} -1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} -2 & 4 \\ 6 & 8 \end{bmatrix}. \]

Step 2: Calculate \( 2A + B \) \[ 2A + B = \begin{bmatrix} -2 & 4 \\ 6 & 8 \end{bmatrix} + \begin{bmatrix} 3 & 1 \\ 5 & -2 \end{bmatrix} = \begin{bmatrix} -2 + 3 & 4 + 1 \\ 6 + 5 & 8 + (-2) \end{bmatrix} = \begin{bmatrix} 1 & 5 \\ 11 & 6 \end{bmatrix}. \]

Step 3: Substitute into \( X = - (2A + B) \) \[ X = - \begin{bmatrix} 1 & 5 \\ 11 & 6 \end{bmatrix} = \begin{bmatrix} -1 & -5 \\ -11 & -6 \end{bmatrix}. \]

Therefore, the required matrix is: \[ \boxed{ X = \begin{bmatrix} -1 & -5 \\ -11 & -6 \end{bmatrix}. } \]

Correct Answer:

(C) \( \begin{bmatrix} -1 & -5 \\ -11 & -6 \end{bmatrix} \)