Question

Let
\[ A = \begin{bmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 1 & 1 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} 2 \\ 1 \\ 7 \end{bmatrix}. \]
For the validity of the result \(AX = B\), \(X\) is:

• A. \(\begin{bmatrix} -1 \\ 1 \\ 7 \end{bmatrix}\)
• B. \(\begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix}\)
• C. \(\begin{bmatrix} 3 \\ -1 \\ -1 \end{bmatrix}\)
• D. \(\begin{bmatrix} 4 \\ 2 \\ 1 \end{bmatrix}\)

Hint:

For matrix equations AX = B, always check if A is invertible (det(A) ̸= 0) before solving for X.

Answer 1

The Correct Option is D

The matrix equation \( AX = B \) can be solved by substituting each option for \( X \) and checking if the equation holds.

Compute \( AX \) for \( X = \begin{bmatrix} 4 \\ 2 \\ 1 \end{bmatrix} \): \[ A = \begin{bmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 1 & 1 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} 4 \\ 2 \\ 1 \end{bmatrix}. \]

Perform the matrix multiplication: \[ AX = \begin{bmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 1 & 1 & 1 \end{bmatrix} \cdot \begin{bmatrix} 4 \\ 2 \\ 1 \end{bmatrix} = \begin{bmatrix} (1)(4) + (-1)(2) + (0)(1) \\ (0)(4) + (1)(2) + (-1)(1) \\ (1)(4) + (1)(2) + (1)(1) \end{bmatrix}. \]

Simplify the resulting vector: \[ AX = \begin{bmatrix} 4 - 2 + 0 \\ 0 + 2 - 1 \\ 4 + 2 + 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ 7 \end{bmatrix}. \]

Since \( AX = B \), the solution \( X = \begin{bmatrix} 4 \\ 2 \\ 1 \end{bmatrix} \) is valid.

Thus, the correct answer is (D).