The system of equations
\(3x - z = 4\)
\(2x + 5y = 1\)
\(y - 2z = 7\)
is written in the form \(AX = B\), where \(X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}\). Then A is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\begin{bmatrix} 3 & 0 & -1 \\ 2 & 5 & 0 \\ 0 & 1 & -2 \end{bmatrix}\)
Row i of the coefficient matrix holds the coefficients of x, y and z, in this order, from equation i. A missing variable has coefficient 0.
\(3x + 0y - 1z = 4\)
\(2x + 5y + 0z = 1\)
\(0x + 1y - 2z = 7\)
So the rows are \((3, 0, -1)\), \((2, 5, 0)\) and \((0, 1, -2)\).
\( \displaystyle \begin{bmatrix} 3 & 0 & -1 \\ 2 & 5 & 0 \\ 0 & 1 & -2 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 4 \\ 1 \\ 7 \end{bmatrix} \)
Why the other options are incorrect
Option (2) \(\begin{bmatrix} 3 & -1 & 0 \\ 2 & 5 & 0 \\ 0 & 1 & -2 \end{bmatrix}\) — Incorrect. Its first row gives \(3x - y = 4\); the −1 belongs to z, not to y.
Option (3) \(\begin{bmatrix} 3 & 0 & -1 \\ 2 & 0 & 5 \\ 0 & 1 & -2 \end{bmatrix}\) — Incorrect. Its second row gives \(2x + 5z = 1\); the 5 belongs to y.
Option (4) \(\begin{bmatrix} 3 & 2 & 0 \\ 0 & 5 & 1 \\ -1 & 0 & -2 \end{bmatrix}\) — Incorrect. This is the transpose of A: the coefficients of each equation have been written in a column instead of a row.