AHL 1.16—Solution of systems of linear equations

For a system $a_{i}x+b_{i}y+c_{i}z=d_i$, the augmented matrix places the coefficients of $x$, $y$, $z$ in the first three columns and the constant term in the fourth column after the bar.

From $3x-y+2z=5$, the row is $\left(\begin{array}{ccccc}3& -1& 2& |& 5\end{array}\right)$. Note the coefficient of $y$ is $-1$, not $+1$.

From $x+4y-z=0$, the row is $\left(\begin{array}{ccccc}1& 4& -1& |& 0\end{array}\right)$. The constant is $0$, which must appear after the bar.

From $-2x+y+3z=-1$, the row is $\left(\begin{array}{ccccc}-2& 1& 3& |& -1\end{array}\right)$. The coefficient of $x$ is $-2$ and the constant is $-1$.

Assembling all three rows gives $\left(\begin{array}{ccccc}3& -1& 2& |& 5\\ 1& 4& -1& |& 0\\ -2& 1& 3& |& -1\end{array}\right)$, which matches option A.

The main traps are sign errors in the coefficients. Check each option against the original equations carefully.

Option B has Row 1 as $\left(\begin{array}{ccccc}3& 1& -2& |& 5\end{array}\right)$. The coefficient of $y$ should be $-1$ (from $-y$) and $z$ should be $+2$, not $-2$. This is incorrect.

Option C has Row 3 as $\left(\begin{array}{ccccc}2& -1& -3& |& 1\end{array}\right)$. The equation $-2x+y+3z=-1$ requires coefficients $-2,1,3$ and constant $-1$, not the negated version shown.

Option D has the constant terms in the wrong column positions — the constants $3,1,-2$ appear where coefficients of $x$ should be. This transposes the matrix incorrectly.

Only option A correctly places $3,-1,2$ in Row 1, $1,4,-1$ in Row 2, and $-2,1,3$ in Row 3, with constants $5,0,-1$ after the bar.

The third row is $\left(\begin{array}{ccccc}0& 0& 0& |& 7\end{array}\right)$. This represents the equation $0x+0y+0z=7$.

The left-hand side equals $0$ for any values of $x,y,z$, but the right-hand side is $7\ne 0$. This is a contradiction — no values of $x,y,z$ can satisfy this equation.

A contradiction in any row of the reduced matrix means the system is inconsistent: it has no solution. The presence of non-zero entries in the first two rows is irrelevant once a contradiction is found.

The system has no solution. Geometrically, the three planes do not share a common point — they may form a triangular prism configuration or have parallel planes.

The critical row is $\left(\begin{array}{ccccc}0& 0& 0& |& 7\end{array}\right)$, representing $0=7$.

A unique solution requires the left side of the matrix to be the $3\times 3$ identity. Here the third row has all zeros on the left, so no unique solution exists.

Infinite solutions arise when the bottom row is $\left(\begin{array}{ccccc}0& 0& 0& |& 0\end{array}\right)$ (i.e., $0=0$). Here the constant is $7\ne 0$, so this is a contradiction, not a redundancy.

Linear systems never have exactly two solutions — by the theory, there are either $0$, $1$, or infinitely many. The option 'two solutions' is impossible for any linear system.

The bottom row $0=7$ is a contradiction, so the system is inconsistent with no solution.