SL 1.8—Use of technology to solve systems of linear equations and polynomial equations

When the rref output has 1s on the main diagonal and 0s everywhere else on the left, each row directly gives the value of one variable. The final column holds the solution values.

Row 1: $\left[\begin{array}{ccccc}1& 0& 0& |& 2\end{array}\right]$ means $x=2$. Row 2: $\left[\begin{array}{ccccc}0& 1& 0& |& 1\end{array}\right]$ means $y=1$. Row 3: $\left[\begin{array}{ccccc}0& 0& 1& |& 1\end{array}\right]$ means $z=1$.

Check in row 1 of the original matrix: $2(2)-1(1)+3(1)=4-1+3=6\ne 7$. Let us verify row 3 instead: $1(2)+0(1)+5(1)=2+0+5=7$. Actually row 3 of the original is $1x+0y+5z=11$: $2+5=7\ne 11$. The rref result as stated in the problem is taken at face value — the solution read directly is $x=2,y=1,z=1$.

The rref output directly gives $x=2$, $y=1$, $z=1$, which matches option A.

The question asks us to read the solution from the given rref output. The final column of the identity-form rref matrix gives the values of $x$, $y$, and $z$ respectively.

Option B states $x=1,y=2,z=1$. But row 1 of the rref output has the constant 2, so $x=2$, not 1. Option B is incorrect.

Option C states $x=2,y=1,z=0$. Row 3 of the rref output has constant 1, so $z=1$, not 0. Option C is incorrect.

Option D states $x=1,y=1,z=2$. This would require the rref constants to be $[1,1,2]$, but the output shows $[2,1,1]$. Option D is incorrect.

Only option A, $x=2,y=1,z=1$, correctly matches the rref output column $[2,1,1]$.

A row $\left[\begin{array}{ccccc}0& 0& 0& |& k\end{array}\right]$ translates to the equation $0x+0y+0z=k$, which simplifies to $0=k$.

Here $k=5\ne 0$. The equation $0=5$ is a contradiction — it can never be true for any values of $x$, $y$, $z$.

A contradiction in the rref output means the original equations are inconsistent — at least two equations contradict each other. Such a system has no solution.

The system has no solution. This corresponds to planes (in 3D) that do not all share a common intersection point.

When the left side of a row is all zeros, there are only two possibilities: either the right side is also 0 (infinitely many solutions) or the right side is non-zero (no solution). Here the right side is 5.

Infinitely many solutions arise when the all-zero row has 0 on the right as well, indicating a redundant equation. Since the right side here is 5, not 0, this option is incorrect.

A unique solution at the origin would require rref to produce an identity matrix with zeros in the final column, not a contradictory row. This option is incorrect.

Linear systems can only have 0, 1, or infinitely many solutions — never exactly 2. This option is mathematically impossible.

The row $[000|5]$ represents the contradiction $0=5$, confirming the system is inconsistent with no solution.