A system of equations may have a unique solution, an infinite number of solutions, or no solution. Use matrices to find the general solution of the following system, if a solution exists.

Question

Quizplus · Accepted Answer

We can represent the system of equations as a matrix equation:

$$\begin{bmatrix}
1&1&1\
3&4&-1\
2&-1&3
\end{bmatrix} \begin{bmatrix}
x\y\z
\end{bmatrix} = \begin{bmatrix}
2\4\-1
\end{bmatrix}$$

Using row reduction, we get the row echelon form:

$$\begin{bmatrix}
1&1&1&|&2\
0&1&-4&|&-2\
0&0&1&|&1
\end{bmatrix}$$

This implies that $z=1$, and substituting this back into the system gives:

$$\begin{aligned}x+y&=1\y-4z&=-2\2x-z&=-2\end{aligned}$$

Solving for $x$ and $y$, we get:

$$\begin{aligned}x&=1-y\y&=2+4z\x&=-2+z\end{aligned}$$

Therefore, the general solution is $x=-2+z$, $y=2+4z$, and $z=z$, for any real number $z$. This matches choice C.

A System of Equations May Have a Unique Solution, an Infinite