Question

use matrices to solve the following system of linear equations.\

$$\begin{cases}x + y = 1\\\\3y = 6\\\\2x + y - 5z = 10\\end{cases}$$

\select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.\\a. there is one solution. the solution is \\((\square,\square,\square)\\).\\(\text{(type integers or simplified fractions.)}\\)\\b. there are infinitely many solutions. the solution set is \\(\\{(x,y,z) | 3y = 6\\}\\).\\c. the solution set is \\(\varnothing\\).

Explanation:

Step1: Analyze the system of equations

We have the system:

$$
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$$

First, solve the second equation \(3y = 6\) for \(y\). Divide both sides by 3: \(y=\frac{6}{3} = 2\).

Step2: Substitute \(y = 2\) into the first equation

Substitute \(y = 2\) into \(x + y = 1\): \(x+2 = 1\). Subtract 2 from both sides: \(x=1 - 2=-1\).

Step3: Substitute \(x=-1\) and \(y = 2\) into the third equation

Substitute into \(2x + y - 5z = 10\): \(2(-1)+2 - 5z = 10\). Simplify the left - hand side: \(-2 + 2-5z=10\), which simplifies to \(-5z = 10\). Divide both sides by \(-5\): \(z=\frac{10}{-5}=-2\).

So there is one solution \((x,y,z)=(-1,2,-2)\).

Answer:

A. There is one solution. The solution is \((-1,2,-2)\)