The augmented matrix is in row-echelon form. Assume that the variables are \( x \) and \( y \) and use back substitution to obtain the solution of the associated system of linear equations.
[tex]\[
\left[\begin{array}{ll|r}
1 & 2 & -12 \\
0 & 1 & -5
\end{array}\right]
\][/tex]

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

A. There is one solution. The solution set is \(\{ \square \}\).
(Simplify your answer. Type an ordered pair, using integers or fractions.)

B. There are infinitely many solutions. The solution set is the set of all ordered pairs \(\{ (\square, y )\}\), where \( y \) is any real number.
(Type an expression using \( y \) as the variable. Simplify your answer.)

C. The system is inconsistent. The solution set is [tex]\(\varnothing\)[/tex].



Answer :

To solve the given system of linear equations using the augmented matrix:
[tex]\[ \begin{bmatrix} 1 & 2 & -12 \\ 0 & 1 & -5 \end{bmatrix} \][/tex]
we can perform back substitution since the matrix is already in row-echelon form.

First, let's interpret the augmented matrix as a system of linear equations:

[tex]\[ \begin{cases} x + 2y = -12, \\ y = -5 \end{cases} \][/tex]

From the second equation, we have:
[tex]\[ y = -5 \][/tex]

With this value of \( y \), we can substitute it back into the first equation to find \( x \):

[tex]\[ x + 2(-5) = -12 \\ x - 10 = -12 \\ x = -12 + 10 \\ x = -2 \][/tex]

So, the values of \( x \) and \( y \) are:
[tex]\[ x = -2 \][/tex]
[tex]\[ y = -5 \][/tex]

Therefore, the solution set is the ordered pair:
[tex]\[ (-2, -5) \][/tex]

Considering the given options, the correct choice is:
A. There is one solution. The solution set is [tex]\(\{ (-2, -5) \} \)[/tex].

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