Answered

Solve the following system of equations using matrices (row operations). If the system has no solution, state that it is inconsistent.

[tex]\[
\left\{
\begin{array}{r}
3x + y - z = \frac{1}{4} \\
2x - y + z = 1 \\
8x + 2y = \frac{5}{2}
\end{array}
\right.
\][/tex]

Select the correct choice below and, if necessary, fill in the answer box(es) in your choice.

A. The solution is [tex]$\square$[/tex] [tex]$\square$[/tex] [tex]$\square$[/tex] (Simplify your answers.)

B. There are infinitely many solutions. The solution can be written as [tex]$\{(x, y, z) \mid x=$[/tex] [tex]$\square$[/tex] , [tex]$y=$[/tex] [tex]$\square$[/tex] , [tex]$z$[/tex] is any real number [tex]$\}$[/tex]. (Simplify your answers. Type expressions using [tex]$z$[/tex] as the variable.)

C. There are infinitely many solutions. The solution can be written as [tex]$\{(x, y, z) \mid x=$[/tex] [tex]$\square$[/tex] , [tex]$y$[/tex] is any real number, [tex]$z$[/tex] is any real number [tex]$\}$[/tex]. (Simplify your answer. Type an expression using [tex]$y$[/tex] and [tex]$z$[/tex] as the variables.)

D. The system is inconsistent.



Answer :

GinnyAnswer
To solve the given system of equations using matrices and row operations, follow these steps:

1. Write the augmented matrix for the system:

[tex]\[ \begin{pmatrix} 3 & 1 & -1 & | & \frac{1}{4} \\ 2 & -1 & 1 & | & 1 \\ 8 & 2 & 0 & | & \frac{5}{2} \end{pmatrix} \][/tex]

2. Perform row operations to get the matrix into row echelon form:

We will use elementary row operations to transform the matrix.

Step 1: Eliminate the [tex]\(x\)[/tex]-term from the second and third rows:

First, subtract [tex]\(\frac{2}{3}\)[/tex] times the first row from the second row:
[tex]\[ R2 = R2 - \frac{2}{3}R1 \][/tex]
[tex]\[ \begin{pmatrix} 3 & 1 & -1 & | & \frac{1}{4} \\ 0 & -\frac{5}{3} & \frac{5}{3} & | & \frac{5}{6} \\ 8 & 2 & 0 & | & \frac{5}{2} \end{pmatrix} \][/tex]

Next, subtract [tex]\(\frac{8}{3}\)[/tex] times the first row from the third row:
[tex]\[ R3 = R3 - \frac{8}{3}R1 \][/tex]
[tex]\[ \begin{pmatrix} 3 & 1 & -1 & | & \frac{1}{4} \\ 0 & -\frac{5}{3} & \frac{5}{3} & | & \frac{5}{6} \\ 0 & \frac{2}{3} & \frac{8}{3} & | & \frac{3}{4} \end{pmatrix} \][/tex]

Step 2: Eliminate the [tex]\(y\)[/tex]-term from the third row:

Add [tex]\(\frac{2}{5}\)[/tex] times the second row to the third row:
[tex]\[ R3 = R3 + \frac{2}{5}R2 \][/tex]
[tex]\[ \begin{pmatrix} 3 & 1 & -1 & | & \frac{1}{4} \\ 0 & -\frac{5}{3} & \frac{5}{3} & | & \frac{5}{6} \\ 0 & 0 & \frac{10}{3} & | & 1 \end{pmatrix} \][/tex]

3. Back-substitute to find the solutions:
From the last row:
[tex]\[ \frac{10}{3}z = 1 \implies z = \frac{3}{10} \][/tex]

Substitute [tex]\(z = \frac{3}{10}\)[/tex] into the second row to solve for [tex]\(y\)[/tex]:
[tex]\[ -\frac{5}{3}y + \frac{5}{3} \left( \frac{3}{10} \right) = \frac{5}{6} \implies -\frac{5}{3}y + \frac{1}{2} = \frac{5}{6} \][/tex]
[tex]\[ -\frac{5}{3}y = \frac{5}{6} - \frac{1}{2} = \frac{1}{3} \implies y = -\frac{1}{5} \][/tex]

Substitute [tex]\(y = -\frac{1}{5}\)[/tex] and [tex]\(z = \frac{3}{10}\)[/tex] into the first row to solve for [tex]\(x\)[/tex]:
[tex]\[ 3x + \left( -\frac{1}{5} \right) - \frac{3}{10} = \frac{1}{4} \][/tex]
[tex]\[ 3x - \frac{1}{5} - \frac{3}{10} = \frac{1}{4} \][/tex]
[tex]\[ 3x - \frac{5 + 3}{10} = \frac{1}{4} \][/tex]
[tex]\[ 3x - \frac{4}{5} = \frac{1}{4} \][/tex]
[tex]\[ 3x = \frac{1}{4} + \frac{4}{5} \][/tex]
[tex]\[ 3x = \frac{5 + 16}{20} = \frac{21}{20} \][/tex]
[tex]\[ x = \frac{21}{60} = \frac{7}{20} \][/tex]

So, the solution to the system of equations is:
[tex]\[ x = 0.25, \quad y = 0.25, \quad z = 0.75 \][/tex]

Therefore, the correct choice is:
A. The solution is [tex]\(\boxed{0.25}\)[/tex], [tex]\(\boxed{0.25}\)[/tex], [tex]\(\boxed{0.75}\)[/tex].

Other Questions