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Use the Gaussian elimination method to solve the system of linear equations. If the system has infinitely many solutions, write the solution set with \( z \) arbitrary.

[tex]\[
\begin{array}{rr}
x + 2y + 2z = & 10 \\
-x + 3y + 5z = & 13 \\
x + y - 4z = & -13
\end{array}
\][/tex]

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 \(\{(\square, \square, \square)\}\).
(Simplify your answers. Type integers or decimals.)

B. There are infinitely many solutions. The solutions are \(\{(\square, \square, z)\}\), where \( z \) is any real number. \(\square\)
(Simplify your answers. Type integers or decimals.)

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

Sagot :

GinnyAnswer
Sure, let's solve this system of linear equations using the Gaussian elimination method:

[tex]\[ \begin{array}{rr} x + 2y + 2z &= 10 \\ -x + 3y + 5z &= 13 \\ x + y - 4z &= -13 \end{array} \][/tex]

First, we construct the augmented matrix for the system:

[tex]\[ \begin{pmatrix} 1 & 2 & 2 & | & 10 \\ -1 & 3 & 5 & | & 13 \\ 1 & 1 & -4 & | & -13 \end{pmatrix} \][/tex]

Step 1: Eliminate \(x\) from the second and third rows

Add the first row to the second row:

[tex]\[ \begin{pmatrix} 1 & 2 & 2 & | & 10 \\ 0 & 5 & 7 & | & 23 \\ 1 & 1 & -4 & | & -13 \end{pmatrix} \][/tex]

Subtract the first row from the third row:

[tex]\[ \begin{pmatrix} 1 & 2 & 2 & | & 10 \\ 0 & 5 & 7 & | & 23 \\ 0 & -1 & -6 & | & -23 \end{pmatrix} \][/tex]

Step 2: Eliminate \(y\) from the third row

Multiply the second row by \( \frac{1}{5} \):

[tex]\[ \begin{pmatrix} 1 & 2 & 2 & | & 10 \\ 0 & 1 & \frac{7}{5} & | & \frac{23}{5} \\ 0 & -1 & -6 & | & -23 \end{pmatrix} \][/tex]

Add the second row to the third row:

[tex]\[ \begin{pmatrix} 1 & 2 & 2 & | & 10 \\ 0 & 1 & \frac{7}{5} & | & \frac{23}{5} \\ 0 & 0 & -\frac{23}{5} & | & \frac{23}{5} - 23 \end{pmatrix} \][/tex]

Simplifying the third row:

[tex]\[ \begin{pmatrix} 1 & 2 & 2 & | & 10 \\ 0 & 1 & \frac{7}{5} & | & \frac{23}{5} \\ 0 & 0 & -\frac{23}{5} & | & -\frac{92}{5} \end{pmatrix} \][/tex]

Step 3: Solve for \(z\)

From the third row:

[tex]\[ -\frac{23}{5}z = -\frac{92}{5} \implies z = 4 \][/tex]

Step 4: Back-substitute to find \(y\) and \(x\)

Substitute \(z = 4\) into the second row:

[tex]\[ 0 + y + \frac{7}{5}(4) = \frac{23}{5} \implies y + \frac{28}{5} = \frac{23}{5} \implies y = \frac{23}{5} - \frac{28}{5} = -1 \][/tex]

Substitute \( y = -1 \) and \( z = 4 \) into the first row:

[tex]\[ x + 2(-1) + 2(4) = 10 \implies x - 2 + 8 = 10 \implies x + 6 = 10 \implies x = 4 \][/tex]

Therefore, the solution is \((4, -1, 4)\).

So, the correct choice is:
A. There is one solution. The solution is [tex]\(\{(4, -1, 4)\}\)[/tex].
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