PLEASE HELP! Write an augmented matrix.

Use elementary row operations in order to solve the following system of equations. Your final matrix should be in reduced row echelon form. In order to get credit you will have to have a correct final answer as accurate steps in each row operation.
All help is appreciated! :)

10x + 3y = 4
2x + y = 8

PLEASE HELP Write an augmented matrix Use elementary row operations in order to solve the following system of equations Your final matrix should be in reduced r class=


Answer :

Answer:

(-5, 18)

Step-by-step explanation:

Augmented Matrix:

[tex]\left[\begin{array}{ccc}10&3&4\\2&1&8\end{array}\right][/tex]

Reduced Row Echelon form:

[tex]\left[\begin{array}{ccc}1&0&x\\0&1&y\end{array}\right][/tex]

In Elementary Row Operations, you can...
1. Multiply an Entire Row by a Constant (just a number);
2. Add constant multiples of Rows together.

First, get "1" in the top left corner by multiplying Row1 by 1/10:
[tex]\frac{1}{10} \left[\begin{array}{ccc}10&3&4\end{array}\right] =\left[\begin{array}{ccc}1&\frac{3}{10}&\frac{2}{5} \end{array}\right][/tex]
Replace the [tex]\left[\begin{array}{ccc}10&3&4\end{array}\right] with \left[\begin{array}{ccc}1&\frac{3}{10}&\frac{2}{5} \end{array}\right],[/tex] and you now have [tex]\left[\begin{array}{ccc}1&\frac{3}{10}&\frac{2}{5}\\2&1&8\end{array}\right][/tex].

Next, get a "0" in the bottom left corner by multiplying Row1 by -2 and adding it to Row2:  
[tex]-2\left[\begin{array}{ccc}1&\frac{3}{10}&\frac{2}{5} \end{array}\right]=\left[\begin{array}{ccc}-2&-\frac{3}{5}&-\frac{4}{5} \end{array}\right][/tex];
[tex]\left[\begin{array}{ccc}-2&-\frac{3}{5}&-\frac{4}{5} \end{array}\right]+\left[\begin{array}{ccc}2&1&8\end{array}\right]=\left[\begin{array}{ccc}0&\frac{2}{5}&\frac{36}{5}\end{array}\right][/tex]

Replace the [tex]\left[\begin{array}{ccc}2&1&8\end{array}\right] with\left[\begin{array}{ccc}0&\frac{2}{5}&\frac{36}{5}\end{array}\right][/tex], and you now have
[tex]\left[\begin{array}{ccc}1&\frac{3}{10}&\frac{2}{5}\\0&\frac{2}{5}&\frac{36}{5}\end{array}\right][/tex].

WITHOUT CHANGING ANYTHING ELSE, get a "1" in Row2Column2 by Multiplying Row2 by 5/2:
[tex]\frac{5}{2} \left[\begin{array}{ccc}0&\frac{2}{5}&\frac{36}{5}\end{array}\right]=\left[\begin{array}{ccc}0&1&18\end{array}\right][/tex]  ⇒  [tex]\left[\begin{array}{ccc}1&\frac{3}{10}&\frac{2}{5}\\0&1&18\end{array}\right][/tex]

Finally, change 2/5 to a "0" by multiplying Row2 by -3/10 and adding that result to Row1:
[tex]-\frac{3}{10}\left[\begin{array}{ccc}0&1&18\end{array}\right] =\left[\begin{array}{ccc}0&-\frac{3}{10}&-\frac{27}{5}\end{array}\right][/tex];
[tex]\left[\begin{array}{ccc}0&-\frac{3}{10}&-\frac{27}{5}\end{array}\right]+\left[\begin{array}{ccc}1&\frac{3}{10}&\frac{2}{5} \end{array}\right]=\left[\begin{array}{ccc}1&0&-5 \end{array}\right][/tex];
⇒[tex]\left[\begin{array}{ccc}1&0&-5\\0&1&18\end{array}\right][/tex].

From this matrix, we now see that:
1x + 0y = -5  ⇒  x = -5    and    0x + 1y = 18 ⇒ y = 18

Answer:

[tex]\left[\begin{array}{cc|c}1&0&-5\\0&1&18\end{array}\right][/tex]

x = -5,  y = 18

Step-by-step explanation:

Given system of equations:

[tex]10x+3y=4\\\\2x+y=8[/tex]

To write a system of linear equations in matrix form, place the coefficients of the variables from the equations on the left side, with each row representing an equation and each column representing a variable, with an additional column on the right side containing the constants from the equations:

[tex]\left[\begin{array}{cc|c}10&3&4\\2&1&8\end{array}\right][/tex]

An augmented matrix is said to be in reduced row echelon form if it satisfies the following conditions:

  • All zero rows, if any, are at the bottom.
  • The leading entry (the first non-zero entry) of each non-zero row is 1.
  • The leading 1 in each non-zero row occurs to the right of the leading 1 in the row above it.
  • All entries in the column above and below a leading 1 are zero.

Divide the first row (R₁) by 10:

[tex]\left[\begin{array}{cc|c}\vphantom{\dfrac12}10&3&4\\2&1&8\end{array}\right]\dfrac{R_1}{10}\rightarrow R_1\left[\begin{array}{cc|c}\vphantom{\dfrac12}1&\frac{3}{10}&\frac{2}{5}\\2&1&8\end{array}\right][/tex]

Subtract 2 times the first row (R₁) from the second row (R₂):

[tex]\left[\begin{array}{cc|c}\vphantom{\dfrac12}1&\frac{3}{10}&\frac{2}{5}\\2&1&8\end{array}\right] R_2-2R_1\rightarrow R_2 \left[\begin{array}{cc|c}\vphantom{\dfrac12}1&\frac{3}{10}&\frac{2}{5}\\0&\frac{2}{5}&\frac{36}{5}\end{array}\right][/tex]

Multiply the second row (R₂) by 5/2:

[tex]\left[\begin{array}{cc|c}\vphantom{\dfrac12}1&\frac{3}{10}&\frac{2}{5}\\0&\frac{2}{5}&\frac{36}{5}\end{array}\right]2R_2\rightarrow R_2 \left[\begin{array}{cc|c}\vphantom{\dfrac12}1&\frac{3}{10}&\frac{2}{5}\\0&1&18\end{array}\right][/tex]

Subtract 3/10 times the second row (R₂) from the first row (R₁):

[tex]\left[\begin{array}{cc|c}\vphantom{\dfrac12}1&\frac{3}{10}&\frac{2}{5}\\0&1&18\end{array}\right] R_1-\frac{3}{10}R_2 \rightarrow R_1 \left[\begin{array}{cc|c}\vphantom{\dfrac12}1&0&-5\\0&1&18\end{array}\right][/tex]

Therefore, the final matrix in reduced row echelon form is:

[tex]\left[\begin{array}{cc|c}1&0&-5\\0&1&18\end{array}\right][/tex]

In reduced row echelon form, each row corresponds to an equation:

  • The first row represents the equation 1x + 0y = -5, which simplifies to x = -5.
  • The second row represents the equation 0x + 1y = 18, which simplifies to y = 18.

So, the solutions to the system of equations are:

  • [tex]x = -5[/tex]
  • [tex]y = 18[/tex]