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]
