Respuesta :
we are given
system of equations
First equation is
[tex]3x-2y=31[/tex]
Second equation is
[tex]3x+2y=-1[/tex]
now, we can find augmented matrix
[tex]A=\begin{pmatrix}3&-2&31\\ 3&2&-1\end{pmatrix}[/tex]
now, we can change it into reduced row echelon form
step-1: multiply the 1st row by 1/3
[tex]A=\begin{pmatrix}1&-2/3&31/3\\ 3&2&-1\end{pmatrix}[/tex]
step-2:add -3 times the 1st row to the 2nd row
[tex]A=\begin{pmatrix}1&-2/3&31/3\\ 0&4&-32\end{pmatrix}[/tex]
step-3:multiply the 2nd row by 1/4
[tex]A=\begin{pmatrix}1&-2/3&31/3\\ 0&1&-8\end{pmatrix}[/tex]
step-4: add 2/3 times the 2nd row to the 1st row
[tex]A=\begin{pmatrix}1&0&5\\ 0&1&-8\end{pmatrix}[/tex]
so, we get
[tex]x=5,y=-8[/tex]
Answer is (5,-8)
Solution is (5,-8)
3x – 2y = 31
3x + 2y = -1
first we write it in matrix form
3 -2 | 31 -------------> Row 1
3 2 | -1 -------------> Row 2
We use row operations and get matrix in 1 0 and 0 1 form
Subtract Row 2 by row 1. R2 -> R2 - R1. So R1 remains the same
3 -2 | 31 -------------> Row 1
0 4 | -32 -------------> Row 2
Divide row 1 by 3, R1-> R1/3. So R2 remains the same
1 [tex]\frac{-2}{3}[/tex] | [tex]\frac{31}{3}[/tex] -------------> Row 1
0 4 | -32 -------------> Row 2
Divide row 2 by 4. R2 -> R2/4
1 [tex]\frac{-2}{3}[/tex] | [tex]\frac{31}{3}[/tex] -------------> Row 1
0 1 | -8 -------------> Row 2
Now we need to get 0 in the place of -2/3
Multiply Row 2 by 2/3 and add it with Row 1. R1 --> 2/3 times R2 + R1
1 [tex]1*\frac{2}{3}+\frac{-2}{3}[/tex] | [tex]-8*\frac{-2}{3}\frac{31}{3}[/tex] -------------> Row 1
0 1 | -8 -------------> Row 2
Now simplify the fraction
1 0 | 5 -------------> Row 1
0 1 | -8 -------------> Row 2
So from this we can see that x=5 and y = -8
Solution is (5,-8)
