Question

Question 8

Answer 1

we are given

system of equations

First equation is

$3x-2y=31$

Second equation is

$3x+2y=-1$

now, we can find augmented matrix

$A=\begin{pmatrix}3&-2&31\\ 3&2&-1\end{pmatrix}$

now, we can change it into reduced row echelon form

step-1: multiply the 1st row by 1/3

$A=\begin{pmatrix}1&-2/3&31/3\\ 3&2&-1\end{pmatrix}$

step-2:add -3 times the 1st row to the 2nd row

$A=\begin{pmatrix}1&-2/3&31/3\\ 0&4&-32\end{pmatrix}$

step-3:multiply the 2nd row by 1/4

$A=\begin{pmatrix}1&-2/3&31/3\\ 0&1&-8\end{pmatrix}$

step-4: add 2/3 times the 2nd row to the 1st row

$A=\begin{pmatrix}1&0&5\\ 0&1&-8\end{pmatrix}$

so, we get

$x=5,y=-8$

Answer is (5,-8)

Answer 2

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   $\frac{-2}{3}$   | $\frac{31}{3}$    -------------> Row 1

0   4   | -32  -------------> Row 2

Divide row 2 by 4. R2 -> R2/4

1   $\frac{-2}{3}$   | $\frac{31}{3}$    -------------> 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   $1*\frac{2}{3}+\frac{-2}{3}$   | $-8*\frac{-2}{3}\frac{31}{3}$    -------------> 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)