Question

solve the system of equation by gauss elimination method x+2y+z=3,
2x+3y+3z=10
3x-y+2z=13​

Answer 1

Step-by-step explanation:

The augmented matrix is A | B

1 2 3 x. 3

2 3 3. y = 10

3 -1 2 z. 13

1 2 3. x 3

0 -1 -4 y = 4 R²→R²-2R¹

0 -7 -4 z 4 R³→ R³-3R¹

1 2 3 x 3

0 -1 -4 y = 4

0 0 24 Z -15 R³→R³-7R²

The given system of equations

x+2y+3z=3 ---------(1)

-y -4z = 4--------(2)

24z= -15 -----(3)

z = -15/24= -5/8

Substitute z = - 5/8 in (2)

-y -4(-5/8)= 4

-y +5/2= 4

-y= 4-5/2

-y = 3/2

y = -3/2

Substitute z & y in (1)

x + 2(-3/2)+3(-5/8)=3

x-3-15/8 = 3

x = 143/8

Answer 2

Answer:

Final Answer

Step-by-step explanation:

Given

System of equations

x+2y+z=3  ---- (i)

2x+3y+3z=10  ---- (ii)

3x-y+2z=13  ---- (iii)

we will solve it by using augmented matrix

Using the method A.I=B

Here, A = $\left[\begin{array}{ccc}1&3&1\\2&3&3\\3&-1&2\end{array}\right]$

         I = $\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$

    &   B = $\left[\begin{array}{ccc}3\\10\\13\end{array}\right]$

So, A.I=B will be:

$\left[\begin{array}{ccc}1&3&1\\2&3&3\\3&-1&2\end{array}\right]$.$\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ = $\left[\begin{array}{ccc}3\\10\\13\end{array}\right]$

Using Row Transformation, $R_{2}$ → $R_{2}$ - $R_{1}$

                                          & $R_{3}$ → $R_{3}$ - $R_{1}$

we get,

$\left[\begin{array}{ccc}1&2&1\\0&-1&1\\0&-7&-1\end{array}\right]$.$\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ = $\left[\begin{array}{ccc}3\\-4\\-4\end{array}\right]$

Row Transformation $R_{2}$ → $-R_{2}$

                                  $R_{3}$ → $-R_{3}$

$\left[\begin{array}{ccc}1&2&1\\0&1&-1\\0&7&1\end{array}\right]$.$\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ = $\left[\begin{array}{ccc}3\\-4\\-4\end{array}\right]$

Row Transformation, $R_{3}$ → $R_{3}$ - $7R_{2}$

$\left[\begin{array}{ccc}1&2&1\\0&1&-1\\0&0&8\end{array}\right]$.$\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ = $\left[\begin{array}{ccc}3\\-4\\24\end{array}\right]$

Row Transformation, $R_{3}$ → $\frac{1}{8}$$R_{3}$

$\left[\begin{array}{ccc}1&2&1\\0&1&-1\\0&0&1\end{array}\right]$.$\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ = $\left[\begin{array}{ccc}3\\-4\\3\end{array}\right]$

From the above matrix, we get the following equations,

z=3

y-z = -4, and

x+2y+z=3

Solving these equations, we get the values of x, y and z

z=3

y-z = -4

⇒ y = -4+z = -1

for x: x+2y+z=3

x = 3-2y-z

x = 3-2(-1)-3

x = 3+2-3 = 2

Hence, the final solution is,

x = 2, y = -1, z = 3

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