Question

solve the equation of cramers rule x+y+z=3,2x+3y+4z=9,4x-y+z=4​

Answer 1

Step-by-step explanation:

x+y+z=3,2x+3y+4z=9,4x-y+z=4

Answer 2

Answer:

Solving by Cramer's rule

The values are: x = 1, y = 1, and z = 1.

Step-by-step explanation:

Given, the linear equations are:

x + y + z = 3

2x + 3y + 4z = 9

4x - y + z = 4

Solving using Cramar's Rule, we get

$\triangle = \left[\begin{array}{ccc}1&1&1\\2&3&4\\4&-1&1\end{array}\right] \\\\\triangle = 1[3\times 1 - 4\times (-1)] - 1[2\times 1 - 4\times4] + 1[2\times (-1) - 4\times 3]\\\\\triangle = 7 + 14 + (-14)\\\\\triangle = 7$

$\triangle_1 = \left[\begin{array}{ccc}3&1&1\\9&3&4\\4&-1&1\end{array}\right] \\\\\triangle_1 = 3[3\times 1 - 4\times(-1)] - 1[9\times1 - 4\times4] + 1[9\times(-1) - 4\times3]\\\\\triangle_1 = 3\times7 - (-7)+ (-21)\\\\\triangle_1 = 7$

$\triangle_2 = \left[\begin{array}{ccc}1&3&1\\2&9&4\\4&4&1\end{array}\right] \\\\\triangle_2 = 1[9\times 1 - 4\times4] - 3[2\times1 - 4\times4] + 1[2\times4 - 4\times9]\\\\\triangle_2 = -7 + 42 -28\\\\\triangle_2 = 7$

$\triangle_3 = \left[\begin{array}{ccc}1&1&3\\2&3&9\\4&-1&4\end{array}\right] \\\\\triangle_3 = 1[3\times 4 - 9\times(-1)] - 1[2\times4 - 4\times9] + 3[2\times(-1) - 4\times9]\\\\\triangle_3 = 21 + 28 -42\\\\\triangle_3 = 7$

∵ Δ≠0, Δ₁≠0, Δ₂≠0, and Δ₃≠0

So, the solutions will be unique.

$x = \frac{\triangle_1}{\triangle}\ \ , \ \ y = \frac{\triangle_2}{\triangle}\ \ , \ \ z = \frac{\triangle_3}{\triangle} \\\\x = \frac{7}{7}\ \ , \ \ y = \frac{7}{7}\ \ , \ \ z = \frac{7}{7}$

Therefore, x= 1, y = 1, and z = 1.

To learn more about the Cramer's rule, click on the link below:

https://brainly.in/question/54131131

To learn more about the linear equations, click on the link below:

https://brainly.in/question/48534636

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