Is it a solution of the linear system?

[frctl]

Determine whether the given 3-tuple is a solution of the linear system.
2x₁ - 4x₂ - x₃ = 1
x₁ - 3x₂ + x₃ = 1
3x₁ - 5x₂ - 3x₃ = 1

a) (3,1,1)

 

[Dr.Peterson]

Hi, frctl. What have you tried? Have you replaced the variables in each equation with their values and checked whether they are all true?

 

[frctl]

2(3) - 4(1) - (1) = 1 therefore true (obtained 1)
(1) - 3(1) + 1 = 1 therefore false (obtained -1)
3(1) - 5(1) - 3(1) = 1 therefore false (obtained -5)

 

[Dr.Peterson]

2(3) - 4(1) - (1) = 1 therefore true (obtained 1)
(1) - 3(1) + 1 = 1 therefore false (obtained -1)
3(1) - 5(1) - 3(1) = 1 therefore false (obtained -5)

Check your check!

 

[frctl]

Do you mean to enter (1, -1, -5) into
2(1) - 4(1) - (1) = 1 therefore true
(-1) - 3(-1) + 1 = -8 therefore false
3(-5) - 5(-5) - 3(-5) = -34 therefore false

 

[ksdhart2]

Do you mean to enter (1, -1, -5)...

No. Why would that make any sense to do? How would plugging in $x_1 = 1, \: x_2 = -1, \: x_3 = -5$ give you any insight into whether or not $x_1 = 3, \: x_2 = 1, \: x_3 = 1$ is a solution?

What Dr. P. meant was to check to make sure you copied down the values correctly (hint: you didn't). Let $x_1 = {\color{red}3}, \: x_2 = 1, \: x_3 = 1$ and plug in:

\(\displaystyle
2({\color{red}3}) - 4(1) - (1) = 1 \: \: \: \checkmark \\
({\color{red}3}) - 3(1) + (1) = 1 \: \: \: \checkmark \\
3({\color{red}3}) - 5(1) - 3(1) = 1 \: \: \: \checkmark
\)
Contrast that with what you wrote:

\(\displaystyle
2({\color{red}3}) - 4(1) - (1) = 1 \\
({\color{red}1}) - 3(1) + (1) = -1 \\
3({\color{red}1}) - 5(1) - 3(1) = -5
\)

 

[frctl]

I understand now, thank you for displaying to me my error and the proper solution.