Studdy AI Logo
Math  /  Algebra

QuestionWrite an augmented matrix and use elementary row operations in order to solve the following system of equations. Your final matrix should be in reduced row echelon form. In order to get credit you will have to have a correct final answer as accurate steps in each row operation. 11x+4y−6z=−4−6x−2y+3z=13x+y−z=2\begin{aligned} 11 x+4 y-6 z & =-4 \\ -6 x-2 y+3 z & =1 \\ 3 x+y-z & =2 \end{aligned}
Write the augmented matrix:

Studdy Solution

STEP 1

What is this asking? We need to solve a system of three equations with three unknowns (xx, yy, and zz) using an augmented matrix and elementary row operations, achieving reduced row echelon form. Watch out! Keep track of your operations and double-check your arithmetic – one small slip-up can lead to a wrong answer!

STEP 2

1. Create the Augmented Matrix
2. Row Reduce to Echelon Form
3. Back Substitution

STEP 3

Let's **create** our augmented matrix!
This matrix combines the coefficients of our variables and the constants from the right-hand side of our equations.
We're doing this so we can perform operations on the entire system simultaneously, which is way more efficient than juggling equations!

STEP 4

The augmented matrix looks like this:
[114−6−4−6−23131−12]\begin{bmatrix} 11 & 4 & -6 & -4 \\ -6 & -2 & 3 & 1 \\ 3 & 1 & -1 & 2 \end{bmatrix}

STEP 5

Our **goal** is to transform this matrix into reduced row echelon form.
This means we want a diagonal of **1**s (with **0**s below them) and then we'll use back-substitution to solve.

STEP 6

First, let's **divide** the first row by **11** to get a leading **1**.
Remember, dividing a row by a non-zero number doesn't change the solution to the system!
1411−611−411−6−23131−12\begin{matrix} 1 & \frac{4}{11} & -\frac{6}{11} & -\frac{4}{11} \\ -6 & -2 & 3 & 1 \\ 3 & 1 & -1 & 2 \end{matrix}

STEP 7

Now, let's **eliminate** the −6-6 and 33 in the first column below the leading **1**.
We'll do this by adding **6** times the first row to the second row, and adding −3-3 times the first row to the third row.
[1411−611−4110211311−13110−1117113411]\begin{bmatrix} 1 & \frac{4}{11} & -\frac{6}{11} & -\frac{4}{11} \\ 0 & \frac{2}{11} & \frac{3}{11} & -\frac{13}{11} \\ 0 & -\frac{1}{11} & \frac{7}{11} & \frac{34}{11} \end{bmatrix}

STEP 8

Next, we'll **multiply** the second row by 112\frac{11}{2} to get a leading **1** in the second row.
[1411−611−4110132−1320−1117113411]\begin{bmatrix} 1 & \frac{4}{11} & -\frac{6}{11} & -\frac{4}{11} \\ 0 & 1 & \frac{3}{2} & -\frac{13}{2} \\ 0 & -\frac{1}{11} & \frac{7}{11} & \frac{34}{11} \end{bmatrix}

STEP 9

Let's **eliminate** the −111-\frac{1}{11} below the leading **1** in the second row by adding 111\frac{1}{11} times the second row to the third row.
[1411−611−4110132−132001252]\begin{bmatrix} 1 & \frac{4}{11} & -\frac{6}{11} & -\frac{4}{11} \\ 0 & 1 & \frac{3}{2} & -\frac{13}{2} \\ 0 & 0 & \frac{1}{2} & \frac{5}{2} \end{bmatrix}

STEP 10

Finally, **multiply** the third row by **2** to get a leading **1** in the third row.
[1411−611−4110132−1320015]\begin{bmatrix} 1 & \frac{4}{11} & -\frac{6}{11} & -\frac{4}{11} \\ 0 & 1 & \frac{3}{2} & -\frac{13}{2} \\ 0 & 0 & 1 & 5 \end{bmatrix}

STEP 11

Now, we can **read** our equations from the matrix and use back-substitution.
The last row tells us z=5z = 5.

STEP 12

The second row tells us y+32z=−132y + \frac{3}{2}z = -\frac{13}{2}.
Substituting z=5z = 5, we get y+152=−132y + \frac{15}{2} = -\frac{13}{2}, so y=−14y = -14.

STEP 13

The first row tells us x+411y−611z=−411x + \frac{4}{11}y - \frac{6}{11}z = -\frac{4}{11}.
Substituting y=−14y = -14 and z=5z = 5, we get x−5611−3011=−411x - \frac{56}{11} - \frac{30}{11} = -\frac{4}{11}, so x=8x = 8.

STEP 14

Our solution is x=8x = 8, y=−14y = -14, and z=5z = 5.

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord