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Math  /  Algebra

QuestionSolve the system by using Gaussian elimination of Gauss-Jordan elimination. −2x−3y+5z=−10x+2y+2z=105x−3y+2z=29\begin{aligned} -2 x-3 y+5 z & =-10 \\ x+2 y+2 z & =10 \\ 5 x-3 y+2 z & =29 \end{aligned}

Studdy Solution

STEP 1

What is this asking? We need to find the values of xx, yy, and zz that satisfy all three equations simultaneously using either Gaussian or Gauss-Jordan elimination. Watch out! Keep those signs straight, and don't rush those calculations!
A single slip-up can lead you down the wrong path.

STEP 2

1. Set up the augmented matrix
2. Perform Gauss-Jordan elimination
3. Extract the solution

STEP 3

Let's **create** our augmented matrix from the coefficients of the variables and the constants on the right-hand side of the equations.
This neatly organizes our problem: [−2−35−10122105−3229]\begin{bmatrix} -2 & -3 & 5 & -10 \\ 1 & 2 & 2 & 10 \\ 5 & -3 & 2 & 29 \end{bmatrix}

STEP 4

We want a **leading 1** in the first row, first column.
Swapping rows 1 and 2 is an easy way to achieve this: [12210−2−35−105−3229]\begin{bmatrix} 1 & 2 & 2 & 10 \\ -2 & -3 & 5 & -10 \\ 5 & -3 & 2 & 29 \end{bmatrix}

STEP 5

Now let's **eliminate** the −2-2 and 55 below the leading 11.
We'll add 22 times the first row to the second row, and add −5-5 times the first row to the third row: [12210019100−13−8−21]\begin{bmatrix} 1 & 2 & 2 & 10 \\ 0 & 1 & 9 & 10 \\ 0 & -13 & -8 & -21 \end{bmatrix}

STEP 6

Great! Now, we want a **leading 1** in the second row, second column, which we already have!
Let's **eliminate** the 22 and −13-13 above and below this leading 11.
We'll add −2-2 times the second row to the first row, and add 1313 times the second row to the third row: [10−16−100191000109109]\begin{bmatrix} 1 & 0 & -16 & -10 \\ 0 & 1 & 9 & 10 \\ 0 & 0 & 109 & 109 \end{bmatrix}

STEP 7

Almost there!
We want a **leading 1** in the third row, third column.
We can achieve this by dividing the third row by 109109: [10−16−10019100011]\begin{bmatrix} 1 & 0 & -16 & -10 \\ 0 & 1 & 9 & 10 \\ 0 & 0 & 1 & 1 \end{bmatrix}

STEP 8

Finally, let's **eliminate** the −16-16 and 99 above our leading 11.
We'll add 1616 times the third row to the first row, and add −9-9 times the third row to the second row: [100601010011]\begin{bmatrix} 1 & 0 & 0 & 6 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \end{bmatrix}

STEP 9

This final matrix tells us that x=6x = 6, y=1y = 1, and z=1z = 1.
Boom!

STEP 10

The solution to the system of equations is x=6x = 6, y=1y = 1, and z=1z = 1.

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