Question

Solve the system of 3×3 linear equations using elementary row operations

Answer 1

We want to solve the following system

`\begin{cases}4x+3y-8z=-19 \\ x+y-2z=-4 \\ 3x+2y-5z=-13\end{cases}`

using the elimination method.

Gaussian elimination is the name of the method we use to perform the three types of matrix row operations on an augmented matrix coming from a linear system of equations in order to find the solutions for such system.

The Gaussian elimination rules are the same as the rules for the three elementary row operations, in other words, you can algebraically operate on the rows of a matrix in the next three ways (or combination of):

1 - Interchanging two rows

2 - Multiplying a row by a constant (any constant which is not zero)

3 - Adding a row to another row

We start our process by converting the system into an augmented matrix.

`\begin{cases}4x+3y-8z=-19 \\ x+y-2z=-4 \\ 3x+2y-5z=-13\end{cases}\Leftrightarrow\begin{bmatrix}4 & 3 & -8 & \bigm| & -19 \\ 1 & 1 & -2 & \bigm| & -4 \\ 3 & 2 & -5 & \bigm| & -13\end{bmatrix}`

Now, starting the elimination process, we can add (-4) times the second row to the first row and add (-3) times the second row to the third row.

`\begin{bmatrix}4 & 3 & -8 & \bigm & -19 \\ 1 & 1 & -2 & \bigm & -4 \\ 3 & 2 & -5 & \bigm & -13\end{bmatrix}\rightarrow\begin{bmatrix}0 & -1 & 0 & \bigm & -3 \\ 1 & 1 & -2 & \bigm & -4 \\ 0 & -1 & 1 & \bigm & -1\end{bmatrix}`

Then, interchanging the first and the second row

`\begin{bmatrix}0 & -1 & 0 & \bigm & -3 \\ 1 & 1 & -2 & \bigm & -4 \\ 0 & -1 & 1 & \bigm & -1\end{bmatrix}\rightarrow\begin{bmatrix}1 & 1 & -2 & \bigm & -4 \\ 0 & -1 & 0 & \bigm & -3 \\ 0 & -1 & 1 & \bigm & -1\end{bmatrix}`

Adding the second row to the first row, we have

`\begin{bmatrix}1 & 1 & -2 & \bigm & -4 \\ 0 & -1 & 0 & \bigm & -3 \\ 0 & -1 & 1 & \bigm & -1\end{bmatrix}\rightarrow\begin{bmatrix}1 & 0 & -2 & \bigm & -7 \\ 0 & -1 & 0 & \bigm & -3 \\ 0 & -1 & 1 & \bigm & -1\end{bmatrix}`

Multiplying the second row by (-1), we have

`\begin{bmatrix}1 & 0 & -2 & \bigm & -7 \\ 0 & -1 & 0 & \bigm & -3 \\ 0 & -1 & 1 & \bigm & -1\end{bmatrix}\rightarrow\begin{bmatrix}1 & 0 & -2 & \bigm & -7 \\ 0 & 1 & 0 & \bigm & 3 \\ 0 & -1 & 1 & \bigm & -1\end{bmatrix}`

Adding the second row to the third row

`\begin{bmatrix}1 & 0 & -2 & \bigm & -7 \\ 0 & 1 & 0 & \bigm & 3 \\ 0 & -1 & 1 & \bigm & -1\end{bmatrix}\rightarrow\begin{bmatrix}1 & 0 & -2 & \bigm & -7 \\ 0 & 1 & 0 & \bigm & 3 \\ 0 & 0 & 1 & \bigm & 2\end{bmatrix}`

And finally, adding 2 times the third row to the first, we have

`\begin{bmatrix}1 & 0 & -2 & \bigm & -7 \\ 0 & 1 & 0 & \bigm & 3 \\ 0 & 0 & 1 & \bigm & 2\end{bmatrix}\rightarrow\begin{bmatrix}1 & 0 & 0 & \bigm & -3 \\ 0 & 1 & 0 & \bigm & 3 \\ 0 & 0 & 1 & \bigm & 2\end{bmatrix}`

Converting this matrix back to a system, we have

`\begin{bmatrix}1 & 0 & 0 & \bigm & -3 \\ 0 & 1 & 0 & \bigm & 3 \\ 0 & 0 & 1 & \bigm & 2\end{bmatrix}\rightarrow\begin{cases}x=-3 \\ y=3 \\ z=2\end{cases}`

Therefore, our solution is

`(x,y,z)=(-3,3,2)`