More Examples, Linear Systems
Matrix & Basic Operations

Example 1.

We are given the system

(1)   \begin{align*} -2x +2y =2 \\ -x +3y =4\end{align*}

The matrix associated to this system is

(2)   \begin{equation*}\left (\begin{array}{ccc} -2 & 2 & 2 \\ -1 & 3 & 4\end{array}\right ) \end{equation*}

We want to find the solution set of the system of linear equations represented by this augmented matrix. The first row does not have a leading 1 so multiply the first row with \frac{ -1}{2} and get the row equivalent matrix

(3)   \begin{equation*}\left (\begin{array}{ccc}1 & -1 & -1 \\ -1 & 3 & 4\end{array}\right ) \end{equation*}

Next, add the first row and the second row and replace the second row with it to get zero below the leading 1.

(4)   \begin{equation*}\left (\begin{array}{ccc}1 & -1 & -1 \\ 0 & 2 & 3\end{array}\right ) \end{equation*}

Next, multiply the second row by \frac{1}{2} to get a leading 1

(5)   \begin{equation*}\left (\begin{array}{ccc}1 & -1 & -1 \\ 0 & 1 & \frac{3}{2}\end{array}\right ) \end{equation*}

Now this matrix is in row-echelon form. We can see that y =\frac{3}{2} and a back substitution can be used to find x from the first row

(6)   \begin{align*}x -\frac{3}{2} = -1 \\ x =\frac{1}{2}\end{align*}

We had the option to continue and get a row reduced echelon form that is add first and second row and replace the first row with it

(7)   \begin{equation*}\left (\begin{array}{ccc}1 & 0 & \frac{1}{2} \\ 0 & 1 & \frac{3}{2}\end{array}\right ) \end{equation*}

This provides the solution \left (\frac{1}{2} ,\frac{3}{2}\right ).

Example 2.

Suppose we are given the matrix

(8)   \begin{equation*}\left (\begin{array}{cccc} -1 & 0 & 1 & 3 \\ 2 & 2 & 2 & 1 \\ -3 & 0 & -1 & 0\end{array}\right ) \end{equation*}

We want to find the solution set of the system of linear equations represented by this augmented matrix. One option is as followed:

 

1. Multiply first row with -1 to get a leading 1

(9)   \begin{equation*}\left (\begin{array}{cccc}1 & 0 & -1 & -3 \\ 2 & 2 & 2 & 1 \\ -3 & 0 & -1 & 0\end{array}\right ) \end{equation*}

 

2. Multiply the second row with \frac{1}{2} and subtract from row one and replace row 2 to get a leading zero

(10)   \begin{equation*}\left (\begin{array}{cccc}1 & 0 & -1 & -3 \\ 0 & -1 & -2 & -\frac{7}{2} \\ -3 & 0 & -1 & 0\end{array}\right ) \end{equation*}

 

3. Multiply the second row with -1

(11)   \begin{equation*}\left (\begin{array}{cccc}1 & 0 & -1 & -3 \\ 0 & 1 & 2 & \frac{7}{2} \\ -3 & 0 & -1 & 0\end{array}\right ) \end{equation*}

 

4. Multiply the third row with -\frac{1}{3} and subtract from the first row to get a leading zero

(12)   \begin{equation*}\left (\begin{array}{cccc}1 & 0 & -1 & -3 \\ 0 & 1 & 2 & \frac{7}{2} \\ 0 & 0 & -\frac{4}{3} & -3\end{array}\right ) \end{equation*}

 

5. Multiply the third row with -\frac{3}{4} to get 1 in the third entry

(13)   \begin{equation*}\left (\begin{array}{cccc}1 & 0 & -1 & -3 \\ 0 & 1 & 2 & \frac{7}{2} \\ 0 & 0 & 1 & -\frac{9}{4}\end{array}\right ) \end{equation*}

Now this matrix is in row echelon form. We see that z = -\frac{9}{4} and could use a back substitution to solve for y and z . However, we continue the reduction process to get a row reduced echelon form

 

6. Multiply the third row by 2 and subtract the third row from the second row and replace the second row

(14)   \begin{equation*}\left (\begin{array}{cccc}1 & 0 & -1 & -3 \\ 0 & 1 & 0 & 8 \\ 0 & 0 & 1 & -\frac{9}{4}\end{array}\right ) \end{equation*}

 

7. Add first and third row and replace the first row with it

(15)   \begin{equation*}\left (\begin{array}{cccc}1 & 0 & 0 & -\frac{21}{4} \\ 0 & 1 & 0 & 8 \\ 0 & 0 & 1 & -\frac{9}{4}\end{array}\right ) \end{equation*}

Therefore, the solution to this system is \left ( -\frac{21}{4} ,8 , -\frac{9}{4}\right ).

Example 3.

Suppose we are given the matrix

(16)   \begin{equation*}\left (\begin{array}{cccc}0 & -1 & 3 & 1 \\ 2 & -1 & 0 & 4 \\ -\frac{1}{2} & 0 & 1 & -1\end{array}\right ) \end{equation*}

We want to find the solution set of the system of linear equations represented by this augmented matrix. One option is as followed:

1. Interchange rows 1 and 2

(17)   \begin{equation*}\left (\begin{array}{cccc}2 & -1 & 0 & 4 \\ 0 & -1 & 3 & 1 \\ -\frac{1}{2} & 0 & 1 & -1\end{array}\right ) \end{equation*}

 

2. Multiply the first row by \frac{1}{2}

(18)   \begin{equation*}\left (\begin{array}{cccc}1 & -\frac{1}{2} & 0 & 2 \\ 0 & -1 & 3 & 1 \\ -\frac{1}{2} & 0 & 1 & -1\end{array}\right ) \end{equation*}

 

3. Multiply the second row by -1

(19)   \begin{equation*}\left (\begin{array}{cccc}1 & -\frac{1}{2} & 0 & 2 \\ 0 & 1 & -3 & -1 \\ -\frac{1}{2} & 0 & 1 & -1\end{array}\right ) \end{equation*}

 

4. Multiply the third row by 2, add to row one and then replace row three

(20)   \begin{equation*}\left (\begin{array}{cccc}1 & -\frac{1}{2} & 0 & 2 \\ 0 & 1 & -3 & -1 \\ 0 & -\frac{1}{2} & 2 & 0\end{array}\right ) \end{equation*}

 

5. Multiply row 2 by 2, add to the second row and replace row three

(21)   \begin{equation*}\left (\begin{array}{cccc}1 & -\frac{1}{2} & 0 & 2 \\ 0 & 1 & -3 & -1 \\ 0 & 0 & 1 & -1\end{array}\right ) \end{equation*}

As we see the matrix is in row echelon form we can use a back substitution to solve the system

(22)   \begin{align*}z = -1 \\ y = -1 +3z = -1 -3 = -4 \\ x =2 +\frac{1}{2}y =2 +\frac{1}{2}\left ( -4\right ) =0\end{align*}

Therefore, the solution is \left (0 , -4 , -1\right ).

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