Section 43
(26) Solve the system:
3x_{1}

+

5x_{2}



x_{3}

=

7

x_{1}

+

x_{2}

+

x_{3}

=

1

2x_{1}

+

11x_{3}

=

7

First, we transform the system to an augmented matrix, and then reduce the matrix. In our first step, we switch two rows to get a 1 at the top left corner.






[ 

] 
This reduced matrix corresponds to the following system, which gives the solution.
x_{1}


=

2


x_{2}


=

0


x_{3}

=

1

(30) Solve the system
2x_{1}

+

4x_{2}



6x_{3}

=

10

3x_{1}

+

3x_{2}



3x_{3}

=

6


1/3R_{2} >R_{2}


R_{1}<>R_{2}


2R_{1}+R_{2}>R_{2}


1/2R_{2}>R_{2}


_{R2+ R1> R1}


(reduced)

Since the matrix is reduced we can now write down the solution to the original system. The reduced matrix corresponds to the system

or 

so the solutions are 

(32) Solve the system
2x_{1}

x_{2}

=

0

3x_{1}

+2x_{2}

=

7

x_{1}

x_{2}

=

2


R_{1} <>R_{3}


3R_{1}+R_{2}>R_{2} 2R_{1}+R_{3}>R_{3} 

5R_{3}+R_{2}>R_{2}



Even though the last matrix is not yet reduced, we can stop, because the second row corresponds to the equation 0x_{1}+0x_{2}=7, or 0=7. Therefore it is impossible for the system to have any solutions. NO SOLUTIONS