Solve the system of equations.
3x + 4y + z = 3
4x + 3y + 32 - 4
5x+67+72-5

First, we write the augmented matrix.

⎡

⎢

⎣

1

−

1

1

2

3

−

1

3

−

2

−

9

|

8

−

2

9

⎤

⎥

⎦

[1−1123−13−2−9 | 8−29]

Next, we perform row operations to obtain row-echelon form.

−

2

R

1

+

R

2

=

R

2

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

3

−

2

−

9

|

8

−

18

9

⎤

⎥

⎦

−

3

R

1

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

0

1

−

12

|

8

−

18

−

15

⎤

⎥

⎦

−2R1+R2=R2→[1−1105−33−2−9|8−189]−3R1+R3=R3→[1−1105−301−12|8−18−15]

The easiest way to obtain a 1 in row 2 of column 1 is to interchange

\displaystyle {R}_{2}

R2

and

\displaystyle {R}_{3}

R3

.

Interchange

R

2

and

R

3

→

⎡

⎢

⎣

1

−

1

1

8

0

1

−

12

−

15

0

5

−

3

−

18

⎤

⎥

⎦

InterchangeR2andR3→[1−11801−12−1505−3−18]

Then

−

5

R

2

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

57

|

8

−

15

57

⎤

⎥

⎦

−

1

57

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

1

|

8

−

15

1

⎤

⎥

⎦

−5R2+R3=R3→[1−1101−120057|8−1557]−157R3=R3→[1−1101−12001|8−151]

The last matrix represents the equivalent system.

x

−

y

+

z

=

8

y

−

12

z

=

−

15

z

=

1

x−y+z=8 y−12z=−15 z=1

Using back-substitution, we obtain the solution as

\displaystyle \left(4,-3,1\right)

(4,−3,1)

l.ln.lln.k.nkl

step-by-step explanation: