Question

prove without actually solving that the following system of equations has a unique solution.
5x+3y+14z=4
y+2z=1
x-y+2z=0​

Answer 1

Step-by-step explanation:

Given set of equations,

5x+3y+7z=4

3x+26y+2z=9

7x+2y+10z=5

Arranging the above equations in form of matrix and finding the coefficient of matrix, we get,

A=

⎣

⎢

⎢

⎡

5

3

7

3

26

2

7

2

10

⎦

⎥

⎥

⎤

∣A∣=

∣

∣

∣

∣

∣

∣

∣

∣

5

3

7

3

26

2

7

2

10

∣

∣

∣

∣

∣

∣

∣

∣

=5(260−4)−3(30−14)+7(6−182)

∣A∣=0

det(A)=0. Therefore the system is consistent.

R

2

→5R

2

−3R

1

R

3

→5R

3

−7R

1

A∼

⎣

⎢

⎢

⎡

5

0

0

3

121

−11

7

−11

1

⎦

⎥

⎥

⎤

as, ρ(A)=2, submatrix is [

5

0

3

121

]

we get,

5x+3y+7z=4

121y−11z=33

Let z=k

121y−11z=33

y=

121

33+11z

=

11

3+k

5x=4−3y−7z

=4−3(

11

3+k

)−7k

55x=44−3(k+3)−77k

x=

55

35−80k

=

11

7−16k