Question

If 4x+3y+6z=25,x+5y+7z=13,2x+9y+z=1 then find the value of z ?

Answer 1

Answer:

Let the matrix A be

⎣

⎢

⎢

⎡

4

1

2

3

5

9

6

7

1

⎦

⎥

⎥

⎤

and matrix B be

⎣

⎢

⎢

⎡

25

13

1

⎦

⎥

⎥

⎤

∣A∣



=0 , therefore the rank of A is 3

Now consider A∣B=

⎣

⎢

⎢

⎡

4

1

2

3

5

9

6

7

1

25

13

1

⎦

⎥

⎥

⎤

, the rank is 3

Therefore ρ(A)=ρ(A∣B)=3 , which implies that the system of equations is consistent.

ρ(A)=ρ(A∣B)= number of rows , therefore the sytem has unique solution.

We will solve it by determinant method.

The value of D=

∣

∣

∣

∣

∣

∣

∣

∣

4

1

2

3

5

9

6

7

1

∣

∣

∣

∣

∣

∣

∣

∣

=−199

D

x

=

∣

∣

∣

∣

∣

∣

∣

∣

3

5

9

6

7

1

25

13

1

∣

∣

∣

∣

∣

∣

∣

∣

=−796

D

y

=

∣

∣

∣

∣

∣

∣

∣

∣

4

25

6

1

13

7

2

1

1

∣

∣

∣

∣

∣

∣

∣

∣

=199 ,

D

z

=

∣

∣

∣

∣

∣

∣

∣

∣

4

3

25

1

5

13

2

9

1

∣

∣

∣

∣

∣

∣

∣

∣

=−398

Therefore x=

−199

−796

=4 , y=

−199

199

=−1 and z=

−199

−398

=2

Hence 4,−1,2 is the solution of the given simultaneous equations.

Step-by-step explanation:

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