Given: - Three equation

2x + y – 2z = 4

x – 2y + z = – 2

5x – 5y + z = – 2

Tip: - We know that

For a system of 3 simultaneous linear equation with 3 unknowns

(i) If D ≠ 0, then the given system of equations is consistent and has a unique solution given by

(ii) If D = 0 and D₁ = D₂ = D₃ = 0, then the given system of equation may or may not be consistent. However if consistent, then it has infinitely many solution.

(iii) If D = 0 and at least one of the determinants D₁, D₂ and D₃ is non – zero, then the system is inconsistent.

Now,

We have,

2x + y – 2z = 4

x – 2y + z = – 2

5x – 5y + z = – 2

Lets find D

⇒

Expanding along 1^st row

⇒ D = 2[ – 2 – ( – 5)(1)] – (1)[(1)1 – 5(1)] + ( – 2)[ – 5 – 5( – 2)]

⇒ D = 2[3] – 1[ – 4] – 2[5]

⇒ D = 0

Again, D₁ by replacing 1^st column by B

Here

⇒

⇒ D₁ = 4[ – 2 – ( – 5)(1)] – (1)[( – 2)1 – ( – 2)(1)] + ( – 2)[( – 2)( – 5) – ( – 2)( – 2)]

⇒ D_{1 =} 4[ – 2 + 5] – [ – 2 + 2] – 2[6]

⇒ D_{1 =} 12 + 0 – 12

⇒ D₁ = 0

Also, D₂ by replacing 2^nd column by B

Here

⇒

⇒ D₂ = 2[ – 2 – ( – 2)(1)] – (4)[(1)1 – (5)] + ( – 2)[ – 2 – 5( – 2)]

⇒ D_{2 =} 2[ – 2 + 2] – 4[ – 4] + ( – 2)[8]

⇒ D_{2 =} 0 + 16 – 16

⇒ D₂ = 0

Again, D₃ by replacing 3^rd column by B

Here

⇒

⇒ D₃ = 2[4 – ( – 2)( – 5)] – (1)[( – 2)1 – 5( – 2)] + 4[1( – 5) – 5( – 2)]

⇒ D_{3 =} 2[ – 6] – [8] + 4[ – 5 + 10]

⇒ D_{3 =} – 12 – 8 + 20

⇒ D₃ = 0

So, here we can see that

D = D₁ = D₂ = D₃ = 0

Thus,

Either the system is consistent with infinitely many solutions or it is inconsistent.

Now, by 1^st two equations, written as

x – 2y = – 2 – z

5x – 5y = – 2 – z

Now by applying Cramer’s rule to solve them,

New D and D₁, D₂

⇒

⇒ D = – 5 + 10

⇒ D = 5

Again, D₁ by replacing 1^st column with

⇒

⇒ D₁ = 10 + 5z – ( – 2)( – 2 – z)

⇒ D₁ = 6 + 3z

Again, D₂ by replacing 2^nd column with

⇒

⇒ D₂ = – 2 – z – 5 ( – 2 – z)

⇒ D₂ = 8 + 4z

Hence, using Cramer’s rule

⇒

⇒

again,

⇒

⇒

Let, z = k

Then

And z = k

By changing value of k you may get infinite solutions