Given: - Three equation

x + y – z = 0

x – 2y + z = 0

3x + 6y – 5z = 0

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,

x + y – z = 0

x – 2y + z = 0

3x + 6y – 5z = 0

Lets find D

⇒

Expanding along 1^st row

⇒ D = 1[10 – (6)1] – (1)[( – 5)1 – (1)3] + ( – 1)[6 – ( – 2)3]

⇒ D = 1[4] – 1[ – 8] – [12]

⇒ D = 0

Again, D₁ by replacing 1^st column by B

Here

⇒

As one column is zero its determinant is zero

⇒ D₁ = 0

Also, D₂ by replacing 2^nd column by B

Here

⇒

As one column is zero its determinant is zero

⇒ D₂ = 0

Again, D₃ by replacing 3^rd column by B

Here

⇒

As one column is zero its determinant is zero

⇒ 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 + y = z

x – 2y = – z

Now by applying Cramer’s rule to solve them,

New D and D₁, D₂

⇒

⇒ D = – 2 – 1

⇒ D = – 3

Again, D₁ by replacing 1^st column with

⇒

⇒ D₁ = – 2z – 1( – z)

⇒ D₁ = – z

Again, D₂ by replacing 2^nd column with

⇒

⇒ D₂ = – z – z

⇒ D₂ = – 2z

Hence, using Cramer’s rule

⇒

⇒

Let, z = k

Then

again,

⇒

⇒

⇒

And z = k

By changing value of k you may get infinite solutions