Answer :

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 D_{1} = D_{2} = D_{3} = 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 D_{1}, D_{2} and D_{3} 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, D_{1} by replacing 1^{st} column by B

Here

⇒

⇒ D_{1} = 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

⇒ D_{1} = 0

Also, D_{2} by replacing 2^{nd} column by B

Here

⇒

⇒ D_{2} = 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

⇒ D_{2} = 0

Again, D_{3} by replacing 3^{rd} column by B

Here

⇒

⇒ D_{3} = 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

⇒ D_{3} = 0

So, here we can see that

D = D_{1} = D_{2} = D_{3} = 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 D_{1}, D_{2}

⇒

⇒ D = – 5 + 10

⇒ D = 5

Again, D_{1} by replacing 1^{st} column with

⇒

⇒ D_{1} = 10 + 5z – ( – 2)( – 2 – z)

⇒ D_{1} = 6 + 3z

Again, D_{2} by replacing 2^{nd} column with

⇒

⇒ D_{2} = – 2 – z – 5 ( – 2 – z)

⇒ D_{2} = 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

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