Show that each of the following systems of linear equations has infinite number of solutions and solve:
x – y + z = 3
2x + y – z = 2
– x – 2y + 2z = 1
Given: - Three equation
x – y + z = 3
2x + y – z = 2
– x – 2y + 2z = 1
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
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(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 = 3
2x + y – z = 2
– x – 2y + 2z = 1
Lets find D
⇒ 
Expanding along 1st row
⇒ D = 1[2 – ( – 1)( – 2)] – ( – 1)[(2)2 – (1)] + 1[ – 4 – ( – 1)]
⇒ D = 1[0] + 1[3] + [ – 3]
⇒ D = 0
Again, D1 by replacing 1st column by B
Here

⇒ 
⇒ D1 = 3[2 – ( – 1)( – 2)] – ( – 1)[(2)2 – ( – 1)] + 1[ – 4 – 1]
⇒ D1 = 3[2 – 2] + [4 + 1] + 1[ – 5]
⇒ D1 = 0 + 5 – 5
⇒ D1 = 0
Also, D2 by replacing 2nd column by B
Here

⇒ 
⇒ D2 = 1[4 – ( – 1)(1)] – (3)[(2)2 – (1)] + 1[2 – ( – 2)]
⇒ D2 = 1[4 + 1] – 3[4 – 1] + 1[4]
⇒ D2 = 5 – 9 + 4
⇒ D2 = 0
Again, D3 by replacing 3rd column by B
Here

⇒ 
⇒ D3 = 1[1 – ( – 2)(2)] – ( – 1)[(2)1 – 2( – 1)] + 3[2( – 2) – 1( – 1)]
⇒ D3 = [1 + 4] + [2 + 2] + 3[ – 4 + 1]
⇒ D3 = 5 + 4 – 9
⇒ 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 1st two equations, written as
x – y = 3 – z
2x + y = 2 + z
Now by applying Cramer’s rule to solve them,
New D and D1, D2
⇒ ![]()
⇒ D = 1 + 2
⇒ D = 3
Again, D1 by replacing 1st column with
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⇒ ![]()
⇒ D1 = 3 – z – ( – 1)(2 + z)
⇒ D1 = 5
Again, D2 by replacing 2nd column with
![]()
⇒ ![]()
⇒ D2 = 2 + z – 2 (3 – z)
⇒ D2 = – 4 + 3z
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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