Question

Q.4)
For what value of (b1,b2,b3) the system of linear equations is inconsistent
x- y+2z =b1, x+2y- z =b2, 2y – 2z = b3
(a) (2,2,0) (b) (0, 3, 2)
(c) (2,2,1)
(d) (2,-1,-2).​

Answer 1

Answer:

x + 2y + 3z = b1, 4y + 5z = b2 and x + 2y + z = b

or

C -x + 2y 5z b1, 2x 4y + 10z =b2 and x - 2y + 5z = b

or

x + 2y + 5z = b1, 2x 3z = b2 and x + 4y - 5z = b

Step-by-step explanation:

We find D=0, where D is the determinant formed by the coefficients of x, y, z in the three equations and since no pair of planes are parallel, so there is an infinite number of solutions.

Let αP

1

+λP

2

=P

3

⇒P

1

+7P

2

=13P

3

⇒b

1

+7b

2

=13b

3

(A) D



=0⇒ unique solution for any b

1

,b

2

,b

3

(B) D=0 but P

1

+7P

2



=13P

3

(C) D=0 Also b

2

=−2b

1

,b

3

=−b

1

Satisfied b

1

+7b

2

=13b

3

(Actually all three planes are co-incident)

(D) D



=0.