Question

Show that each of the following systems of linear equations has infinite number of solutions and solve : 

x – y + 3z = 6 

x + 3y – 3z = – 4 

5x + 3y + 3z = 10

Answer 1

Given : - 

Three equation

x – y + 3z = 6 

x + 3y – 3z = – 4 

5x + 3y + 3z = 10

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

  x = \(\frac{D_1}{D}\), y = \(\frac{D_2}{D}\) and z = \(\frac{D_3}{D}\)

(ii) If D = 0 and D₁ = D₂ = D₃ = 0, then the system is consistent and has infinitely many solution. 

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

Now, 

We have, 

x – y + 3z = 6 

x + 3y – 3z = – 4 

5x + 3y + 3z = 10

Lets find D

⇒ D = \(\begin{vmatrix} 1 & -1 & 3\\[0.3em] 1 & 3& -3 \\[0.3em] 5 &3 &3 \end{vmatrix}\)

Expanding along 1^st row 

⇒ D = 1[9 – ( – 3)(3)] – ( – 1)[(3)1 – 5( – 3)] + 3[3 – 5(3)] 

⇒ D = 1[18] + 1[18] + 3[12] 

⇒ D = 0 

Again, 

D₁ by replacing 1^st column by B 

Here,

B = \(\begin{vmatrix} 6 \\[0.3em] -4\\[0.3em]10 \end{vmatrix}\)

⇒ D₁ = \(\begin{vmatrix} 6& -1 & 3 \\[0.3em]-4 &3& -3 \\[0.3em] 10 &3 &3 \end{vmatrix}\)

⇒ D₁ = 6[9 – ( – 3)(3)] – ( – 1)[( – 4)3 – 10( – 3)] + 3[ – 12 – 30]

⇒ D₁ = 6[9 + 9] + [ – 12 + 30] + 3[ – 42] 

⇒ D₁ = 6[18] + 18 – 3[42] 

⇒ D₁ = 0

Also, 

D₂ by replacing 2^nd column by B 

Here,

B = \(\begin{vmatrix} 6 \\[0.3em] -4\\[0.3em]10 \end{vmatrix}\)

⇒ D₂ = \(\begin{vmatrix} 1& 6 & 3 \\[0.3em]1 &-4& -3 \\[0.3em] 5 &10 &3 \end{vmatrix}\)

⇒ D₂ = 1[ – 12 – ( – 3)10] – 6[3 – 5( – 3)] + 3[10 – 5( – 4)] 

⇒ D₂ = [ – 12 + 30] – 6[3 + 15] + 3[10 + 20] 

⇒ D₂ = 18 – 6[18] + 3[30] 

⇒ D₂ = 0 

Again, 

D₃ by replacing 3^rd column by B 

Here,

B = \(\begin{vmatrix} 6 \\[0.3em] -4\\[0.3em]10 \end{vmatrix}\)

⇒ D₃ = \(\begin{vmatrix} 1& -1 & 6 \\[0.3em]1 &3& -4 \\[0.3em] 5 &3 &10 \end{vmatrix}\)

⇒ D₃ = 1[30 – ( – 4)(3)] – ( – 1)[(10 – 5( – 4)] + 6[3 – 15] 

⇒ D₃ = 1[30 + 12] + 1[10 + 20] + 6[ – 12] 

⇒ D₃ = 42 + 30 – 72 

⇒ 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 = 6 – 3z 

x + 3y = – 4 + 3z 

Now, 

by applying Cramer’s rule to solve them, 

New D and D₁, D₂

⇒ D = \(\begin{vmatrix}1&-1 \\[0.3em]1&3 \\[0.3em] \end{vmatrix}\)

⇒ D = 3 + 1 

⇒ D = 4

Again, 

D₁ by replacing 1^st column with

⇒ B = \(\begin{vmatrix}6-3z \\[0.3em]-4+3z \\[0.3em] \end{vmatrix}\)

⇒ D₁ = \(\begin{vmatrix}6-3z&-1 \\[0.3em]-4+3z&3 \\[0.3em] \end{vmatrix}\)

⇒ D₁ = 18 – 9z – ( – 1)( – 4 + 3z) 

⇒ D₁ = 14 – 5z

Again, 

D₂ by replacing 2^nd column with

⇒ B = \(\begin{vmatrix}6-3z \\[0.3em]-4+3z \\[0.3em] \end{vmatrix}\)

⇒ D₂ = \(\begin{vmatrix}1&6-3z\\[0.3em]1&-4+3z \\[0.3em] \end{vmatrix}\)

⇒ D₂ = – 4 + 3z – (6 – 3z)

⇒ D₂ = – 10 + 6z

Hence, using Cramer’s rule

And z = k 

By changing value of k you may get infinite solutions.