Given: - Equations are: –

2y – 3z = 0

X + 3y = – 4

3x + 4y = 3

Tip: - Theorem – Cramer’s Rule

Let there be a system of n simultaneous linear equations and with n unknown given by

and let D_j be the determinant obtained from D after replacing the j^th column by

Then,

provided that D ≠ 0

Now, here we have

2y – 3z = 0

x + 3y = – 4

3x + 4y = 3

So by comparing with theorem, lets find D , D₁ and D₂

Solving determinant, expanding along 1^st Row

⇒ D = 0[0] – 2[(0)(1) – 0] – 3[1(4) – 3(3)]

⇒ D = 0 – 0 – 3[4 – 9]

⇒ D = 0 – 0 + 15

⇒ D = 15

Again, Solve D₁ formed by replacing 1^st column by B matrices

Here

Solving determinant, expanding along 1^st Row

⇒ D₁ = 0[0] – 2[(0)( – 4) – 0] – 3[4( – 4) – 3(3)]

⇒ D₁ = 0 – 0 – 3[ – 16 – 9]

⇒ D₁ = 0 – 0 – 3( – 25)

⇒ D₁ = 0 – 0 + 75

⇒ D₁ = 75

Again, Solve D₂ formed by replacing 2^nd column by B matrices

Here

Solving determinant

⇒ D₂ = 0[0] – 0[(0)(1) – 0] – 3[1(3) – 3( – 4)]

⇒ D₂ = 0 – 0 + ( – 3)(3 + 12)

⇒ D₂ = – 45

And, Solve D₃ formed by replacing 3^rd column by B matrices

Here

⇒

Solving determinant, expanding along 1^st Row

⇒ D₃ = 0[9 – ( – 4)4] – 2[(3)(1) – ( – 4)(3)] + 0[1(4) – 3(3)]

⇒ D₃ = 0[25] – 2(3 + 12) + 0(4 – 9)

⇒ D₃ = 0 – 30 + 0

⇒ D₃ = – 30

Thus by Cramer’s Rule, we have

⇒

⇒

⇒ x = 5

again,

⇒

⇒

⇒ y = – 3

and,

⇒

⇒

⇒ z = – 2