Given: - Equations are: –

x + y = 5

y + z = 3

x + z = 4

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

x + y = 5

y + z = 3

x + z = 4

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

Solving determinant, expanding along 1^st Row

⇒ D = 1[1] – 1[ – 1] + 0[ – 1]

⇒ D = 1 + 1 + 0

⇒ D = 2

⇒ D = 2

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

Here

Solving determinant, expanding along 1^st Row

⇒ D₁ = 5[1] – 1[(3)(1) – (4)(1)] + 0[0 – (4)(1)]

⇒ D₁ = 5 – 1[3 – 4] + 0[– 4]

⇒ D₁ = 5 – 1[ – 1] + 0

⇒ D₁ = 5 + 1 + 0

⇒ D₁ = 6

Again, Solve D₂ formed by replacing 1^st column by B matrices

Here

Solving determinant

⇒ D₂ = 1[3 – 4] – 5[ – 1] + 0[0 – 3]

⇒ D₂ = 1[ – 1] + 5 + 0

⇒ D₂ = 4

And, Solve D₃ formed by replacing 1^st column by B matrices

Here

⇒

Solving determinant, expanding along 1^st Row

⇒ D₃ = 1[4 – 0] – 1[0 – 3] + 5[0 – 1]

⇒ D₃ = 1[4] – 1( – 3) + 5( – 1)

⇒ D₃ = 4 + 3 – 5

⇒ D₃ = 2

Thus by Cramer’s Rule, we have

⇒

⇒

⇒ x = 3

again,

⇒

⇒

⇒ y = 2

and,

⇒

⇒

⇒ z = 1