Given: - Equations are: –

2x – 3z + w = 1

x – y + 2w = 1

– 3y + z + w = 1

x + y + z = 1

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

2x – 3z + w = 1

x – y + 2w = 1

– 3y + z + w = 1

x + y + z = 1

So by comparing with theorem, lets find D, D₁, D₂,D₃ and D₄

applying,

⇒

Solving determinant, expanding along 4^th Row

⇒

applying,

⇒

expanding along 3^rd row

⇒ D = – 1[ – 3 – ( – 6)4]

⇒ D = – 21

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

Here

applying,

⇒

Solving determinant, expanding along 1^st Row

⇒

⇒ D₁ = ( – 1)[(4)( – 1) – 0(4)] – (3)[( – 3)( – 1) – 0] + 1[ – 12 – 4]

⇒ D₁ = – 1[ – 4 – 0] – 3[3 – 0] – 16

⇒ D₁ = 4 – 9 – 16

⇒ D₁ = – 21

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

Here

applying,

⇒

Solving determinant, expanding along 4^th Row

⇒

⇒ D₂ = – 1{( – 1)[1( – 1) – 1(2)] – ( – 5)[0 – 1(2)] + 1[0 – ( – 1)]}

⇒ D₂ = – 1{ – 1[ – 1 – 2] + 5( – 2) + 1}

⇒ D₂ = 6

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

Here

⇒

applying,

⇒

Solving determinant, expanding along 4^th Row

⇒

⇒ D₃ = – 1{( – 2)[0 – (1)2] – ( – 1)[ – 2 – ( – 3)(2)] + 1[ – 2 – 0]}

⇒ D₃ = – 1{ – 2[ – 2] + 1( – 2 + 6) + 1( – 2)}

⇒ D₃ = – 1{4 + 4 – 2}

⇒ D₃ = – 6

And, Solve D₄ formed by replacing 4^th column by B matrices

Here

⇒

applying,

⇒

Solving determinant, expanding along 4^th Row

⇒

⇒ D₄ = ( – 1){( – 2)[( – 1)1 – 0] – ( – 5)[ – 2 – 0] + ( – 1)[ – 2 – 3]}

⇒ D₄ = ( – 1){2 – 10 + 5}

⇒ D₄ = 3

⇒ D₄ = 3

Thus by Cramer’s Rule, we have

⇒

⇒

⇒ x = 1

again,

⇒

⇒

⇒

again,

⇒

⇒

⇒

And,

⇒

⇒

⇒