Given: - Equations are: –

x + y + z + w = 2

x – 2y + 2z + 2w = – 6

2x + y – 2z + 2w = – 5

3x – y + 3z – 3w = – 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

x + y + z + w = 2

x – 2y + 2z + 2w = – 6

2x + y – 2z + 2w = – 5

3x – y + 3z – 3w = – 3

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

applying,

⇒

Solving determinant, expanding along 1^st Row

⇒

applying,

⇒

⇒ D = 1[ – 6 – 88]

⇒ D = – 94

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

Here

applying,

⇒

Solving determinant, expanding along 1^st Row

⇒

⇒ D₁ = – 1{( – 10)[6( – 1) – 2( – 4)] – ( – 4)[( – 9)6 – ( – 4)3] + 0}

⇒ D₁ = – 1{ – 10[ – 6 + 8] + 4[ – 54 + 12]}

⇒ D₁ = – 1{ – 10[2] + 4[ – 42] }

⇒ D₁ = 188

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

Here

applying,

⇒

Solving determinant, expanding along 1^st Row

⇒

⇒ D₂ = – 1{( – 1)[6( – 9) – 3( – 4)] – ( – 10)[0 – 6( – 4)] + 0[0 + 54]}

⇒ D₂ = – 1{ – 1[ – 54 + 12] + 10(24) + 0}

⇒ D₂ = – 282

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

Here

⇒

applying,

⇒

Solving determinant, expanding along 1^st Row

⇒

⇒ D₃ = – 1{( – 1)[ – 3 – ( – 9)2] – ( – 4)[0 – 6( – 9)] + ( – 10)[0 + 6]}

⇒ D₃ = – 1{ – 1[15] + 4(54) – 10(6)}

⇒ D₃ = – 1{ – 15 + 216 – 60}

⇒ D₃ = – 141

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

Here

⇒

applying,

⇒

Solving determinant, expanding along 1^st Row

⇒

⇒ D₄ = ( – 3)[( – 9)( – 4) – 0] – 1[9 – ( – 4)( – 9)] + ( – 8)[0 – 16]

⇒ D₄ = – 3[36] – 1(9 – 36) – 8( – 16)

⇒ D₄ = – 108 + 27 + 128

⇒ D₄ = 47

Thus by Cramer’s Rule, we have

⇒

⇒

⇒ x = – 2

again,

⇒

⇒

⇒ y = 3

again,

⇒

⇒

⇒

And,

⇒

⇒

⇒