Using matrices, solve the following system of equations:

2x + 3y + 3z = 5, x – 2y + z = – 4, 3x – y – 2z = 3

Given equations are –

2x + 3y + 3z = 5 …(1)

x – 2y + z = – 4 …(2)

3x – y – 2z = 3 …(3)

Writing equation as AX = B, we have -

=

Therefore,

A = X = B =

We know that solution of AX = B is given by –

X = A^-1B and the solution is unique if |A| ≠ 0

So first we need to check the value of |A|

∴

⇒ |A| = (2) – (3) + (3)

⇒ |A| = 2 (4 – (-1) – 3 (-2 – 3) + 3 (-1 – (-6))

⇒ |A| = 2 (5) – 3 (-5) + 3 (5)

⇒ |A| = 10 + 15 + 15

⇒ |A| = 40 0

A is non – singular, its inverse exists and the system of equations is consistent and has a unique solution.

Hence, X=A^-1B

We know that: A^-1 = adj (A)

Adj (A) =

Where C_ij is called the co-factor of the elemnt of i^th row and j^th column of matrix A

A =

= (-1)² (4+1) = 5, = 5 = 5 {Taken from previous calculations}

= 3

= -13

= 11

= 9

= 1

= -7

Co-factor Matrix C =

Adj(A) = C^T =

A⁻¹ =

A^-1 =

∵ X = A⁻¹B

∴ X =

⇒ X=

Or =

x = 1, y = 2 & z = -1.