Question

Use the product of matrices `[(1,2,-3),(3,2,-2),(2,-1,1)][(0,1,2),(-7,7,-7),(-7,5,-4)]` to solve the following system of equations:

x + 2y – 3z = 6

3x + 2y – 2z = 3

2x  –  y + z = 2

Answer 1

Let `A = [(1,2,-3),(3,2,-2),(2,-1,1)]`

and `B = [(0,1,2),(-7,7,-7),(-7,5,-4)]`

`AB = [(1,2,-3),(3,2,-2),(2,-1,1)][(0,1,2),(-7,7,-7),(-7,5,-4)]`

`= [(0-14+21,1+14-15,2-14+12),(0-14+14,3+14-10,6-14+8),(0+7-7,2-7+5,4+7-4)]`

`= [(7,0,0),(0,7,0),(0,0,7)]`

`= 7[(1,0,0),(0,1,0),(0,0,1)]`

AB = 7I

`1/7 (AB) = I`

`1/7B = A^-1`

`A^-1 = 1/7[(0,1,2),(-7,7,-7),(-7,5,-4)]`

Now, The given system of equation is

x + 2y − 3z = 6

3x + 2y − 2z = 3

2x − y + z = 2

`[(1,2,-3),(3,2,-2),(2,-1,1)][(x),(y),(z)]=[(6),(3),(2)]`

AX = C

X = A^-1 C

`x= 1/7 [(0,1,2),(-7,7,-7),(-7,5,-4)][(6),(3),(2)]`   ...(from eqn 1)

`x= 1/7 [(0+3+4),(-42+21-14),(-42+15-8)]`

`x = 1/7[(7),(-35),(-35)]`

`[(x),(y),(z)]=[(1),(-5),(-5)]`

⇒ x = 1, y = −5, z = −5,