Question

Find A^-1, if `A = [(1,2,1),(2,3,-1),(1,0,1)]`. Hence, solve the following system of equations:

x + 2y + z = 5

2x + 3y = 1

x – y + z = 8

Answer 1

Given `A = [(1,2,1),(2,3,-1),(1,0,1)]`

Here, |A| = 1(3 − 0) −2(2 + 1) + 1 (0 − 3)

= 3 − 6 − 3

= −6 ≠ 0

So, A^-1 exists.

Now, `M_11 = |(3,-1),(0,1)| = 3`

`M_12 = |(2,-1),(1,1)| = 3`

`M_13 = |(2,3),(1,0)| = −3`

`M_21 = |(2,1),(0,1)| = 2`

`M_22 = |(1,1),(1,1)| = 0`

`M_23 = |(1,2),(1,0)| = −2`

`M_31 = |(2,1),(3,-1)| = −5`

`M_32 = |(1,1),(2,-1)| = −3`

`M_33 = |(1,2),(2,3)| = −1`

Thus the minor matrix of `A = [(3,3,-3),(2,0,-2),(-5,-3,-1)]`

co-factor matrix of `[(3,-3,-3),(-2,0,2),(-5,3,-1)]`

Also, adj `A = [(3,-2,-5),(-3,0,3),(-3,2,-1)]`

Therefore, `A^-1 = (adj A)/|A|`

`= 1/-6 [(3,-2,-5),(-3,0,3),(-3,2,-1)]`

`= [(-1/2,1/3,5/6),(1/2,0,-1/2),(1/2,-1/3,1/6)]`

Now, Given x + 2y + z = 5

2x + 3y = 1

x − y + z = B

writing equation as AX = B

`[(1,2,1),(2,3,0),(1,-1,1)] [(x),(y),(z)] = [(5),(1),(8)]`

i.e., A'X = B

⇒ X = (A')^-1B

= (A^-1)'B

⇒ `[(x),(y),(z)][(-1/2,1/3,5/6),(1/2,0,-1/2),(1/2,-1/3,1/6)] [(5),(1),(8)]`

`= [(-1/2,1/2,1/2),(1/3,0,-1/3),(5/6,-1/2,1/6)] [(5),(1),(8)]`

`= [(-5/2+1/2+8/2),(5/3+0-8/3),(25/6-1/2+8/6)] [(2),(-1),(5)]`

x = 2, y = −1, z = 5