Question

For two matrices A = `[(3, -6, -1),(2, -5, -1),(-2, 4, 1)]` and B = `[(1, -2, -1),(0, -1, -1),(2, 0, 3)]`, find the product AB and hence solve the system of equations:

3x – 6y – z = 3

2x – 5y − z + 2 = 0

–2x + 4y + z = 5

Answer 1

Finding the product:

AB = `[(3, -6, -1),(2, -5, -1),(-2, 4, 1)][(1, -2, -1),(0, -1, -1),(2, 0, 3)]`

= `[(3(1) + (-6)(0) + (-1)(2), 3(-2) + (-6)(-1) + (-1)(0), 3(-1) + (-6)(-1) + (-1)(3)),(2(1)+(-5)(0) + (-1)(2), 2(-2)+(-5)(-1)+(-1)(0), 2(-1) + (-5)(-1) + (-1)(3)),((-2)(1) + 4(0) + 1(2), (-2)(-2) + 4(-1) + 1(0), (-2)(-1) + 4(-1) + 1(3))]`

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

Thus, AB = I

We know that AA^–1 = I

So B is the inverse of A.

Now, solving the equation,

Given equations are

3x – 6y – z = 3

2x – 5y – z = −2

–2x + 4y + z = 5

Writing the equation as AX = D

`[(3, -6, -1),(2, -5, -1),(-2, 4, 1)] [(x),(y),(z)] = [(3),(-2),(5)]`

Here A = `[(3, -6, -1),(2, -5, -1),(-2, 4, 1)]`, X = `[(x),(y),(z)]` and D = `[(3),(-2),(5)]`

Now AX = D

X = A^–1 D

Putting A^–1 = B = `[(1, -2, -1),(0, -1, -1),(2, 0, 3)]`

So equation become

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

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

`[(x),(y),(z)] = [(3+4-5),(0+2-5),(6+0+15)]`

`[(x),(y),(z)] = [(2),(-3),(21)]` 

Hence x = 2, y = –3 and z = 21.