Question

If A = `[(3,1),(-1,2)]`  show that  `A^2 - 5A + 7I = 0`.

Answer 1

We have to proved that A² -  5A + 7I = 0

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

`= [(9 - 1,3 + 2),(-3 -2,-1 + 4)] = [(8,5), (-5,3)]`

`5A = 5 = [(3,1), (-1,2)] = [(15,5), (-5, 10)]`

Now, substituting the values in A² - 5A + 71, we have,

`A^2 - 5A + 7I = [(8,5), (-5, 3)] - [(15,5),(-5,10)] + [(7,0),(0,7)]`

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

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