Question

If A = `((1, 1, -1),(2, -3, 4),(3, -2, 3))` and B = `((1, -2, 3),(-2, 4, -6),(5, 1, -1))`, then find the rank of AB and the rank of BA.

Answer 1

A = `[(1, 1, -1),(2, -3, 4),(3, -2, 3)]` and B = `[(1, -2, 3),(-2, 4, -6),(5, 1, -1)]`

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

= `[(1 - 2 - 5, -2 + 4 + 1, 3 - 6 + 1),(2 + 6 + 20, -4 - 12 + 4, 6 + 18 - 4),(3 + 4 + 15, -6 - 8 + 3, 9 + 12 - 3)]`

AB = `[(-6, 1, -2),(28, -12, 20),(22, -11, 18)]`

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

= `[(1 - 4 + 9, 1 + 6 - 6, -1 - 8 + 9),(-2 + 8 - 18, -2 - 12 + 12, 2 + 16 - 18),(5 + 2 - 3, 5 - 3 + 2, -5 + 4 - 3)]`

BA = `[(6, 1, 0),(-12, -2, 0),(4, 4, -4)]`

To find the rank of AB

Order of AB is 3 × 3

∴ p(AB) ≤ 3

Consider the third order minor |AB| = `|(-6, 1, -2),(28, -12, 20),(22, -11, 18)|`

= – 6(– 216 + 220) – 3(504 – 440) – 2(– 308 + 264)

= – 6(4) – 3(64) – 2(– 44)

= – 24 – 192 + 88

= 128 ≠

There is a minor of order 3, which is not zero.

∴ p(AB) = 3

To find the rank of BA

Order of BA is 3 × 3

∴ p(BA) ≤ 3

Consider the third order minor |BA| = `|(6, 1, 0),(-12, -2, 0),(4, 4, -4)|`

= 6(8 – 0) – 1(48 – 0) + 0(– 48 + 8)

= 6(8) – 1(48) + 0(– 40)

= 48 – 48 + 0

= 0

Since the third order minor vanishes, therefore P(BA) ≠ 3

Consider a second-order minor = `|(-12, -2),(4, 4)|`

= – 48 + 8

= – 40 ≠ 0

There is a minor of order 2 which is not zero

∴ p(BA) = 2