Show that satisfies the equation x2–12 x + 1 = 0. Thus, find A – 1

A = We have A2 – 12A + I = 0A2 = = Now, A2 – 12A + 1 = 0= = Hence, = Also, A2 – 12A + 1 = 0= A – 12I + A – 1 = 0= A – 1 = 12I – A= 12= Hence, A – 1 =

Q23 of 74 Page 7

Show that satisfies the equation x²–12 x + 1 = 0. Thus, find A ^{– 1}

A =

We have A² – 12A + I = 0

A^{2 =}

Now, A² – 12A + 1 = 0

Hence, =

Also, A² – 12A + 1 = 0

= A – 12I + A ^{– 1} = 0

= A ^{– 1} = 12I – A

= 12

Hence, A ^{– 1} =

If , find the value of λ so that A² = λA – 2I. Hence, find A ^{– 1}.

Show that satisfies the equation x² – 3A – 7 = 0. Thus, find A ^{– 1}.

For the matrix . Show that A³ – 6A² + 5A + 11I₃ = O.Hence, find A ^{– 1}.

Show that the matrix, satisfies the equation, A³ – A² – 3A – I₃ = O. Hence, find A^–1.