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If A = `[(5, 3),(-1, -2)]`, show that A^{2} – 3A – 7I_{2} = O_{2}. Hence find A^{–1}

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#### Solution

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

A^{2} = A × A

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

= `[(25 - 3, 15 - 6),(-5 + 2, -3 + 4)]`

= `[(22, 9),(-3, 1)]`

– 3A = `-3[(5, 3),(-1, -2)]`

= `[(-15, -9),(3, 6)]`

– 7I_{2} = `-7[(1, 0),(0, 1)]`

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

A^{2} – 3A – 7I_{2}

= `[(22, 9),(-3, 1)] + [(15, -9),(3, 6)] + [(-7, 0),(0, -7)]`

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

∴ A^{2} – 3A – 7I_{2} = O_{2}

Post multiply this equation by A^{–1}

A^{2}A^{–}^{1 }– 3A A^{–}^{1 }– 7I_{2 }A^{–}^{1} = 0

A – 3I – 7A^{–}^{1} = 0

A – 3I = 7 A^{–}^{1 }

A^{–}^{1 }= `1/7 ("A" - 3"I")`

= `1/7 [((5, 3),(-1, -2)), -3((1, 0),(0, 1))]`

= `1/7 [((5, 3),(-1, -2)) + ((-3, 0),(0, -3))]`

A^{–}^{1 }= `1/7[(2, 3),(-1, -5)]`

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