# Show that the matrix A = satisfies the equation 𝒳2 + 4𝒳 – 42 = 0 and hence find A-1.

Given: To show: Matrix A satisfies the equation x2 + 4x – 42 = 0

If Matrix A satisfies the given equation then

A2 + 4A – 42 = 0

Firstly, we find the A2   Taking LHS of the given equation .i.e.

A2 + 4A – 42     = O

= RHS

LHS = RHS

Hence matrix A satisfies the given equation x2 + 4x – 42 = 0

Now, we have to find A-1

Finding A-1 using given equation

A2 + 4A – 42 = O

Post multiplying by A-1 both sides, we get

(A2 + 4A – 42)A-1 = OA-1

A2.A-1 + 4A.A-1 – 42.A-1 = O [OA-1 = O]

A.(AA-1) + 4I – 42A-1 = O [AA-1 = I]

A(I) + 4I – 42A-1 = O

A + 4I – 42A-1 = O

A + 4I – O = 42A-1     Ans. .

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