Q. 324.3( 3 Votes )

# Show that the matrix A = satisfies the equation π³^{2} + 4π³ β 42 = 0 and hence find A^{-1}.

Answer :

Given:

To show: Matrix A satisfies the equation x^{2} + 4x β 42 = 0

If Matrix A satisfies the given equation then

A^{2} + 4A β 42 = 0

Firstly, we find the A^{2}

Taking LHS of the given equation .i.e.

A^{2} + 4A β 42

= O

= RHS

β΄ LHS = RHS

Hence matrix A satisfies the given equation x^{2} + 4x β 42 = 0

Now, we have to find A^{-1}

Finding A^{-1} using given equation

A^{2} + 4A β 42 = O

Post multiplying by A^{-1} both sides, we get

(A^{2} + 4A β 42)A^{-1} = OA^{-1}

β A^{2}.A^{-1} + 4A.A^{-1} β 42.A^{-1} = O [OA^{-1} = O]

β A.(AA^{-1}) + 4I β 42A^{-1} = O [AA^{-1} = I]

β A(I) + 4I β 42A^{-1} = O

β A + 4I β 42A^{-1} = O

β A + 4I β O = 42A^{-1}

Ans. .

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