Q. 324.3( 3 Votes )

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

Answer :

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


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


β‡’ A.(AA-1) + 4I – 42A-1 = O [AA-1 = I]


β‡’ A(I) + 4I – 42A-1 = O


β‡’ A + 4I – 42A-1 = O


β‡’ A + 4I – O = 42A-1







Ans. .


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