Question

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let A be the square of matrix of order 3 satisfies the relation A³-6A²+7A-8I=o and B=A-2I. Also detA=8 then​

Answer 1

Step-by-step explanation:

A^2–7A+4I=0

A^2–4A+4(I*I)=3A [I*I=I]

A^2–2*A*2I+(2I)^2=3A

(A-2I)^2=3A

Det (A-2I)^2= Det (3A)

Det(A-2I)*Det(A-2I)= 27Det(A) [Det(cA) = c^n*Det(A), n= order of matrix.]

Det (A-2I)=(27*3)^0.5

Det (A-2I) = 9 [Det (A-2I) >=Det (A)+ Det (-2I) , Det (A-2I) >= Det (A)- 8 Det of (I), Det (A-2I) >=3–8,Det (A-2I) >=-5]

Det (A-2I) =9.

Answer 2

Answer:

$\huge\rm\underline\purple{Answer:-}$

If A is a square matrix of order n , then we can take k common from each of the n rows of kA

hence,|KA|=K^n|A|

So, for a square matrix of of order 3

$ka = k {}^{3} a$