Suppose A and B are two non singular matrices such that B=I,A6=I and AB2=BA. Find the least value of k for Bk=I.

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Filo · Accepted Answer

Step 1: Given:∣A∣=∣B∣=0A6=IAB2=BABk=IStep 2: To find:Value of k for which Bk=IStep 3: Formula used:AA−1=A−1A=IStep 4: Solution:Consider the given expressionAB2=BA    ... (1)Step 5: Aim to make RHS as independent of APre-multiply (1) from both sides by A5, we getA6B2=A5BAPost-multiply above equation from both sides by A5, we getA6B2A5=A5BA6Since, it is given that A6=I Expression (3) reduces toB2A5=A5BPre-multiply this on both sides by A, we getAB2A5=A6B⇒AB2A5=B        (A6=I)    .....(2) Step 8:Now substituting the value of B from (2) in LHS of (2) itself, we getA(AB2A5)2A5=B⇒A AB2A5 AB2A5 A5=B⇒A2B2A6B2A10=BSince, we know A6=I⇒A2B4A6A4=B⇒A2B4A4=B    ... (3)Step 9:Now substituting the value of B from (2) in LHS of (3), we getA2(AB2A5)4A4=B⇒A2 AB2A5 AB2A5 AB2A5 AB2A5 A4=B⇒A3B2A6B2A6B2A6B2A5A4=BSince, we know A6=I⇒A3B8A3=B    ...(4)Step 10:Looking at the pattern of (2), (3) and (4), we can say that ApBqAr=Bp is increasing by 1, q is forming a GP as 2,4,8,..., andr is decreasing by 1∴ In the same manner, we find that when p=6, we get q=64 and r=0So, the expression becomesA6B64A0=BSince, we know A6=I⇒B64=B    ...(5)Step 11:Now, we have B64=BPost-Multiplying this by B−1 both sides, we getB63BB−1=BB−1    ... (6)Since, BB−1=I, (6) reduces toB63 =I .... (7)Step 12:Comparing (7) with the given condition where Bk=I, we find thatB63=Bk⇒k=63Step 13: Result:k=63

Question Text	Suppose $A$ and $B$ are two non singular matrices such that $B \neq = I, A^{6} = I$ and $A B^{2} = B A$ . Find the least value of $k$ for $B^{k} = I$ .
Answer Type	Text solution:1
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Suppose $A$ and $B$ are two non singular matrices such that $B \neq = I, A^{6} = I$ and $A B^{2} = B A$ . Find the least value of $k$ for $B^{k} = I$ .

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