If A = [ 31 −4−1] then Ak=[1+2k k −4k1−2k]where k is any +ve integer .

Question

If A = [ 31 ​−4−1​] then Ak=[1+2k  k  ​ −4k1−2k​]where k is any +ve integer .

Filo · Accepted Answer

A = [ 31 ​−4−1​] then A1=[1+2k  k  ​ −4k1−2k​]k=1​Also A2 = AA=[ 31 ​−4−1​]   [ 31 ​−4−1​] = [ 9−43−1 ​−12+4−4+1​]or A2 = [52 ​−8−3​] = [1+222 ​−4.21−2.2​]k=2​Now assume that Ak=[1+2k  k  ​ −4k1−2k​]k=k​∴Ak+1=AAk= [ 31 ​−4−1​] [1+2k  k  ​ −4k1−2k​]= [3+6k−4  1+2k−k  ​−12k−4+8k  −4k−1+2k​]= [3+2k  1+k ​−4k−4  −2k−1​]= [1+2(k+1)  1+k ​−4(k+1)1−2k+1​]we observe that our assumption is true for k = k + 1 and it was true when k = 1 or 2. Hence it is true for all +ve integral values for k .

Question Text	If A = $[31 - 4 - 1]$ then $A^{k} = [1 + 2 k k - 4 k 1 - 2 k]$ where k is any +ve integer .
Answer Type	Text solution:1
Upvotes	150

If A = $[31 - 4 - 1]$ then $A^{k} = [1 + 2 k k - 4 k 1 - 2 k]$
where k is any +ve integer .

Text solutionVerified

Practice questions from similar books

If A = [ 31 ​−4−1​] then Ak=[1+2k k ​ −4k1−2k​]where k is any +ve integer .

Text solutionVerified

Practice questions from similar books

If A = $[31 - 4 - 1]$ then $A^{k} = [1 + 2 k k - 4 k 1 - 2 k]$
where k is any +ve integer .