Let A=011k,k∈R and A3=abcd.If d = 228 then b+c=

A=(011k)A2=(1kk1+k2) andA3=(k1+k21+k2k3+2k)=(abcd) Givend=228⇒k3+2k=228 ⇒k(k2+2)=6(38) ⇒k=6Nowb+c=2(1+k2)=2(37)=74

Let A=011k,k∈R and A3=abcd.If d = 228 then b+c=

A=(011k)A2=(1kk1+k2) andA3=(k1+k21+k2k3+2k)=(abcd) Givend=228⇒k3+2k=228 ⇒k(k2+2)=6(38) ⇒k=6Nowb+c=2(1+k2)=2(37)=74

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Let $A = [\begin{matrix} 0 & 1 \\ 1 & k \end{matrix}], k \in R$ and $A^{3} = [\begin{matrix} a & b \\ c & d \end{matrix}] .$ If d = 228 then $b + c =$

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Detailed Solution

$A = (\begin{array}{ll} 0 & 1 \\ 1 & k \end{array})$
$A^{2} = (\begin{array}{cc} 1 & k \\ k & 1 + k^{2} \end{array}) and$
$A^{3} = (\begin{array}{cc} k & 1 + k^{2} \\ 1 + k^{2} & k^{3} + 2 k \end{array}) = (\begin{array}{ll} a & b \\ c & d \end{array})$
$\begin{aligned} Given d = 228 \Rightarrow k^{3} + 2 k = 228 \\ \Rightarrow k (k^{2} + 2) = 6 (38) \\ \Rightarrow k = 6 \end{aligned}$
$Now b + c = 2 (1 + k^{2}) = 2 (37) = 74$