Question

$A=\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]$. Trace $\left(A\right)=9$ and $a,b,c$ are positive integers such that $ab+bc+ca=26$ Let $A$, denotes the adjoin of $A_1{A}_2$ represent the adjoin of $A_3\dots$ and so on. If value of det $A_4$ is $M$, then

• A. $M=3^{48}$
• B. $M=3^{24}$
• C. Last two digits of $M$ are $6$ and $1$
• D. Last three digits of $M$ are $8,6$ and $1$

Answer 1

The correct option is C Last two digits of $M$ are $6$ and $1$

Given : $A=\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]$

$Trace\left(A\right)=9:a,b,c$ are positive integers such that $ab+bc+ca=26$.

$A_1$ is adjoint of A

$A_2$ is adjoint of $A_1$ and so on.

Since, $a+b+c=Tr\left(A\right)=9$

and $ab+bc+ca=26$

$\therefore (a+b+c)(ab+bc+ca)=9\times 26$

$\Rightarrow (a+b)(b+c)(c+a)-abc=9\times 26$

$\Rightarrow (a+b)(b+c)(c+a)-abc=234$

$|A|=a(bc-a^2)+b(ac-b^2)+c(ab-c^2)$

$=abc-a^3+abc-b^2+abc-c^3$

$|A|=3abc-(a^3+b^3+c^3)$ .... (1)

since, $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$

$=\left(9\right)(a^2+b^2+c^2-26)$ ....(2)

$|A_1|=Adj\left(A\right)=|det\left(A\right){|}^{n-1}$

$=|A{|}^{n-1}$

$|A_2|=Adj\left(A_1\right)=\left[det\right(A_1){]}^{n-1}$

$=|A_1{|}^{n-1}$

$|A_3|=Adj\left(A_2\right)=|A_2{|}^{n-1}$

$|A_4|=Adj\left(A_3\right)=|A_3{|}^{n-1}$

$=||A_2{|}^{n-1}{|}^{n-1}$

$=|||A_1{|}^{n-1}{|}^{n-1}{|}^{n-1}$

$|A_4|=||||A{|}^{n-1}{|}^{n-1}{|}^{n-1}{|}^{n-1}$ ...(3)

$\because$ we know that,

$a+b^2+c^2=(a+b+c{)}^2-2ab-2bc-2ac$

$=(9{)}^2=2(ab+bc+ac)$

$=81-2\left(26\right)$

$=81-52$

$a^2+b^2+c^2=29$ subtitute in (2)

$\Rightarrow a^3+b^3+c^3-3abc=9(29-26)$

$=9\left(3\right)$

$$\Rightarrow a^3 + b^3 + c^3 - 3abc =27$ ...(4)

$\therefore$ from (1) we have

$|A|=3abc-(a^3+b^3+c63)$

$\Rightarrow |A|=-(A^3+b^3+c^3-3abc)$

$\Rightarrow |A|=-\left(27\right)$ (from (4))

$\therefore M=|A_4|=||||A{|}^{n-1}{|}^{n-1}{|}^{n-1}{|}^{n-1}$ (from (3))

$\Rightarrow M=|A_{3+1}|=||||A{|}^{3-1}{|}^{3-1}{|}^{3-1}$

$\Rightarrow M=|A_4|=||||A{|}^2{|}^2{|}^2{|}^2$

$\Rightarrow M=|A_4|=A^{16}$

$\Rightarrow M=|A_4|=|27{|}^{16}$

$\Rightarrow M=|A_4|=3^{3\times 16}$

$\Rightarrow M=|A_4|=3^{48}=7.176644308\times 1022$

$\therefore M=3^{48}$ is correct answer.