Q44 / 160 · TS EAMCET 2024 (10 May Shift 2) · Mathematics
$A=\left[a_{i j}\right]$ is a $3 \times 3$ matrix with positive integers as its elements. Elements of A are such that the sum of all the elements of each row is equal to 6 and $a_{22}=2$.
If $a_{i i}=\left\{\begin{array}{ll}a_{i j}+a_{j i}, & j=i+1 \text { when } i \lt 3 \\ a_{i j}+a_{j i}, & j=4-i \text { when } i=3\end{array}\right.$ for $i=1,2,3$, then $|A|=$
- A6
- B18
- C3
- Correct answerD12
Solution
$\operatorname{Consider} \mathbf{A}=\left[\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right]$
Each row sum is 6
$\begin{aligned} & a_{11}+a_{12}+a_{13}=6 \ldots(i)\\ & a_{21}+a_{22}+a_{23}=6 \ldots(ii)\\ & a_{31}+a_{32}+a_{33}=6 \ldots(iii)\end{aligned}$
From given condition on diagonal entries
$\begin{aligned} & a_{11}=a_{12}+a_{21} \ldots(iv)\\ & a_{22}=a_{23}+a_{32}=2 \ldots(v)\\ & a_{33}=a_{13}+a_{31}\ldots(vi)\end{aligned}$
Since entries are the integers only $\therefore$ From (v)
$a_{23}=a_{32}=1 \Rightarrow a_{21}=3$ [From(ii)]
From (iv)
$\begin{aligned} & a_{11}=a_{12}+3 \Rightarrow a_{11}-a_{12}=3 \\ & \Rightarrow a_{11}=4, a_{12}=1 \Rightarrow a_{13}=1\text{[From (i)]}\end{aligned}$
From (iii) and (vi)
$a_{31}+a_{33}=5 \Rightarrow a_{33}-a_{31}=1 \Rightarrow a_{33}=3, a_{31}=2$
Hence, $A=\left[\begin{array}{lll}4 & 1 & 1 \\ 3 & 2 & 1 \\ 2 & 1 & 3\end{array}\right]$ and $|A|=12$
Hence, the correct answer is option: 'D'.