Q44 / 160 · TS EAMCET 2018 (05 May Shift 1) · Mathematics
$$
\text { If } A=\left[\left.\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{array} \right\rvert\, \text {, then }\left(A A^{\prime}\right)^{\prime}=\right.
$$
- A$$ \left[\begin{array}{ccc} 14 & 32 & 50 \\ 32 & 122 & 194 \\ 50 & 194 & 256 \end{array}\right] $$
- B$$ \text { }\left[\begin{array}{ccc} 14 & 50 & 32 \\ 32 & 122 & 194 \\ 50 & 194 & 122 \end{array}\right] $$
- C$$ \left[\begin{array}{ccc} 14 & 32 & 50 \\ 32 & 194 & 122 \\ 32 & 122 & 77 \end{array}\right] $$
- Correct answerD$$ \left[\begin{array}{ccc} 14 & 32 & 50 \\ 32 & 77 & 122 \\ 50 & 122 & 194 \end{array}\right] $$
Solution
We have,
$$
A=\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{array}\right] \Rightarrow A^{\prime}=\left[\begin{array}{lll}
1 & 4 & 7 \\
2 & 5 & 8 \\
3 & 6 & 9
\end{array}\right]
$$
Now,
$$
\begin{aligned}
& \text { Now, } A \cdot A^{\prime}=\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{array}\right]\left[\begin{array}{lll}
1 & 4 & 7 \\
2 & 5 & 8 \\
3 & 6 & 9
\end{array}\right] \\
& =\left[\begin{array}{ccc}
1+4+9 & 4+10+18 & 7+16+27 \\
4+10+18 & 16+25+36 & 28+40+54 \\
7+16+27 & 28+40+54 & 49+64+81
\end{array}\right] \\
& \text { So, } A A^{\prime}=\left[\begin{array}{ccc}
14 & 32 & 50 \\
32 & 77 & 122 \\
50 & 122 & 194
\end{array}\right]\left(A A^{\prime}\right)^{\prime}=\left[\begin{array}{ccc}
14 & 32 & 50 \\
32 & 77 & 122 \\
50 & 122 & 194
\end{array}\right]
\end{aligned}
$$
So, $A A^{\prime}=\left[\begin{array}{ccc}14 & 32 & 50 \\ 32 & 77 & 122 \\ 50 & 122 & 194\end{array}\right]\left(A A^{\prime}\right)^{\prime}=\left[\begin{array}{ccc}14 & 32 & 50 \\ 32 & 77 & 122 \\ 50 & 122 & 194\end{array}\right]$
Hence, the correct answer is option: 'D'.