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Question: Answered & Verified by Expert
$$
\text { If } A=\left[\left.\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{array} \right\rvert\, \text {, then }\left(A A^{\prime}\right)^{\prime}=\right.
$$
MathematicsMatricesTS EAMCETTS EAMCET 2018 (05 May Shift 1)
Options:
  • 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 $$
    \left[\begin{array}{ccc}
    14 & 32 & 50 \\
    32 & 194 & 122 \\
    32 & 122 & 77
    \end{array}\right]
    $$
  • D $$
    \left[\begin{array}{ccc}
    14 & 32 & 50 \\
    32 & 77 & 122 \\
    50 & 122 & 194
    \end{array}\right]
    $$
Solution:
1354 Upvotes Verified Answer
The correct answer is: $$
\left[\begin{array}{ccc}
14 & 32 & 50 \\
32 & 77 & 122 \\
50 & 122 & 194
\end{array}\right]
$$
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, $A A^{\prime}=\left[\begin{array}{ccc}14 & 32 & 50 \\ 32 & 77 & 122 \\ 50 & 122 & 194\end{array}\right]\left(A A^{\prime}\right)^{\prime}=\left[\begin{array}{ccc}14 & 32 & 50 \\ 32 & 77 & 122 \\ 50 & 122 & 194\end{array}\right]$

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