Q46 / 160 · AP EAMCET 2022 (06 Jul Shift 1) · Mathematics
If $A=\left[\begin{array}{lll}1 & 1 & 3 \\ 1 & 7 & 9 \\ 2 & 3 & 7\end{array}\right]$ then $\operatorname{Tr}\left(A^2-A\right)=$
- A0
- B-12
- Correct answerC152
- D125
Solution
Given $A=\left[\begin{array}{lll}1 & 1 & 3 \\ 1 & 7 & 9 \\ 2 & 3 & 7\end{array}\right]$
$$
\text { Now }\left(\mathrm{A}^2-\mathrm{A}\right)=\mathrm{A}(\mathrm{A}-\mathrm{I})
$$
$$
\Rightarrow A^2-A=\left[\begin{array}{lll}
1 & 1 & 3 \\
1 & 7 & 9 \\
2 & 3 & 7
\end{array}\right]\left(\left[\begin{array}{lll}
1 & 1 & 3 \\
1 & 7 & 9 \\
2 & 3 & 7
\end{array}\right]-\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\right)
$$
$$
\Rightarrow A^2-A=\left[\begin{array}{lll}
1 & 1 & 3 \\
1 & 7 & 9 \\
2 & 3 & 7
\end{array}\right]\left[\begin{array}{lll}
0 & 1 & 3 \\
1 & 6 & 9 \\
2 & 3 & 6
\end{array}\right]
$$
Thus $\operatorname{Tr}\left(A^2-A\right)=\left[\begin{array}{ccc}7 & x & y \\ a & 70 & Z \\ b & c & 75\end{array}\right]$
$$
\begin{aligned}
& =7+70+75 \\
& =152
\end{aligned}
$$
Hence, the correct answer is option: 'C'.