If A= [ 9 3 0 1 5 8 7 6 2 ] and AA^T-A^2= [ a_11 a_12 a_13 a_21 a_22 a_23 a_31 a_32 a_33 ] , then ∑_substack1 ≤ i ≤ 3 1 ≤ j ≤ 3 a_i j=

Correct answer: (A) 35. A A^T= [ 9 3 0 1 5 8 7 6 2 ] [ 9 1 7 3 5 6 0 8 2 ]= [ 90 24 81 24 90 53 81 53 89 ] A^2= [ 9 3 0 1 5 8 7 6 2 ] [ 9 3 0 1 5 8 7 6 2 ]= [ 84 42 24 70 76 56 83 63 52 ] A A^T-A^2= [ 6 -18 57 -46 14 -3 -2 -10 37 ]= [ a_11 a_12 a_13 a_21 a_22 a_23 a_31 a_32 a_33 ] ∑_1 ≤ i ≤ 3 a_i j=6-18+57-46+14-3-2-10+37=35 .

Q45 / 160 · TS EAMCET 2024 (10 May Shift 1) · Mathematics

If $A=\left[\begin{array}{lll}9 & 3 & 0 \\ 1 & 5 & 8 \\ 7 & 6 & 2\end{array}\right]$ and $\mathrm{AA}^T-\mathrm{A}^2=\left[\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right]$, then $\sum_{\substack{1 \leq i \leq 3 \\ 1 \leq j \leq 3}} a_{i j}=$

Correct answer
A
35
B
0
C
33
D
1

Solution

$\begin{aligned} & \text { } A A^T=\left[\begin{array}{lll}9 & 3 & 0 \\ 1 & 5 & 8 \\ 7 & 6 & 2\end{array}\right]\left[\begin{array}{lll}9 & 1 & 7 \\ 3 & 5 & 6 \\ 0 & 8 & 2\end{array}\right]=\left[\begin{array}{lll}90 & 24 & 81 \\ 24 & 90 & 53 \\ 81 & 53 & 89\end{array}\right] \\ & A^2=\left[\begin{array}{lll}9 & 3 & 0 \\ 1 & 5 & 8 \\ 7 & 6 & 2\end{array}\right]\left[\begin{array}{lll}9 & 3 & 0 \\ 1 & 5 & 8 \\ 7 & 6 & 2\end{array}\right]=\left[\begin{array}{lll}84 & 42 & 24 \\ 70 & 76 & 56 \\ 83 & 63 & 52\end{array}\right] \\ & A A^T-A^2=\left[\begin{array}{ccc}6 & -18 & 57 \\ -46 & 14 & -3 \\ -2 & -10 & 37\end{array}\right]=\left[\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right] \\ & \sum_{1 \leq i \leq 3} a_{i j}=6-18+57-46+14-3-2-10+37=35 .\end{aligned}$

Hence, the correct answer is option: 'A'.

Solution

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