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TS EAMCET 10 May 2024 Shift 1 Solved Paper

Section: Mathematics

Question : 5 of 160

Marks: +1, -0

A = \begin{pmatrix} 9 & 3 & 0 \\ 1 & 5 & 8 \\ 7 & 6 & 2 \end{pmatrix}

AA^{\top} - A^2 = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}

\sum\limits_{\substack{1 \le i \le 3 \\ 1 \le j \le 3}} a_{ij} =

Solution:

\text{Given, } A = \begin{pmatrix} 9 & 3 & 0 \\ 1 & 5 & 8 \\ 7 & 6 & 2 \end{pmatrix}

\text{ and } A A^T - A^2 = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}

Now,

A^2 = \begin{pmatrix} 9 & 3 & 0 \\ 1 & 5 & 8 \\ 7 & 6 & 2 \end{pmatrix} \begin{pmatrix} 9 & 3 & 0 \\ 1 & 5 & 8 \\ 7 & 6 & 2 \end{pmatrix} = \begin{pmatrix} 84 & 42 & 24 \\ 70 & 76 & 56 \\ 83 & 63 & 52 \end{pmatrix}

A A^T - A^2 = \begin{pmatrix} 9 & 3 & 0 \\ 1 & 5 & 8 \\ 7 & 6 & 2 \end{pmatrix} \begin{pmatrix} 9 & 1 & 7 \\ 3 & 5 & 6 \\ 0 & 8 & 2 \end{pmatrix} - \begin{pmatrix} 84 & 42 & 24 \\ 70 & 76 & 56 \\ 83 & 63 & 52 \end{pmatrix}

= \begin{pmatrix} 90 & 24 & 81 \\ 24 & 90 & 53 \\ 81 & 53 & 89 \end{pmatrix} - \begin{pmatrix} 84 & 42 & 24 \\ 70 & 76 & 56 \\ 83 & 63 & 52 \end{pmatrix}

= \begin{pmatrix} 6 & -18 & 57 \\ -46 & 14 & -3 \\ -2 & -10 & 37 \end{pmatrix}

\;

\sum a_{ij} = 6 - 18 + 57 - 46 + 14

1 \le i \le 3

1 \le j \le 3

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