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TS EAMCET 9 May 2024 Shift 2 Solved Paper

Section: Mathematics

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Question : 5 of 160

Marks: +1, -0

If

A = \begin{pmatrix} 1 & 2 & 2 \\ 3 & 2 & 3 \\ 1 & 1 & 2 \end{pmatrix}

A^{-1} = \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} =

[TG EAPCET 9 May 2024 Shift 2]

$\; \frac{2}{3}$
$\; \frac{1}{3}$
$1$
$17$

Solution:

We have,

\; A = \begin{pmatrix} 1 & 2 & 2 \\ 3 & 2 & 3 \\ 1 & 1 & 2 \end{pmatrix} \; \text{ and } \; A^{-1} = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}

\; |A| = 1(4-3)-2(6-3)+2(3-2)=1-6+2=-3

\; C_{11} = \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} = 1

\; C_{12} = -\begin{vmatrix} 3 & 3 \\ 1 & 2 \end{vmatrix} = -3

\; C_{13} = \begin{vmatrix} 3 & 2 \\ 1 & 1 \end{vmatrix} = 1

\; C_{21} = -\begin{vmatrix} 2 & 2 \\ 1 & 2 \end{vmatrix} = -2

\; C_{22} = \begin{vmatrix} 1 & 2 \\ 1 & 2 \end{vmatrix} = 0

\; C_{23} = -\begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix} = 1

\; C_{31} = \begin{vmatrix} 2 & 2 \\ 2 & 3 \end{vmatrix} = 2, C_{32} = -\begin{vmatrix} 1 & 2 \\ 3 & 3 \end{vmatrix} = 3

\; C_{33} = \begin{vmatrix} 1 & 2 \\ 3 & 2 \end{vmatrix} = -4

\; \operatorname{adj} A = \begin{pmatrix} 1 & -3 & 1 \\ -2 & 0 & 1 \\ 2 & 3 & -4 \end{pmatrix}^T = \begin{pmatrix} 1 & -2 & 2 \\ -3 & 0 & 3 \\ 1 & 1 & -4 \end{pmatrix}

\; \therefore A^{-1} = \frac{1}{|A|} (\operatorname{adj} A) = \frac{1}{-3} \begin{pmatrix} 1 & -2 & 2 \\ -3 & 0 & 3 \\ 1 & 1 & -4 \end{pmatrix}

Now,

\; \sum\limits_{\substack{1 \le i \le 3 \\ 1 \le j \le 3}} a_{ij} = \left(-\;\frac{1}{3}\right)

\; [1+(-2)+2+(-3)+0+3+1+1+(-4)]

\; \; \; = \left(-\;\frac{1}{3}\right)(-1) = \frac{1}{3}

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