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TG EAMCET 3 May 2025 Shift 2 Paper

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Question : 7 of 160

Marks: +1, -0

If

A = \begin{pmatrix} 1 & 5 & 2 \\ 4 & 1 & 3 \\ 2 & 6 & 3 \end{pmatrix}

\left| (\operatorname{Adj} A)^{-1} \right| =

$-1$
$1$
$4$
$-4$

Solution:

Step 1: Find the determinant of

A

Matrix

A

is:

A = \begin{pmatrix} 1 & 5 & 2 \\ 4 & 1 & 3 \\ 2 & 6 & 3 \end{pmatrix}

So,

|A| = 1(3-18) - 5(12-6) + 2(24-2)

Calculate each part:

\;1 \times (3-18) = 1 \times (-15) = -15

\; -5 \times (12-6) = -5 \times 6 = -30

\; 2 \times (24-2) = 2 \times 22 = 44

Add them up:

|A| = -15 - 30 + 44 = -1

Step 2: Find the value of

\left| (\operatorname{adj} A)^{-1} \right|

\left| (\operatorname{adj} A)^{-1} \right| = \;\frac{1}{|\operatorname{adj} A|}

The order of matrix

A

is 3 , so:

|\operatorname{adj} A| = |A|^{3-1} = |A|^2

So,

\left| (\operatorname{adj} A)^{-1} \right| = \;\frac{1}{|A|^2}

Since

|A| = -1, \left| (\operatorname{adj} A)^{-1} \right| = \;\frac{1}{(-1)^2} = \;\frac{1}{1} = 1

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