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TG EAMCET 2 May 2025 Shift 1 Paper

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Question : 6 of 160

Marks: +1, -0

A = \begin{pmatrix} 1 & -3 & -5 \\ -2 & 4 & -6 \\ 7 & -11 & 13 \end{pmatrix}

\sqrt{|\operatorname{adj} A|} =

64
16
36
216

Solution:

\;\;\text{ Given, }\; A = \begin{pmatrix} 1 & -3 & -5 \\ -2 & 4 & -6 \\ 7 & -11 & 13 \end{pmatrix}

|A| = \begin{vmatrix} 1 & -3 & -5 \\ -2 & 4 & -6 \\ 7 & -11 & 13 \end{vmatrix}

= 1(4 \times 13 - (-6) \times (-11)) - (-3)(-2 \times 13 - (-6) \times 7) - 5(-2 \times (-11) - 7 \times 4)

= 1(52 - 66) + 3(-26 + 42) - 5(22 - 28)

= -14 + 48 + 30 = 64

Since,

|\operatorname{adj} A| = |A|^{n-1}

, for

n \times n

matrix

A

So, for

3 \times 3

matrix,

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

\therefore \;\; \therefore \;\; |\operatorname{adj} A|\; = |A|^2 = (64)^2 = 4096

\sqrt{|\operatorname{adj} A|}\; = \sqrt{4096} = 64

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