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Question Number 175193 by Best1 last updated on 22/Aug/22

	Answered by Rasheed.Sindhi last updated on 22/Aug/22
	$A= [(2,1,0),(1,2,1),(3,3,2) ], A^T = [(2,1,3),(1,2,3),(0,1,2) ] AA^T = [((4+1+0),(2+2+0),(6+3+0)),((2+2+0),(1+4+1),(3+6+2)),((6+3+0),(3+6+2),(9+9+4)) ] AA^T = [(5,4,9),(4,6,(11)),(9,(11),(22)) ] 2AA^T = [((10),8,(18)),(8,(12),(22)),((18),(22),(44)) ] determinant (((2AA^T )))=2∙2 determinant ((5,8,9),(4,(12),(11)),(9,(22),(22))) =4 determinant ((5,8,9),(4,(12),(11)),(9,(22),(22))) =4 determinant ((5,8,( 1)),(4,(12),(−1)),(9,(22),( 0))) C3−C2 =4 determinant ((5,8,( 1)),(9,(20),( 0)),(9,(22),( 0))) R2+R1 =4{1(9∙22−9∙20)} = 4∙9(22−20)=4∙9∙2=72$
	$A = [\begin{matrix} 2 & 1 & 0 \\ 1 & 2 & 1 \\ 3 & 3 & 2 \end{matrix}], A^{T} = [\begin{matrix} 2 & 1 & 3 \\ 1 & 2 & 3 \\ 0 & 1 & 2 \end{matrix}]$ ${A A}^{T} = [\begin{matrix} 4 + 1 + 0 & 2 + 2 + 0 & 6 + 3 + 0 \\ 2 + 2 + 0 & 1 + 4 + 1 & 3 + 6 + 2 \\ 6 + 3 + 0 & 3 + 6 + 2 & 9 + 9 + 4 \end{matrix}]$ ${A A}^{T} = [\begin{matrix} 5 & 4 & 9 \\ 4 & 6 & 11 \\ 9 & 11 & 22 \end{matrix}]$ $2 {A A}^{T} = [\begin{matrix} 10 & 8 & 18 \\ 8 & 12 & 22 \\ 18 & 22 & 44 \end{matrix}]$ $\| \begin{matrix} 2 {A A}^{T} \end{matrix} \| = 2 \cdot 2 \| \begin{matrix} 5 & 8 & 9 \\ 4 & 12 & 11 \\ 9 & 22 & 22 \end{matrix} \|$ $= 4 \| \begin{matrix} 5 & 8 & 9 \\ 4 & 12 & 11 \\ 9 & 22 & 22 \end{matrix} \|$ $= 4 \| \begin{matrix} 5 & 8 & 1 \\ 4 & 12 & - 1 \\ 9 & 22 & 0 \end{matrix} \| C 3 - C 2$ $= 4 \| \begin{matrix} 5 & 8 & 1 \\ 9 & 20 & 0 \\ 9 & 22 & 0 \end{matrix} \| R 2 + R 1$ $= 4 {1 (9 \cdot 22 - 9 \cdot 20)}$ $= 4 \cdot 9 (22 - 20) = 4 \cdot 9 \cdot 2 = 72$
	Answered by kapoorshah last updated on 23/Aug/22

	$∣ A ∣ = 3$ $∣ 2 {A A}^{T} ∣ = 2^{3} ∣ A ∣∣ A^{T} ∣$ $= 8 ∣ A ∣∣ A ∣$ $= 8.3.3$ $= 72$
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