If A = begin{bmatrix} 1 & 2 & 2 2 & 1 & -2 | vedclass

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(C) Given $A = \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & -2 \ a & 2 & b \end{bmatrix}$.Then $A^T = \begin{bmatrix} 1 & 2 & a \ 2 & 1 & 2 \ 2 & -2 & b \e

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$5$

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(C) Given $A = \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & -2 \ a & 2 & b \end{bmatrix}$.Then $A^T = \begin{bmatrix} 1 & 2 & a \ 2 & 1 & 2 \ 2 & -2 & b \end{bmatrix}$.The condition is $A A^T = 9 I = \begin{bmatrix} 9 & 0 & 0 \ 0 & 9 & 0 \ 0 & 0 & 9 \end{bmatrix}$.Calculating the product $A A^T$:Row $1 	imes$ Column $1$: $1(1) + 2(2) + 2(2) = 1 + 4 + 4 = 9$.Row $2 	imes$ Column $2$: $2(2) + 1(1) + (-2)(-2) = 4 + 1 + 4 = 9$.Row $3 	imes$ Column $3$: $a(a) + 2(2) + b(b) = a^2 + 4 + b^2$.Since $A A^T = 9 I$,the diagonal elements must be $9$. Thus,$a^2 + 4 + b^2 = 9$,which implies $a^2 + b^2 = 5$.Also,checking off-diagonal elements:Row $1 	imes$ Column $2$: $1(2) + 2(1) + 2(-2) = 2 + 2 - 4 = 0$.Row $1 	imes$ Column $3$: $1(a) + 2(2) + 2(b) = a + 4 + 2b = 0$.Row $2 	imes$ Column $3$: $2(a) + 1(2) + (-2)(b) = 2a + 2 - 2b = 0 \implies a - b = -1$.From $a + 2b = -4$ and $a - b = -1$,subtracting gives $3b = -3 \implies b = -1$. Then $a = -2$.Checking $a^2 + b^2 = (-2)^2 + (-1)^2 = 4 + 1 = 5$.

If $A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{bmatrix}$ is a matrix satisfying the equation $A A^T = 9 I$,where $I$ is the identity matrix,then $a^2 + b^2 =$

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