If 3 A = begin{bmatrix} 1 & 2 & 2 2 & 1 & -2 | vedclass

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(D) Given $3 A = \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & -2 \ a & 2 & b \end{bmatrix}$. Since $A A^{T} = I$,we have $(3 A)(3 A)^{T} = 9 I$.$(3 A)(3 A)^{

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$\frac{5}{2}$

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(D) Given $3 A = \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & -2 \ a & 2 & b \end{bmatrix}$. Since $A A^{T} = I$,we have $(3 A)(3 A)^{T} = 9 I$.$(3 A)(3 A)^{T} = \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & -2 \ a & 2 & b \end{bmatrix} \begin{bmatrix} 1 & 2 & a \ 2 & 1 & 2 \ 2 & -2 & b \end{bmatrix} = \begin{bmatrix} 9 & 0 & 0 \ 0 & 9 & 0 \ 0 & 0 & 9 \end{bmatrix}$.Multiplying the matrices,the element at $(1, 3)$ is $1(a) + 2(2) + 2(b) = a + 4 + 2b = 0$.The element at $(2, 3)$ is $2(a) + 1(2) - 2(b) = 2a + 2 - 2b = 0$.From $2a - 2b = -2$,we get $a - b = -1$,so $a = b - 1$.Substituting into $a + 2b = -4$,we get $(b - 1) + 2b = -4 \Rightarrow 3b = -3 \Rightarrow b = -1$.Then $a = -1 - 1 = -2$.Thus,$\frac{a}{b} + \frac{b}{a} = \frac{-2}{-1} + \frac{-1}{-2} = 2 + \frac{1}{2} = \frac{5}{2}$.

If $3 A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{bmatrix}$ and $A A^{T} = I$,then find the value of $\frac{a}{b} + \frac{b}{a}$.

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