If A = frac{1}{7} begin{bmatrix} 3 & -2 & 6 | vedclass

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(B) Given $A = \frac{1}{7} \begin{bmatrix} 3 & -2 & 6 \ -6 & -3 & 2 \ -2 & 6 & 3 \end{bmatrix}$.First,we check if $A$ is an orthogonal matrix by cal

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$A^{-1} = A^T$

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(B) Given $A = \frac{1}{7} \begin{bmatrix} 3 & -2 & 6 \ -6 & -3 & 2 \ -2 & 6 & 3 \end{bmatrix}$.First,we check if $A$ is an orthogonal matrix by calculating $AA^T$.$A^T = \frac{1}{7} \begin{bmatrix} 3 & -6 & -2 \ -2 & -3 & 6 \ 6 & 2 & 3 \end{bmatrix}$.$AA^T = \frac{1}{49} \begin{bmatrix} 3 & -2 & 6 \ -6 & -3 & 2 \ -2 & 6 & 3 \end{bmatrix} \begin{bmatrix} 3 & -6 & -2 \ -2 & -3 & 6 \ 6 & 2 & 3 \end{bmatrix}$.Calculating the product:Row $1 	imes$ Col $1 = (3)(3) + (-2)(-2) + (6)(6) = 9 + 4 + 36 = 49$.Row $1 	imes$ Col $2 = (3)(-6) + (-2)(-3) + (6)(2) = -18 + 6 + 12 = 0$.Row $1 	imes$ Col $3 = (3)(-2) + (-2)(6) + (6)(3) = -6 - 12 + 18 = 0$.Similarly,all off-diagonal elements result in $0$ and diagonal elements result in $49$.Thus,$AA^T = \frac{1}{49} (49I) = I$.Since $AA^T = I$,it follows that $A^{-1} = A^T$.

If $A = \frac{1}{7} \begin{bmatrix} 3 & -2 & 6 \\ -6 & -3 & 2 \\ -2 & 6 & 3 \end{bmatrix}$,then:

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