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(B) Given $A U_{1} = \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix}$,$A U_{2} = \begin{bmatrix} 2 \ 3 \ 0 \end{bmatrix}$,and $A U_{3} = \begin{bmatrix} 2

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(B) Given $A U_{1} = \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix}$,$A U_{2} = \begin{bmatrix} 2 \ 3 \ 0 \end{bmatrix}$,and $A U_{3} = \begin{bmatrix} 2 \ 3 \ 1 \end{bmatrix}$.This can be written as $A U = \begin{bmatrix} 1 & 2 & 2 \ 0 & 3 & 3 \ 0 & 0 & 1 \end{bmatrix}$,where $U = [U_{1} U_{2} U_{3}]$.We know that $A U = B$,where $B = \begin{bmatrix} 1 & 2 & 2 \ 0 & 3 & 3 \ 0 & 0 & 1 \end{bmatrix}$.Thus,$U = A^{-1} B$.Taking the inverse of both sides,$U^{-1} = (A^{-1} B)^{-1} = B^{-1} A$.First,find $A^{-1}$. Since $A = \begin{bmatrix} 1 & 0 & 0 \ 2 & 1 & 0 \ 3 & 2 & 1 \end{bmatrix}$,it is a lower triangular matrix.$A^{-1} = \begin{bmatrix} 1 & 0 & 0 \ -2 & 1 & 0 \ 1 & -2 & 1 \end{bmatrix}$.Next,find $B^{-1}$. Since $B = \begin{bmatrix} 1 & 2 & 2 \ 0 & 3 & 3 \ 0 & 0 & 1 \end{bmatrix}$,$|B| = 1(3-0) = 3$.$B^{-1} = \frac{1}{3} \begin{bmatrix} 3 & -2 & 0 \ 0 & 1 & -3 \ 0 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 1 & -2/3 & 0 \ 0 & 1/3 & -1 \ 0 & 0 & 1 \end{bmatrix}$.Now,$U^{-1} = B^{-1} A = \begin{bmatrix} 1 & -2/3 & 0 \ 0 & 1/3 & -1 \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \ 2 & 1 & 0 \ 3 & 2 & 1 \end{bmatrix} = \begin{bmatrix} -1/3 & -2/3 & 0 \ -7/3 & -5/3 & -1 \ 3 & 2 & 1 \end{bmatrix}$.Sum of elements = $(-1/3 - 2/3 + 0) + (-7/3 - 5/3 - 1) + (3 + 2 + 1) = -1 - 5 + 6 = 0$.

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