Using the matrices $$ \begin{array}{r}D=\left[\begin{array}{ccc}4 & -4 & 0 \ -4 & -2 & 3 \ 3 & 3 & 1\end{array}\right], E=\left[\begin{array}{ccc}4 & 0 & -2 \ -4 & -3 & -1 \ -1 & -4 & 1\end{array}\right] \ \text { compute the following. } \ \qquad \begin{array}{ll}\operatorname{tr}\left(6 E^{T}-D\right)\end{array} \ \operatorname{tr}\left(6 E^{T}-D\right)=\mathbf{i}\end{array} $$

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To compute $$\operatorname{tr}(6E^{T} - D)$$, follow these steps:

1. **Transpose Matrix $$E$$:**
   $$
   E^T = \begin{bmatrix}
   4 & -4 & -1 \
   0 & -3 & -4 \
   -2 & -1 & 1 
   \end{bmatrix}
   $$

2. **Multiply $$E^T$$ by 6:**
   $$
   6E^T = \begin{bmatrix}
   24 & -24 & -6 \
   0 & -18 & -24 \
   -12 & -6 & 6 
   \end{bmatrix}
   $$

3. **Subtract Matrix $$D$$ from $$6E^T$$:**
   $$
   6E^T - D = \begin{bmatrix}
   24 - 4 & -24 - (-4) & -6 - 0 \
   0 - (-4) & -18 - (-2) & -24 - 3 \
   -12 - 3 & -6 - 3 & 6 - 1 
   \end{bmatrix}
   = \begin{bmatrix}
   20 & -20 & -6 \
   4 & -16 & -27 \
   -15 & -9 & 5 
   \end{bmatrix}
   $$

4. **Compute the Trace:**
   The trace of a matrix is the sum of its diagonal elements.
   $$
   \operatorname{tr}(6E^T - D) = 20 + (-16) + 5 = 9
   $$

**Final Answer:**  
$$\operatorname{tr}\left(6 E^{T}-D\right) = 9$$
$$\operatorname{tr}\left(6 E^{T}-D\right) = 9$$

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