Let's solve the problem step by step. --- ### **Given Diffusion Tensor:** \[ D = \begin{pmatrix} 7 & 4 & 6 \\ -3 & -1 & -8 \\ 0 & 0 & 1 \end{pmatrix} \] --- ## **a. Calculate a unit vector that points in the direction of the myocardial fibers at that spot** The direction of the myocardial fibers is given by the **principal eigenvector** of the diffusion tensor (i.e., the eigenvector associated with the largest eigenvalue). Let's find the eigenvalues and the corresponding eigenvectors. #### **Step 1: Find the Eigenvalues** Find \( \lambda \) such that: \[ \det(D - \lambda I) = 0 \] \[ \begin{pmatrix} 7 - \lambda & 4 & 6 \\ -3 & -1 - \lambda & -8 \\ 0 & 0 & 1 - \lambda \end{pmatrix} \] The determinant is: \[ \det(D - \lambda I) = (1 - \lambda)\begin{vmatrix}7 - \lambda & 4 \\ -3 & -1 - \lambda\end{vmatrix} \] Calculate the 2x2 determinant: \[ (7 - \lambda)(-1 - \lambda) - (4)(-3) = -(7 - \lambda)(1 + \lambda) + 12 \] So, \[ \det(D - \lambda I) = (1 - \lambda)\left[(7 - \lambda)(-1 - \lambda) + 12\right] \] Expand \((7-\lambda)(-1-\lambda)\): \[ (7-\lambda)(-1-\lambda) = -7 - 7\lambda + \lambda + \lambda^2 = -7 - 6\lambda + \lambda^2 \] So, \[ \det(D - \lambda I) = (1-\lambda)\left[(-7 - 6\lambda + \lambda^2) + 12\right] \] \[ = (1-\lambda)(\lambda^2 - 6\lambda + 5) \] Factor inside the parentheses: \[ \lambda^2 - 6\lambda + 5 = (\lambda-5)(\lambda-1) \] So the eigenvalues are: \[ (1-\lambda)(\lambda-5)(\lambda-1) = 0 \] Thus, the eigenvalues are \( \lambda = 1 \) (with multiplicity 2), and \( \lambda = 5 \). #### **Step 2: Find the Corresponding Eigenvector for the Largest Eigenvalue (\(\lambda = 5\))** Plugging \(\lambda = 5\) into \(D - \lambda I\): \[ D - 5I = \begin{pmatrix} 2 & 4 & 6 \\ -3 & -6 & -8 \\ 0 & 0 & -4 \end{pmatrix} \] Set up the equation: \[ \begin{pmatrix} 2 & 4 & 6 \\ -3 & -6 & -8 \\ 0 & 0 & -4 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \] From the third row: \(-4z = 0 \implies z = 0\) Now substitute \(z = 0\) into the first two equations: 1. \(2x + 4y = 0 \implies x + 2y = 0 \implies x = -2y\) 2. \(-3x - 6y = 0 \implies -3x = 6y \implies x = -2y\) (which matches!) So the eigenvector is: \[ \vec{v} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -2y \\ y \\ 0 \end{pmatrix} \] Let \( y = 1 \): \[ \vec{v} = \begin{pmatrix} -2 \\ 1 \\ 0 \end{pmatrix} \] #### **Step 3: Normalize the Eigenvector** Calculate the magnitude: \[ \|\vec{v}\| = \sqrt{(-2)^2 + 1^2 + 0^2} = \sqrt{4 + 1} = \sqrt{5} \] So the unit vector is: \[ \vec{u} = \frac{1}{\sqrt{5}}\begin{pmatrix} -2 \\ 1 \\ 0 \end{pmatrix} \] --- ### **b. Calculate the trace of the diffusion tensor** The trace is the sum of the diagonal elements: \[ \text{Trace}(D) = 7 + (-1) + 1 = 7 \] --- ## **Final Answers:** ### a. **Unit vector in the direction of myocardial fibers:** \[ \boxed{ \frac{1}{\sqrt{5}} \begin{pmatrix} -2 \\ 1 \\ 0 \end{pmatrix} } \] ### b. **Trace of the diffusion tensor:** \[ \boxed{7} \]

Certainly! Let's go through the detailed calculation for part **b**, which involves computing the trace of the diffusion tensor \( D \). --- ### **Given diffusion tensor:** \[ D = \begin{pmatrix} 7 & 4 & 6 \\ -3 & -1 & -8 \\ 0 & 0 & 1 \end{pmatrix} \] --- ### **Step-by-step calculation of the trace** The **trace** of a matrix is defined as the sum of its **diagonal elements**: \[ \text{Trace}(D) = D_{11} + D_{22} + D_{33} \] From the matrix: - \( D_{11} = 7 \) - \( D_{22} = -1 \) - \( D_{33} = 1 \) Therefore: \[ \boxed{ \text{Trace}(D) = 7 + (-1) + 1 } \] ### **Perform the addition:** \[ 7 - 1 + 1 = (7 - 1) + 1 = 6 + 1 = 7 \] --- ### **Final answer:** \[ \boxed{ \text{Trace}(D) = \mathbf{7} } \] --- ### **Summary:** - The trace is simply the sum of the diagonal elements. - For this tensor, that sum is \(7 + (-1) + 1 = 7\). This completes the detailed calculation for part **b**.