When using strain gages during a mechanical test of a linearly elastic material, you measure strains in three mutually orthogonal directions (1, 2, and 3): ɛ1, Ɛ2, and ɛ3. You

know the Young's modulus (E) of the material is 750 MPa, and the Poisson's ratio (v) of the material is 0.4. Hooke's law for a linear elastic material can be expressed in matrix form as: \left\{\begin{array}{c} \varepsilon_{1} \\ \varepsilon_{2} \\ \varepsilon_{3} \end{array}\right\}=\frac{1}{E}\left[\begin{array}{ccc} 1 & -v & -v \\ -v & 1 & -v \\ -v & -v & 1 \end{array}\right]\left\{\begin{array}{c} \sigma_{1} \\ \sigma_{2} \\ \sigma_{3} \end{array}\right\} \text { where } \sigma_{1}, \sigma_{2} \text { , and } \sigma_{3} \text { , are the stresses in directions } 1,2, \text { and } 3, \text { respectively. } Use Cholesky decomposition to find the upper triangular matrix U such thatа. U^{\boldsymbol{T}} U=\left[\begin{array}{ccc} \mathbf{1} & -\boldsymbol{v} & -\boldsymbol{V} \\ -\boldsymbol{v} & 1 & -\boldsymbol{V} \\ -\boldsymbol{V} & -\boldsymbol{v} & 1 \end{array}\right] b. Use U to solve for the stresses (ơ1, o2, and o3) in the material for three different experiments: \text { Experiment } 1: \vec{\varepsilon}=\left\{\begin{array}{l} 0.001 \\ 0.003 \\ 0.020 \end{array}\right\} \quad \text { Experiment } 2: \vec{\varepsilon}=\left\{\begin{array}{l} 0.004 \\ 0.008 \\ 0.006 \end{array}\right\} \quad \text { Experiment } 3: \vec{\varepsilon}=\left\{\begin{array}{l} 0.050 \\ 0.070 \\ 0.010 \end{array}\right\}