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Problems #14 on pg.372 and #17 on pg.384 are graded.

#14,pg.372: Compute the inverse of the given matrix or show that it is singular: $\begin{bmatrix} 1&2&1 \\ -2&1&8 \\ 1&-2&-7 \end{bmatrix}$.
Solution: This matrix is singular because $$\det \left( \begin{bmatrix} 1&2&1 \\ -2&1&8 \\ 1&-2&-7 \end{bmatrix} \right)=1(-7-(-16))-2(14-8)+1(4-1) = 0.$$

#17,pg.384: Find the eigenvalues and eigenvectors of the following matrix: $\begin{bmatrix} -2 & 1 \\ 1& -2 \end{bmatrix}$.
Solution: To find the eigenvalues we need to solve the characteristic equation $\det (A - \lambda I)=0$. Cmpute $$\begin{array}{ll} \det(A - \lambda I) &= \det \left( \begin{bmatrix} -2-\lambda & 1 \\ 1 & -2-\lambda \end{bmatrix} \right) \\ &=(2+\lambda)^2 - 1 \\ &= 4+4\lambda+\lambda^2 - 1 \\ &= \lambda^2 + 4 \lambda + 3 \\ &= (\lambda+3)(\lambda+1), \end{array}$$ hence we have eigenvalues $\lambda_1=-3$ and $\lambda_2=-1$. To each eigenvalue is associated an eigenvector which we will find. Let's find the eigenvector associated with $\lambda_1$: we must find a vector $\vec{x} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$ such that $$\begin{bmatrix} -2 & 1 \\ 1 & -2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = -3 \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}.$$ This leads us to $$\begin{bmatrix} -2x_1 + x_2 \\ x_1 - 2x_2 \end{bmatrix} = \begin{bmatrix} -3x_1 \\ -3x_2 \end{bmatrix},$$ which encodes the system $$\left\{ \begin{array}{ll} x_1 + x_2 &= 0 \\ x_1 + x_2 &= 0 \end{array} \right.$$ Thus any vector $\vec{x}=\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$ for which the formula $x_1=-x_2$ holds is an eigenvector associated with $\lambda_1$. For a particular example, we could take, say, $\vec{x} = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$.

Let's find the eigenvector associated with $\lambda_2$: we must find a vector $\vec{x}=\begin{bmatrix}x_1 \\x_2 \end{bmatrix}$ such that $$\begin{bmatrix} -2 & 1 \\ 1 & -2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = -\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}.$$ This leads us to $$\begin{bmatrix} -2x_1 + x_2 \\ x_1-2x_2 \end{bmatrix} = \begin{bmatrix} -x_1 \\ -x_2 \end{bmatrix}.$$ which encodes the system $$\left\{ \begin{array}{ll} -x_1 + x_2 &= 0 \\ x_1 - x_2 &= 0. \end{array} \right.$$ Just like before, we now see that any vector $\vec{x} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$ for which the formula $x_1=x_2$ holds is an eigenvector associated with $\lambda_2$. For a particular example, we could take, say $\vec{x} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$.