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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 $\vec{x} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$.
