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Problems #7 on pg.130 and #29, pg.176 are graded.
#7,pg.130: Find an LU factorization of $\begin{bmatrix} 2 & 5 \\ -3 & -4 \end{bmatrix}$.
Solution: We will reduce this matrix to an echelon form and call that $U$ and then use the method outlined in Example 2, pg.126 to construct $L$ from our row reduction, then we will check the answer to make sure it is correct.
Compute
$$\begin{array}{ll}
\begin{bmatrix} 2 & 5 \\ -3 & -4 \end{bmatrix} &\stackrel{r_2^*=r_2+\frac{3}{2}r_1}{\sim} \begin{bmatrix} 2 & 5 \\ 0 & \frac{7}{2}, \end{bmatrix}
\end{array}$$
yielding $U=\begin{bmatrix} 2 & 5 \\ 0 & \frac{7}{2} \end{bmatrix}$. We now see that
$$L=\begin{bmatrix} 1 & 0 \\ -\frac{3}{2} & 1 \\ \end{bmatrix}.$$
Clearly $L$ is lower triangular and $U$ is upper triangular and this calculation confirms that it is indeed a correct factorization.
#29,pg.176: Compute $\mathrm{det} B^5$ where $B=\begin{bmatrix} 1&0&1 \\ 1&1&2 \\ 1&2&1 \end{bmatrix}$.
Solution: Calcluating $B^5$ and then taking the determinant is an inefficient way to answer this question. Using the multiplicative property in Theorem 6, pg.173 we see that
$$\det B^5 = (\det B)^5$$
and so calculate
$$\begin{array}{ll}
\det B &= 1 \det \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} - 0 \det \begin{bmatrix} 1&2 \\ 1&1 \end{bmatrix} + 1 \det \begin{bmatrix} 1 & 1 \\ 1&2 \end{bmatrix} \\
&= 1(1-4) - 0(1-2) + 1(2-1) \\
&= -3 - 0 + 1 \\
&= -2.
\end{array}$$
Hence we see
$$\det B^5 = (\det B)^5 = (-2)^5 = -32.$$
