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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

$\det B^5 = (\det B)^5 = (-2)^5 = -32.$