For Problem V-4: (a) You should multiply by the matrix obtained by scaling the rows of $I$ by the same factors, or in other words, by $\left[\begin{array}{rrr}10&0&0\\ 0&5&0\\ 0&0&2\end{array}\right]$.

(b) You should multiply on the right by the diagonal matrix from (a).

For Problem V-5: Since $P$ can be obtained by switching the first and third rows of $I$ and also the second and fourth rows, $Px$ is $x$ with the first and third entries switched and also the second and fourth entries.

For Problem V-6: (a) Since $N$ is obtained from the identity matrix by ``moving the rows down 1'', meaning replacing the first row by 0's and the second row by the first, taking $NH$ produces the same effect on $H$, so $NH = \left[\begin{array}{ccccccc}0&0&0&0&0&0&0\\ H&E&L&P&I&A&M\end{array}\right]$.

(b) To make $NH = \left[\begin{array}{ccccccc}0&0&0&0&0&0&0\\ H&E&L&P&I&A&M\end{array}\right]$ from $H$ we need to ``move the rows down 1'' by replacing the first row by 0's and second row by the first, so we do that to the identity matrix to deduce that it's $N$ that we want.

(c) $N^2$ is obtained from $N$ by moving the rows down 1 again, or in other words moving the rows of $I$ down twice, producing the zero matrix.

(d) In the same way, $N _ 4 ^ 4$ is otained from the identity matrix by moving the rows down four times, so we get the zero matrix.

For Problem V-7:

Let $N = w v ^ t$. Then $N^2 = (w v ^ t)^2 = w v ^ t w v ^ t = w (v ^ t w)v ^ t = w 0 v ^ t = 0 w v ^ t = 0$ times a $2 \times 2$ matrix = the $2 \times 2$ zero matrix. In other words, $N^2 = 0$. (Remember, a matrix is nipotent if some power of it is the zero matrix.)

But this derivation may seem a bit mysterious. So consider the example of $(1,2)$ and $(2,-1)$, which are orthogonal (perpendicular) since their dot product is 0. Write $v = \left[\begin{array}{c}1\\ 2\end{array}\right]$ and $w = \left[\begin{array}{r}2\\ -1\end{array}\right]$. Then $v ^ t w = \left[\begin{array}{cc}1&2\end{array}\right]\left[\begin{array}{r}2\\ -1\end{array}\right] = 0$ (since we can identify $1 \times 1$ matrices with scalars). The problem says that $w v _ t$, which is $\left[\begin{array}{r}2\\ -1\end{array}\right] \left[\begin{array}{cc}1&2\end{array}\right] = \left[\begin{array}{rr}2&4\\ -1&-2\end{array}\right]$, is nilpotent. In fact, $\left[\begin{array}{rr}2&4\\ -1&-2\end{array}\right]\left[\begin{array}{rr}2&4\\ -1&-2\end{array}\right] = \left[\begin{array}{cc}0&0\\ 0&0\end{array}\right]$ so it is nilpotent with only the second power needed to get to the zero matrix.

The solution to the problem is that $(w v ^ t)^2 = w v ^ t w v ^ t = w (v ^ t w)v ^ t = w 0 v ^ t = 0 w v ^ t = 0$ times a $2 \times 2$ matrix = the $2 \times 2$ zero matrix.

For Problem V-9: We have $\left[\begin{array}{cc}a&b\\ c&d\end{array}\right]\left[\begin{array}{rr}d&-b... ...right] = \left[\begin{array}{cc}ad-bc&0\\ 0&ad-bc\end{array}\right] = \Delta I$, where $\Delta =$   det$A$.

For Problem V-10:

This problem incorrectly referred to a $3 \times 3$ problem. Let's just substitute two examples:

Problem. Using the formula, find the inverse of

(a) $\left[\begin{array}{rr}1&2\\ 3&4\end{array}\right]$; (b) $\left[\begin{array}{rr}1&-1\\ 1&1\end{array}\right]$.

Solutions.

(a) $\Delta = -2$ so the inverse is $\frac 1{-2} \left[\begin{array}{rr}4&-2\\ -3&1\end{array}\right] = \frac 12 \left[\begin{array}{rr}-4&2\\ 3&-1\end{array}\right]$, or $\left[\begin{array}{rr}-2&1\\ \frac 32&-\frac 12\end{array}\right]$.

(b) $\Delta = 2$ so the inverse is $\frac 12 \left[\begin{array}{rr}1&1\\ -1&1\end{array}\right]$.