Problems 16-23
16 (a) If $X$ is twice as old as $Y$ and their ages add to $33$, then
$$ X + Y = 2Y + Y = 33. $$The equation can be expressed as linear system
$$ \mathbf{Ab} = \begin{bmatrix} 2 & 1 \\ 0 & 0 \\ \end{bmatrix} \begin{bmatrix} Y \\ Y \end{bmatrix} = \begin{bmatrix} 3Y \\ 0 \end{bmatrix} = \begin{bmatrix} 33 \\ 0 \end{bmatrix} = \mathbf{b}. $$The first component of $\mathbf{Ab}$ is $3Y$ $=$ $33$, so $Y$ $=$ $11$. Substituting this back to the original equation means that
$$ X + 11 = 33, $$so $X$ $=$ $22$.
(b) When $(x,y)$ $=$ $(2,5)$ or $(3,7)$ and the slope-intersect form of a line is the well known $y$ $=$ $mx$ $+$ $c$, a linear equation represented as
$$ \left[ \begin{array}{c c | c} 2 & 5 & m \\ 3 & 7 & c \\ \end{array} \right] \Longleftrightarrow \left\{ \begin{array}{l} 5 = 2m + c \\ 7 = 3m + c. \end{array} \right. $$Subtracting first row $\frac{3}{2}$ times from the second leads to
$$ \left\{ \begin{array}{l l l} 5 & = & 2m + c \\ 7 - \frac{3}{2}(5) & = & 3m - \frac{3}{2}(3m) + c - \frac{3}{2}(c) \\ \end{array} \right. \ = \ \left\{ \begin{array}{l l l} 5 & = & 2m + c \\ - \frac{1}{2} & = & -\frac{1}{2}(c) \\ \end{array} \right. \quad , $$so $c$ $=$ $1$ and substituting it back to, say, $5$ $=$ $2m$ $+$ $c$, shows that $2m$ $+$ $1$ $=$ $5$, so $m$ $=$ $2$.
17 When parabola $y$ $=$ $a$ $+$ $bx$ $+$ $cx^2$ goes through points $(1,4)$, $(2,8)$, $(3,14)$ and a given unknown solution is $(a,b,c)$, the problem can be modeled as a system of linear functions
and then just applying elimination operations
$$ \begin{array}{c c} \xrightarrow{\mathbf{E}_{31} \cdot} & \left[ \begin{array}{c c c | c} 1 & 3 & 9 & 14 \\ 1 & 2 & 4 & 8 \\ 1 & 1 & 1 & 4 \\ \end{array} \right] \\[1em] \xrightarrow{2_3 - R_1} & \left[ \begin{array}{c c c | c} 1 & 3 & 9 & 14 \\ 0 & -1 & -5 & -6 \\ 1 & 1 & 1 & 4 \\ \end{array} \right] \\[1em] \xrightarrow{R_3 - R_1} & \left[ \begin{array}{c c c | c} 1 & 3 & 9 & 14 \\ 0 & -1 & -5 & -6 \\ 0 & -2 & -8 & -10 \\ \end{array} \right] \\[1em] \xrightarrow{R_3 + 2R_2} & \ \phantom{.} \left[ \begin{array}{c c c | c} 1 & 3 & 9 & 14 \\ 0 & -1 & -5 & -6 \\ 0 & 0 & 2 & 2 \\ \end{array} \right] \ . \end{array} $$Third row shows that $2c$ $=$ $2$, so $c$ $=$ $1$. Substituting $c$ back to second row, leads to $-b$ $-$ $5c$ $=$ $-$ $b$ $-$ $5$ = $-6$, so multiplying with $-1$ and subtracting $5$, shows that $b$ $=$ $1$. Substituting $b$ and $c$ to first row shows that, $a$ $+$ $3b$ $+$ $9c$ $=$ $a$ $+$ $3$ $+$ $9$ $=$ $a$ $+$ $12$ $=$ $14$, so $a$ $=$ $2$.
Substituting $a$ $=$ $2$, $b$ $=$ $1$ and $c$ $=$ $1$ to the first matrix equation and evaluating the product shows that
$$ \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 3 & 9 \\ \end{bmatrix} \begin{bmatrix} 2 \\ 1 \\ 1 \\ \end{bmatrix} = \begin{bmatrix} 2(1) + 1(1) + 1(1) \\ 2(1) + 1(2) + 1(4) \\ 2(1) + 1(3) + 1(9) \\ \end{bmatrix} = \begin{bmatrix} 4 \\ 8 \\ 14 \\ \end{bmatrix} $$as was to be expected.
18 When given matrices are
$$ \mathbf{E} = \begin{bmatrix} 1 & 0 & 0 \\ a & 1 & 0 \\ b & 0 & 1 \\ \end{bmatrix} \quad , \quad \mathbf{F} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & c & 1 \\ \end{bmatrix} , $$then
$$ \begin{array}{c c c c} \mathbf{EF} & = & \begin{bmatrix} 1(1) + 0(0) + 0(0) & 0(1) + 1(0) + c(0) & 0(1) + 0(0) + 1(0) \\ 1(a) + 0(1) + 0(0) & 0(a) + 1(1) + c(0) & 0(a) + 0(1) + 1(0) \\ 1(b) + 0(0) + 0(1) & 0(b) + 1(0) + c(1) & 0(b) + 0(0) + 1(1) \\ \end{bmatrix} & \\[1em] & = & \begin{bmatrix} 1 & 0 & 0 \\ a & 1 & 0 \\ b & c & 1 \\ \end{bmatrix} & \\[1em] & = & \begin{bmatrix} 1(1) + a(0) + b(0) & 0(1) + 1(0) + 0(0) & 0(1) + 0(0) + 1(0) \\ 1(0) + a(1) + b(0) & 0(0) + 1(1) + 0(0) & 0(0) + 0(1) + 1(0) \\ 1(0) + a(c) + b(1) & 0(0) + 1(c) + 0(1) & 0(0) + 0(c) + 1(1) \\ \end{bmatrix} & = & \mathbf{FE}. \end{array} $$Also
$$ \begin{array}{r c l} \mathbf{E}^2 & = & \begin{bmatrix} 1 & 0 & 0 \\ a & 1 & 0 \\ b & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ a & 1 & 0 \\ b & 0 & 1 \\ \end{bmatrix} \\[1em] & = & \begin{bmatrix} 1(1) + a(0) + b(0) & 0(1) + 1(0) + 0(0) & 0(1) + 0(0) + 1(0) \\ 1(a) + a(1) + b(0) & 0(a) + 1(1) + 0(0) & 0(a) + 0(1) + 1(0) \\ 1(b) + a(0) + b(1) & 0(b) + 1(0) + 0(1) & 0(b) + 0(0) + 1(1) \\ \end{bmatrix} \\[1em] & = & \begin{bmatrix} 1 & 0 & 0 \\ a + a & 1 & 0 \\ b + b & 0 & 1 \\ \end{bmatrix} \end{array} $$and
$$ \begin{array}{r c l} \mathbf{F}^3 & = & \mathbf{FFF} \\[1em] & = & \mathbf{F(FF)} \\[1em] & = & \mathbf{F} \left( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & c & 1 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & c & 1 \\ \end{bmatrix} \right) \\[1em] & = & \mathbf{F} \left( \begin{bmatrix} 1(1) + 0(0) + 0(0) & 0(1) + 1(0) + c(0) & 0(1) + 0(0) + 1(0) \\ 1(0) + 0(1) + 0(0) & 0(0) + 1(1) + c(0) & 0(0) + 0(1) + 1(0) \\ 1(0) + 0(c) + 0(1) & 0(0) + 1(c) + c(1) & 0(0) + 0(c) + 1(1) \\ \end{bmatrix} \right) \\[1em] & = & \mathbf{F} \left( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & c + c & 1 \\ \end{bmatrix} \right) = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & c & 1 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & c + c & 1 \\ \end{bmatrix} \\[1em] & = & \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & c + c + c & 1 \\ \end{bmatrix}. \end{array} $$So then it’s possible to guess that
$$ \mathbf{F}^{100} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & c^{100} & 1 \end{bmatrix}. $$