Problems 26-35
26
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27
The row picture will show $2$ planes in $3$-dimensional space. The column picture is in $2$-dimensional space. The solutions normally lie on a line in $\mathbf{3}$-D space.
28
For four linear equations in two unknowns $x$ and $y$, the row picture shows four in a lines in 2-D plane. The column picture is in $2-dimensional space$. The equations have no solution unless the vector on the right side is a combination of of the vectors on the left side.
29
If
$$ \phantom{.} \ \mathbf{u}_1 = \begin{bmatrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0.8 \\ 0.2 \end{bmatrix} \ , $$then
$$ \mathbf{u}_2 = \mathbf{Au}_1 = \begin{bmatrix} - \\ - \end{bmatrix} $$29 – 29
Solved in a blog post.
31
Example of such matrix is
Step for finding such a matrix is starting with the given first row
$$ \begin{array}{c | c} \begin{bmatrix} 8 & 3 & 4 \\ & & \\ & & \end{bmatrix} & \begin{array}{l} \text{Remaining:} \\ \{ 1, 2, 5, 6, 7, 9 \} \ . \end{array} \end{array} $$Then taking $5$ and $2$ from the “$\text{Remaining}$” numbers and adding the to diagonal leads to $8 + 5 + 2 = 15$, so the matrix can be updated into form
$$ \begin{array}{c | c} \begin{bmatrix} 8 & 3 & 4 \\ & \textcolor{#FF6868}{5} & \\ & & \textcolor{#FF6868}{2} \end{bmatrix} & \begin{array}{l} \text{Remaining:} \\ \{ 1, 6, 7, 9 \} \ . \end{array} \end{array} $$Additionally from the anti-diagonal, it’s possible to see a simple eqution $4 + 5 + x = 15$, and by solving for $x$, we see that $x = 6$, so matrix can be updated
$$ \begin{array}{c | c} \begin{bmatrix} 8 & 3 & 4 \\ & 5 & \\ \textcolor{#FF6868}{6} & & 2 \end{bmatrix} & \begin{array}{l} \text{Remaining:} \\ \{ 1, 7, 9 \} \ . \end{array} \end{array} $$Now because
$$ \begin{array}{r c l} 15 & = & 8 + 3 + 4 \\ & = & 8 + 5 + 2 \\ & = & 4 + 5 + 6 \ , \end{array} $$everything looks okay. Then because $6 + 7 + 2 = 15$ and $3 + 5 + 7 = 15$, the matrix can be updated as
$$ \begin{array}{c | c} \begin{bmatrix} 8 & 3 & 4 \\ & 5 & \\ 6 & \textcolor{#FF6868}{7} & 2 \end{bmatrix} & \begin{array}{l} \text{Remaining:} \\ \{ 1, 9 \} \ . \end{array} \end{array} $$From the current matrix and the numbers “$\text{Remaining}$”, it should be relatively easy to see three new simple arithmetic equations
$$ \begin{array}{r c l} 8 + x + 6 = 15 \ , \\[0.5em] x + 5 + y = 15 \ , \\[0.5em] 4 + y + 2 = 15 \ , \end{array} $$and by letting $x = 1$, then the equations are
$$ \begin{array}{r c l} 8 + 1 + 6 = 15 \ , & \textcolor{green}{\checkmark} \\[0.5em] 1 + 5 + y = 15 \ , & \\[0.5em] 4 + y + 2 = 15 \ . & \end{array} $$and remainig there’s only $9$, so letting $y$ to have that value, the set of equations become
$$ \begin{array}{r c l} 8 + 1 + 6 = 15 \ , & \textcolor{green}{\checkmark} \\[0.5em] 1 + 5 + 9 = 15 \ , & \textcolor{green}{\checkmark} \\[0.5em] 4 + 0 + 2 = 15 \ , & \textcolor{green}{\checkmark} \end{array} $$so everything checks. Now updating the matrix with this info, the matrix becomes
$$ \begin{array}{c | c} \begin{bmatrix} 8 & 3 & 4 \\ \textcolor{#FF6868}{1} & 5 & \textcolor{#FF6868}{9}\\ 6 & 7 & 2 \end{bmatrix} & \begin{array}{l} \text{Remaining:} \\ \varnothing \ . \end{array} \end{array} $$Now chcking the values of all rows, columns, and the diagonal and the anti-diagonal add up to 15, a table can be written such that
from where it can be seen that the matrix the exercise is asking for is
$$ \phantom{.} \ \begin{bmatrix} 8 & 3 & 4 \\ 1 & 5 & 9 \\ 6 & 7 & 2 \end{bmatrix} = \mathbf{M}_3 \ . $$The
32
If the third column is $\mathbf{u} + \mathbf{v}$. Column picture has $\mathbf{b}$ outside the plane of $\mathbf{u}$, $\mathbf{v}$, and $\mathbf{w}$. Rowpicture is an intersection of two planes parallel to a third plane.
33
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34
The matrix equation form is
$$ \phantom{.} \ \mathbf{Ax} = \begin{bmatrix} \phantom{-}2 & -1 & \phantom{-}0 & \phantom{-}0 \\ -1 & \phantom{-}2 & -1 & \phantom{-}0 \\ \phantom{-}0 & -1 & \phantom{-}2 & -1 \\ \phantom{-}0 & \phantom{-}0 & -1 & \phantom{-}2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix} \ . $$35
Such a Sudoku-matrixj is
$$ \phantom{.} \ \mathbf{S} = \left[ \ \begin{array}{ | c c c | c c c | c c c | } \hline 1 & 5 & 2 & 4 & 8 & 9 & 3 & 7 & 6 \\ 7 & 3 & 9 & 2 & 5 & 6 & 8 & 4 & 1 \\ 4 & 6 & 8 & 3 & 7 & 1 & 2 & 9 & 5 \\ \hline 3 & 8 & 7 & 1 & 2 & 4 & 6 & 5 & 9 \\ 5 & 9 & 1 & 7 & 6 & 3 & 4 & 2 & 8 \\ 2 & 4 & 6 & 8 & 9 & 5 & 7 & 1 & 3 \\ \hline 9 & 1 & 4 & 6 & 3 & 7 & 5 & 8 & 2 \\ 6 & 2 & 5 & 9 & 4 & 8 & 1 & 3 & 7 \\ 8 & 7 & 3 & 5 & 1 & 2 & 9 & 6 & 4 \\ \hline \end{array} \ \right] \ . $$The matrix multiplication
$$ \mathbf{Sx} $$when $\mathbf{x} = [ \ 1_1, 1_2, \ldots, 1_9 \ ]^\top$ will produce a vector, where each $i$th component is sum of $i$th row of $S$, i.e.
$$ \phantom{.} \ 1 + 2 + \cdots + 9 = 45 \ . $$The following is lazy answer, but changing the first and the second row will not break the requirement for $\mathbf{S}$. So if the row exchanges are done within the boundaries of a horizontal blocks, then it is possible to have multiple Sudoku-matrices, but there might be other ways too.