Problems 1-15
1 (a) When matrix
$$ \mathbf{E}_{21} = \begin{bmatrix} \phantom{-}1 & 0 & 0 \\ -5 & 1 & 0 \\ \phantom{-}0 & 0 & 1 \end{bmatrix} $$the product
so the first row of $\mathbf{A}$ gets subtracted from the second five times.
(b) For matrix $\mathbf{E}_{32}$ that subtracts $-7$, that is, adds $7$ times the second row from the third row the logic is similar, i.e., when
$$ \mathbf{E}_{32} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 7 & 1 \end{bmatrix} $$the product
(c) Matrix
$$ \mathbf{P}_{12} = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$and
$$ \mathbf{P}_{23} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} $$the multiplication
$$ \begin{array}{r c l} \mathbf{P}_{21} \mathbf{A} & = & \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \textcolor{red}{a_{11}} & \textcolor{red}{a_{12}} & \textcolor{red}{a_{13}} \\ \textcolor{blue}{a_{21}} & \textcolor{blue}{a_{22}} & \textcolor{blue}{a_{23}} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \\[0.5em] & = & \begin{bmatrix} 0\textcolor{red}{a_{11}} + 1\textcolor{blue}{a_{21}} + 0a_{31} & 0\textcolor{red}{a_{12}} + 1\textcolor{blue}{a_{22}} + 0a_{32} & 0\textcolor{red}{a_{13}} + 1\textcolor{blue}{a_{23}} + 0a_{33}\\ 1\textcolor{red}{a_{11}} + 0\textcolor{blue}{a_{21}} + 0a_{31} & 1\textcolor{red}{a_{12}} + 0\textcolor{blue}{a_{22}} + 0a_{32} & 1\textcolor{red}{a_{13}} + 0\textcolor{blue}{a_{23}} + 0a_{33}\\ 0\textcolor{red}{a_{11}} + 0\textcolor{blue}{a_{21}} + 1a_{31} & 0\textcolor{red}{a_{12}} + 0\textcolor{blue}{a_{22}} + 1a_{32} & 0\textcolor{red}{a_{13}} + 0\textcolor{blue}{a_{23}} + 1a_{33} \end{bmatrix} \\[0.5em] & = & \begin{bmatrix} \textcolor{blue}{a_{21}} & \textcolor{blue}{a_{22}} & \textcolor{blue}{a_{23}} \\ \textcolor{red}{a_{11}} & \textcolor{red}{a_{12}} & \textcolor{red}{a_{13}} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \end{array} $$and so multiplying the product further with $\mathbf{P}_{23}$ means that
$$ \phantom{.} \mathbf{P}_{23} \mathbf{P}_{21} \mathbf{A} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} \textcolor{blue}{a_{21}} & \textcolor{blue}{a_{22}} & \textcolor{blue}{a_{23}} \\ \textcolor{red}{a_{11}} & \textcolor{red}{a_{12}} & \textcolor{red}{a_{13}} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} = \begin{bmatrix} \textcolor{blue}{a_{21}} & \textcolor{blue}{a_{22}} & \textcolor{blue}{a_{23}} \\ a_{31} & a_{32} & a_{33} \\ \textcolor{red}{a_{11}} & \textcolor{red}{a_{12}} & \textcolor{red}{a_{13}} \end{bmatrix} . $$2 The matrix multiplication
$$ \begin{array}{r c l} \mathbf{E}_{32} \mathbf{E}_{21} \mathbf{b} & = & \mathbf{E}_{32} \left( \begin{bmatrix} \phantom{-}1 & 0 & 0 \\ -5 & 1 & 0 \\ \phantom{-}0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix} \right) \\[0.5em] & = & \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 7 & 1 \\ \end{bmatrix} \begin{bmatrix} b_1 \\ b_2 - 5b_1 \\ b_3 \end{bmatrix} \\[0.5em] & = & \begin{bmatrix} b_1 \\ b_2 - 5b_1 \\ b_3 + 7(b_2 - 5b_1) \end{bmatrix} \end{array} $$while the matrix multiplication
$$ \begin{array}{r c l} \mathbf{E}_{21} \mathbf{E}_{32} \mathbf{b} & = & \mathbf{E}_{21} \left( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 7 & 1 \end{bmatrix} \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix} \right) \\[0.5em] & = & \begin{bmatrix} \phantom{-}1 & 0 & 0 \\ -5 & 1 & 0 \\ \phantom{-}0 & 0 & 1 \end{bmatrix} \begin{bmatrix} b_1 \\ b_2 \\ b_3 + 7b_2 \end{bmatrix} \\[0.5em] & = & \begin{bmatrix} b_1 \\ b_2 - 5b_1 \\ b_3 + 7_b2 \end{bmatrix}, \end{array} $$so the associativity does now hold between $\mathbf{E}_{21}$ and $\mathbf{E}_{32}$. When $\mathbf{E}_{32}$ comes first, row three feels no eddect from row two.
3 Solving the $\mathbf{A}$ with the elimination method reveals the matrices. The steps are as follows, the first row gets subtracts from the second two times, so
$$ \mathbf{E}_{21} = \begin{bmatrix} \phantom{-}1 & 0 & 0 \\ -4 & 1 & 0 \\ \phantom{-}0 & 0 & 1 \end{bmatrix}. $$The matrix multiplication
Adding first row to third twice brings the $-2$ to zero. This equates to matrix
$$ \mathbf{E}_{31} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & 0 & 1 \end{bmatrix} $$so
Last matrix should be of the form that subtracts row two from row three twice. A matrix
$$ \mathbf{E}_{32} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -2 & 1 \end{bmatrix} $$is such a matrix because
By itself,
$$ \mathbf{E}_{21} \mathbf{E}_{31} \mathbf{E}_32 = \begin{bmatrix} 1 & 0 & 0 \\ -4 & 1 & 0 \\ 10 & -2 & 1 \end{bmatrix}. $$4 Augmenting matrix $\mathbf{A}$ of 3, call it $\mathbf{A}^\prime$, with $\mathbf{b}$ $=$ $\begin{bmatrix}1 & 0 & 0\end{bmatrix}^\top$ leads to augmented matrix
$$ \mathbf{A}^\prime = \left[ \begin{array}{c c c | c} \phantom{-}1 & 1 & 0 & 1 \\ \phantom{-}4 & 6 & 1 & 0 \\ -2 & 2 & 0 & 0 \\ \end{array} \right] $$and doing the multiplication $\mathbf{E}_{21} \mathbf{E}_{31} \mathbf{E}_{32} \mathbf{A}$ leads to
$$ \begin{bmatrix} 1 & 1 & 0 & 1 \\ 0 & 2 & 1 & -4 \\ 0 & 0 & -2 & 10 \end{bmatrix}. $$5
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6 A matrix whose every column is multiple of $\begin{bmatrix} 1 & 1 & 1 \end{bmatrix}^\top$ is a matrix like
$$ \mathbf{A} = \begin{bmatrix} 1 & 3 & 5 \\ 1 & 3 & 5 \\ 1 & 3 & 5 \end{bmatrix}. $$Applying elimination to this matrix leads to matrix
$$ \mathbf{A} = \begin{bmatrix} 1 & 3 & 5 \\ 0 & 0 & 0 \\ 1 & 3 & 5 \end{bmatrix}, $$and then with one row exchange to matrix
$$ \mathbf{A} = \begin{bmatrix} 1 & 3 & 5 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, $$so it is possible to only one pivot, though technically it is found more then once.
7 Matrix that subracts $7$ times row $1$ from row $3$ is matrix
$$ \mathbf{E} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -7 & 0 & 1 \\ \end{bmatrix}. $$(a) To invert the operation that $\mathbf{E}$ does when it’s left multplying some matrix, a matrix that adds $7$ times the row $3$ to row $1$ is needed. Such matrix is
$$ \mathbf{E}^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 7 & 0 & 1 \end{bmatrix}. $$(b) The matrix product
$$ \mathbf{E}^{-1} \mathbf{E} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -7 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 7 & 0 & 1 \end{bmatrix}, $$so the product’s first column is
$$ \begin{bmatrix} 1(1) & + & 0(0) & + & 0(0) \\ 1(0) & + & 0(1) & + & 0(0) \\ 1(-7) & + & 0(0) & + & 7(1) \\ \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} $$and the second and third are obviously
$$ \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} $$and
$$ \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}, $$respctively, so
$$ \mathbf{E}^{-1} \mathbf{E} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \mathbf{I}. $$(c) If the reverse step is applied first, then
$$ \mathbf{E} \mathbf{E}^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 7 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -7 & 0 & 1 \\ \end{bmatrix} $$the first column of the product is still $\begin{bmatrix} 1 & 0 & 0\end{bmatrix}^\top$, and the second and third row are as above, so the product $\mathbf{E^{-1}} \mathbf{E}$ $=$ $\mathbf{I}$.
8
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9 (a) The
$$ \mathbf{E}_{21} = \begin{bmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$and
$$ \mathbf{P}_{23} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} $$so
$$ \mathbf{P}_{23} \mathbf{E}_{21} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ -1 & 1 & 0 \end{bmatrix}. $$(b) The matrix $\mathbf{P}_{23}$ is as above, but
$$ \mathbf{E}_{31} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}, $$so
$$ \mathbf{E}_{31} \mathbf{P}_{23} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ -1 & 1 & 0 \end{bmatrix}. $$10
(a) Exchange matrix
$$ \mathbf{E}_{13} = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$will add row $3$ to row $1$. For example, let matrix $\mathbf{A}$ be a $3 \times 3$ matrix. Then
$$ \begin{array}{r c l} \mathbf{E}_{13} \mathbf{A} & = & \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \\[0.5em] & = & \begin{bmatrix} a_{11} + a_{31} & a_{12} + a_{32} & a_{13} + a_{33} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}. \end{array} $$(b) A matrix that adds row $1$ to row $3$ and at the same time, vice versa, is matrix
$$ \mathbf{E} = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}. $$because
$$ \begin{array}{r c l} \mathbf{E} \mathbf{A} & = & \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \\[0.5em] & = & \begin{bmatrix} a_{11} + a_{31} & a_{12} + a_{32} & a_{13} + a_{33} \\ a_{21} & a_{22} & a_{23} \\ a_{31} + a_{11} & a_{32} + a_{12} & a_{33} + a_{13} \end{bmatrix}. \end{array} $$(c) A matrix that adds row $1$ to row $3$ and then vice versa, is a matrix
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11 Matrix that has $a_{11}$ $=$ $a_{22}$ $=$ $a_{33}$ $=$ $1$ and that produces two negative pivots after elimination when no row exchange are not allowed is
$$ \begin{bmatrix} 1 & a & b \\ m_{21} & 1 & c \\ m_{31} & d & 1 \\ \end{bmatrix} $$where $m_{21}$ $<$ $0$ and $m_{31}$ $\leq$ $m_{21}$, because subtracting first row from the second leads to pivot $m_{21}$ $-$ $\frac{m_{21}}{1}$ $=$ $m_{21}$ $-$ $m_{21}$ $<$ $0$. The same goes for the row of $m_{31}$ no matter if the first or the second row is subtracted.
12 The first product is
$$ \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} = \begin{bmatrix} 7 & 8 & 9 \\ 4 & 5 & 6 \\ 1 & 2 & 3 \end{bmatrix} $$and the second
$$ \begin{bmatrix} 7 & 8 & 9 \\ 4 & 5 & 6 \\ 1 & 2 & 3 \end{bmatrix} \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 9 & 8 & 7 \\ 6 & 5 & 4 \\ 3 & 2 & 1 \end{bmatrix}. $$The multiplication resulted in a permutation of the middle matrix where first the top and bottom rows swapped, and then left and right columns.
The second product is
$$ \begin{bmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 1 & 3 & 1 \\ 1 & 4 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & -2 \\ 0 & 2 & -3 \end{bmatrix}. $$13 If all of the elements of third column of $\mathbf{B}$ are all zeros, then multiplying $\mathbf{B}$ with $\mathbf{E}$ yields the product $\mathbf{EB}$ which third column is always zeros because linear combination between the column and matrix $\mathbf{E}$ always leads to $0$, or in other words
$$ b_{31} \mathbf{e}_1 + b_{32} \mathbf{e}_2 + b_{33} \mathbf{e}_3 \cdots + b_{n3} \mathbf{e}_n = \begin{bmatrix} 0 \\ 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix} $$where $b_{i3}$ is the $i$th zero of $B$ and $\mathbf{e}_i$, is $i$th column of matrix $\mathbf{E}$.
On the other hand, if the third row of $\mathbf{B}$ are all zero, the coefficients $b_{i3}$, might not all be zero, so the dor product leads vector that might have a non-zero component.
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