Question

Consider the matrices

A = −2 7 1

3 4 1

8 1 5,

B =8 1 5

3 4 1

−2 7 1,

C = −2 7 1

3 4 1

2 −7 3

Find elementary matrices E1, E2 and E3 such that

(5.1) E1A = B,

(5.2) E1B = A,

(5.3) E2A = C,

(5.4) E3C = A.

Answer 1

$\textbf{Solution:}$

Let $E$ be the identity matrix:

$E=\left( \begin{array}{cccc} 1 & 0 & \ldots & 0\\ 0 & 1 & \ldots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \ldots & 1 \end{array} \right)=\left( \begin{array}{cccc} \ldots \\ E_{i} \\ \ldots \\ E_{j} \\ \ldots \end{array} \right)$.

Let $P_{ij}$ be the permutation matrix:

$P_{ij}=\left( \begin{array}{cccc} \ldots \\ E_{j} \\ \ldots \\ E_{i} \\ \ldots \end{array} \right)$.

Then the result of left multiplication of the matrix $P_{ij}$ by the matrix

$A=\left( \begin{array}{cccc} \ldots \\ A_{i} \\ \ldots \\ A_{j} \\ \ldots \end{array} \right)$

is the matrix obtained from the original matrix $A$ by permuting its $i$-th and $j$-th rows:

$P_{ij}A=\left( \begin{array}{cccc} \ldots \\ A_{j} \\ \ldots \\ A_{i} \\ \ldots \end{array} \right)$.

For example,

$P_{23}A=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}\begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}=\begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{31} & a_{32} & a_{33} \\ a_{21} & a_{22} & a_{23} \end{pmatrix}$.

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5.1) $E_{1}A=B$

$E_{1}\begin{pmatrix} -2\ & 7\ & 1 \\ 3 & 4\ & 1 \\ 8 &1\ & 5 \end{pmatrix}=\begin{pmatrix} 8 & 1\ & 5 \\ 3 & 4\ & 1 \\ -2\ & 7\ & 1 \end{pmatrix}$.

$P_{ij}=P_{13}\Rightarrow E_{1}=P_{13}=\begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}$.

$\textbf{Check:}$

$\begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}\begin{pmatrix} -2\ & 7\ & 1 \\ 3 & 4\ & 1 \\ 8 &1\ & 5 \end{pmatrix}=\begin{pmatrix} 8 & 1\ & 5 \\ 3 & 4\ & 1 \\ -2\ & 7\ & 1 \end{pmatrix}$.

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5.2) $E_{1}B=A$

$E_{1}\begin{pmatrix} 8 & 1\ & 5 \\ 3 & 4\ & 1 \\ -2\ & 7\ & 1 \end{pmatrix}=\begin{pmatrix} -2\ & 7\ & 1 \\ 3 & 4\ & 1 \\ 8 &1\ & 5 \end{pmatrix}$.

$P_{ij}=P_{13}\Rightarrow E_{1}=P_{13}=\begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}$.

$\textbf{Check:}$

$\begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}\begin{pmatrix} 8 & 1\ & 5 \\ 3 & 4\ & 1 \\ -2\ & 7\ & 1 \end{pmatrix}=\begin{pmatrix} -2\ & 7\ & 1 \\ 3 & 4\ & 1 \\ 8 &1\ & 5 \end{pmatrix}$.

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Let $A=\left( \begin{array}{cccc} \ldots \\ A_{i} \\ \ldots \\ A_{j} \\ \ldots \end{array} \right), \,\widetilde{A}=\left( \begin{array}{cccc} \ldots \\ A_{i}+\lambda A_{j} \\ \ldots \\ A_{j} \\ \ldots \end{array} \right)$, then

$\widetilde{A}=N_{ij}(\lambda)\left( \begin{array}{cccc} \ldots \\ A_{i} \\ \ldots \\ A_{j} \\ \ldots \end{array} \right)=\left( \begin{array}{cccc} 1 && 0 & \ldots & 0 & \ldots & 0 &\ldots & 0\\ 0 && 1 & \ldots & 0 & \ldots & 0 &\ldots & 0\\ \vdots && \vdots & \ddots & \vdots & \ddots & \vdots & \ddots & \vdots\\ 0 && 0 & \ldots & 1 & \ldots & \lambda & \ldots & 0\\ \vdots && \vdots & \ddots & \vdots & \ddots & \vdots & \ddots & \vdots\\ 0 && 0 & \ldots & 0 & \ldots & 0 & \ldots & 1 \end{array} \right)\left( \begin{array}{cccc} \ldots \\ A_{i} \\ \ldots \\ A_{j} \\ \ldots \end{array} \right)=\left( \begin{array}{cccc} \ldots \\ A_{i}+\lambda A_{j} \\ \ldots \\ A_{j} \\ \ldots \end{array} \right).$ --------------------------------------------------------------------------------------------------------------------

5.3) $E_{2}A=C$

$E_{2}\begin{pmatrix} -2\ & 7\ & 1 \\ 3 & 4\ & 1 \\ 8 &1\ & 5 \end{pmatrix}=\begin{pmatrix} -2\ & 7 & 1 \\ 3 & 4 & 1 \\ 2 &-7\ & 3 \end{pmatrix}$.

1-st occasion:

$\begin{pmatrix} 2 & -7 & 3 \end{pmatrix}=\begin{pmatrix} 8 & 1 & 5 \end{pmatrix}+\lambda\begin{pmatrix} -2 & 7 & 1 \end{pmatrix}$.

$\begin{cases} 2=8-2\lambda, \\ -7=1+7\lambda, \\ 3=5+\lambda. \end{cases}$

$\lambda\in\varnothing$.

2-nd occasion:

$\begin{pmatrix} 2 & -7 & 3 \end{pmatrix}=\begin{pmatrix} 8 & 1 & 5 \end{pmatrix}+\lambda\begin{pmatrix} 3 & 4 & 1 \end{pmatrix}$.

$\begin{cases} 2=8+3\lambda, \\ -7=1+4\lambda, \\ 3=5+\lambda. \end{cases}$

$\lambda=-2$.

$i=3, j=2, \lambda=-2\Rightarrow N_{ij}(\lambda)=N_{32}(-2).$

$E_{2}=N_{32}(-2)=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -2 & 1 \end{pmatrix}$.

$\textbf{Check:}$

$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -2 & 1 \end{pmatrix}\begin{pmatrix} -2\ & 7\ & 1 \\ 3 & 4\ & 1 \\ 8 &1\ & 5 \end{pmatrix}=\begin{pmatrix} -2\ & 7 & 1 \\ 3 & 4 & 1 \\ 2 &-7\ & 3 \end{pmatrix}$.

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5.4) $E_{3}C=A$

$E_{3}\begin{pmatrix} -2 & 7 & 1 \\ 3 & 4 & 1\\ 2 & -7 & 3 \end{pmatrix}=\begin{pmatrix} -2 & 7 & 1 \\ 3 & 4 & 1 \\ 8 & 1 & 5 \end{pmatrix}$.

1-st occasion:

$\begin{pmatrix} 8 & 1 & 5 \end{pmatrix}=\begin{pmatrix} 2 & -7 & 3 \end{pmatrix}+\lambda\begin{pmatrix} -2 & 7 & 1 \end{pmatrix}$.

$\begin{cases} 8=2-2\lambda, \\ 1=-7+7\lambda, \\ 5=3+\lambda. \end{cases}$

$\lambda\in\varnothing$.

2-nd occasion:

$\begin{pmatrix} 8 & 1 & 5 \end{pmatrix}=\begin{pmatrix} 2 & -7 & 3 \end{pmatrix}+\lambda\begin{pmatrix} 3 & 4 & 1 \end{pmatrix}$.

$\begin{cases} 8=2+3\lambda, \\ 1=-7+4\lambda, \\ 5=3+\lambda. \end{cases}$

$\lambda=2$.

$i=3, j=2, \lambda=2\Rightarrow N_{ij}(\lambda)=N_{32}(2).$

$E_{3}=N_{32}(2)=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end{pmatrix}$.

$\textbf{Check:}$

$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end{pmatrix}\begin{pmatrix} -2 & 7 & 1 \\ 3 & 4 & 1\\ 2 & -7 & 3 \end{pmatrix}=\begin{pmatrix} -2 & 7 & 1 \\ 3 & 4 & 1 \\ 8 & 1 & 5 \end{pmatrix}$.

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$\textbf{Answer:}$

$\boxed{E_{1}=\begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}, \, E_{2}=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -2 & 1 \end{pmatrix}, \, E_{3}=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end{pmatrix}}$