Question #196702

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.




1
Expert's answer
2021-05-24T13:47:51-0400

Solution:\textbf{Solution:}

Let EE be the identity matrix:

E=(100010001)=(EiEj)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 PijP_{ij} be the permutation matrix:

Pij=(EjEi)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 PijP_{ij} by the matrix

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

is the matrix obtained from the original matrix AA by permuting its ii-th and jj-th rows:

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

For example,

P23A=(100001010)(a11a12a13a21a22a23a31a32a33)=(a11a12a13a31a32a33a21a22a23)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) E1A=BE_{1}A=B

E1(2 7 134 181 5)=(81 534 12 7 1)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}.


Pij=P13E1=P13=(001010100)P_{ij}=P_{13}\Rightarrow E_{1}=P_{13}=\begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}.

Check:\textbf{Check:}

(001010100)(2 7 134 181 5)=(81 534 12 7 1)\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) E1B=AE_{1}B=A

E1(81 534 12 7 1)=(2 7 134 181 5)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}.


Pij=P13E1=P13=(001010100)P_{ij}=P_{13}\Rightarrow E_{1}=P_{13}=\begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}.

Check:\textbf{Check:}

(001010100)(81 534 12 7 1)=(2 7 134 181 5)\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=(AiAj),A~=(Ai+λAjAj)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


A~=Nij(λ)(AiAj)=(1000001000001λ000001)(AiAj)=(Ai+λAjAj).\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) E2A=CE_{2}A=C

E2(2 7 134 181 5)=(2 7134127 3)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:

(273)=(815)+λ(271)\begin{pmatrix} 2 & -7 & 3 \end{pmatrix}=\begin{pmatrix} 8 & 1 & 5 \end{pmatrix}+\lambda\begin{pmatrix} -2 & 7 & 1 \end{pmatrix}.

{2=82λ,7=1+7λ,3=5+λ.\begin{cases} 2=8-2\lambda, \\ -7=1+7\lambda, \\ 3=5+\lambda. \end{cases}

λ\lambda\in\varnothing.

2-nd occasion:

(273)=(815)+λ(341)\begin{pmatrix} 2 & -7 & 3 \end{pmatrix}=\begin{pmatrix} 8 & 1 & 5 \end{pmatrix}+\lambda\begin{pmatrix} 3 & 4 & 1 \end{pmatrix}.

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

λ=2\lambda=-2.

i=3,j=2,λ=2Nij(λ)=N32(2).i=3, j=2, \lambda=-2\Rightarrow N_{ij}(\lambda)=N_{32}(-2).

E2=N32(2)=(100010021)E_{2}=N_{32}(-2)=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -2 & 1 \end{pmatrix}.

Check:\textbf{Check:}

(100010021)(2 7 134 181 5)=(2 7134127 3)\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) E3C=AE_{3}C=A

E3(271341273)=(271341815)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:

(815)=(273)+λ(271)\begin{pmatrix} 8 & 1 & 5 \end{pmatrix}=\begin{pmatrix} 2 & -7 & 3 \end{pmatrix}+\lambda\begin{pmatrix} -2 & 7 & 1 \end{pmatrix}.

{8=22λ,1=7+7λ,5=3+λ.\begin{cases} 8=2-2\lambda, \\ 1=-7+7\lambda, \\ 5=3+\lambda. \end{cases}

λ\lambda\in\varnothing.

2-nd occasion:

(815)=(273)+λ(341)\begin{pmatrix} 8 & 1 & 5 \end{pmatrix}=\begin{pmatrix} 2 & -7 & 3 \end{pmatrix}+\lambda\begin{pmatrix} 3 & 4 & 1 \end{pmatrix}.

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

λ=2\lambda=2.

i=3,j=2,λ=2Nij(λ)=N32(2).i=3, j=2, \lambda=2\Rightarrow N_{ij}(\lambda)=N_{32}(2).

E3=N32(2)=(100010021)E_{3}=N_{32}(2)=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end{pmatrix}.

Check:\textbf{Check:}

(100010021)(271341273)=(271341815)\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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Answer:\textbf{Answer:}

E1=(001010100),E2=(100010021),E3=(100010021)\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}}


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