Answer to Question #203612 in Linear Algebra for khosa

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, (1) (5.2) E1B = A, (1) (5.3) E2A = C, (1) (5.4) E3C = A.

Answer 1

(5.1)

$$A=\begin{pmatrix} -2 & 7 & 1 \\ 3 & 4 & 1 \\ 8 & 1 & 5 \\ \end{pmatrix}, B=\begin{pmatrix} 8 &1 & 5 \\ 3 & 4 & 1 \\ -2 & 7 & 1 \\ \end{pmatrix}$$

A matrix $A$ switches all matrix elements on row $1$ with their counterparts on row $3.$

The corresponding elementary matrix is obtained by swapping row $1$  and row $3$  of the identity matrix.

$$E_1=\begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ \end{pmatrix}$$

$$E_1A=\begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ \end{pmatrix}\begin{pmatrix} -2 & 7 & 1 \\ 3 & 4 & 1 \\ 8 & 1 & 5 \\ \end{pmatrix}$$

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

$E_1=\begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ \end{pmatrix}$

(5.2)

A matrix $B$ switches all matrix elements on row $1$ with their counterparts on row $3.$

The corresponding elementary matrix is obtained by swapping row $1$ and row $3$ of the identity matrix.

$$E_1=\begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ \end{pmatrix}$$

$$E_1B=\begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ \end{pmatrix}\begin{pmatrix} 8 &1 & 5 \\ 3 & 4 & 1 \\ -2 & 7 & 1 \\ \end{pmatrix}$$

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

$E_1=\begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ \end{pmatrix}$

(5.3)

$$A=\begin{pmatrix} -2 & 7 & 1 \\ 3 & 4 & 1 \\ 8 & 1 & 5 \\ \end{pmatrix}, C=\begin{pmatrix} -2 & 7 & 1 \\ 3 & 4 & 1 \\ 2 & -7 & 3 \\ \end{pmatrix}$$

A matrix $C$ is the matrix produced from $A$ by adding $(-2)$ times row $2$ to row $3.$

The corresponding elementary matrix is the identity matrix but with $(-2)$ in the $(2,3)$

position.

$$E_2=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -2 & 1\\ \end{pmatrix}$$

$$E_2A=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -2 & 1 \\ \end{pmatrix}\begin{pmatrix} -2 & 7 & 1 \\ 3 & 4 & 1 \\ 8 & 1 & 5 \\ \end{pmatrix}$$

$$=\begin{pmatrix} -2 & 7 & 1 \\ 3 & 4 & 1 \\ 2 & -7 & 3 \\ \end{pmatrix}=C$$

$E_2=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -2 & 1\\ \end{pmatrix}$

(5.4)

A matrix $A$ is the matrix produced from $C$ by adding $(2)$ times row $2$ to row $3.$

The corresponding elementary matrix is the identity matrix but with $(2)$ in the $(2,3)$

position.

$$E_3=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & 1\\ \end{pmatrix}$$

$$E_3C=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \\ \end{pmatrix}\begin{pmatrix} -2 & 7 & 1 \\ 3 & 4 & 1 \\ 2 & -7 & 3 \\ \end{pmatrix}$$

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

$E_3=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & 1\\ \end{pmatrix}$