$\begingroup$

Let matrix

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

and

$$B = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 7 & 1 \\ 0 & 4 & 2 \end{pmatrix} $$

Find an elementary matrix $E$ such that $EA= B$.

I try with many

$$E = \begin{pmatrix} -1 & -2 & 0 \\ 1 & 2 & 0 \\ 3 & -1 & 1 \end{pmatrix} $$

but not correct

$\endgroup$ 4

1 Answer

$\begingroup$

An elementary matrix is one which differs from the identity matrix by one elementary row operation.

Note that $B$ is the matrix $A$ with three times the first row added to the second. So if we take the matrix

$$E=\begin{pmatrix}1 & 0 & 0 \\ 3 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$$

and now consider

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

$E$ was the identity matrix with three times the first row added to the second. This is easy to figure out because the matrix $B$ is linear combinations of the rows of $A$ and linear combinations of the columns of $E$.

$\endgroup$

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy