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I am given two matrices, and I have to find an elementary matrix $A$ such that $EA=B$.

$$E = \begin{bmatrix}2&4\\2&-6\end{bmatrix}$$

$$B = \begin{bmatrix}10&4\\-10&-6\end{bmatrix}$$

I tried "transposing" the equation, meaning $(EA)^T = B^T$. The equation given would then be $(A^T)(E^T) = B^T$.

I, however, can't manage to end up with the matrix $B$.

Théophile
  • 24,627

1 Answers1

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You don't need to go through transposing.
Just find the inverse matrix of $E$

Then $A = E^{-1} B$

How do you find the inverse of $E$? See e.g. here:

https://www.mathsisfun.com/algebra/matrix-inverse.html

See in particular the section 2x2 Matrix

It says that:

$$\begin{bmatrix}a&b\\c&d\end{bmatrix}^{-1} = \frac{1}{(ad-bc)} \begin{bmatrix}d&-b\\-c&a\end{bmatrix}$$

OK, so you have

$$E = \begin{bmatrix}2&4\\2&-6\end{bmatrix}$$

Therefore

$$E^{-1} = \frac{1}{(-20)} \begin{bmatrix}-6&-4\\-2&2\end{bmatrix} $$

and then

$$A = \frac{-1}{20} \begin{bmatrix}-6&-4\\-2&2\end{bmatrix} \begin{bmatrix}10&4\\-10&-6\end{bmatrix} = \begin{bmatrix}1&0\\2&1\end{bmatrix}$$

peter.petrov
  • 12,568