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Given $$X = \begin{bmatrix} 0 & 1\\ -2 & -18\end{bmatrix}$$ find elementary matrices $E_1$, $E_2$ and $E_3$ such that $X = E_1 E_2 E_3$.


My attempt

I did 3 row operations from $X$ to get to $I_2$

  1. Swapping row 1 and row 2

  2. Row 1 becomes $-\frac12$ of row 1

  3. Row 1 becomes Row 1 - 9 Row 2

So then

$$E_1 = \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}, \qquad E_2 = \begin{bmatrix} -1/2 & 0\\ 0 & 1 \end{bmatrix}, \qquad E_3 = \begin{bmatrix} 1 & -9\\ 0 & 1 \end{bmatrix}$$

However, when I multiply the $E_1$, $E_2$ and $E_3$ it doesn't give $X$. Can someone please tell me where I have made a mistake or if I've approached this question incorrectly?

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2 Answers 2

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From the row operations you've performed, we can say that $E_3E_2E_1 X=I$. So, $X=E_1^{-1}E_2^{-1}E_3^{-1}$.

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  • $\begingroup$ I multiplied 𝐸3𝐸2𝐸1𝑋 but it does not give me I. Can you tell if my elementary matrices are incorrect? $\endgroup$
    – Hayley
    Aug 26, 2021 at 6:44
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To complement Umesh's answer, using SymPy:

>>> from sympy import *
>>> X = Matrix([[  0,   1],
                [ -2, -18]])
>>> E1 = Matrix([[0, 1],
                 [1, 0]])
>>> E2 = Matrix([[-Rational(1,2), 0],
                 [             0, 1]])
>>> E3 = Matrix([[1, -9],
                 [0,  1]])
>>> E3 * E2 * E1 * X
Matrix([[1, 0],
        [0, 1]])
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  • $\begingroup$ so I should just find the inverse of E1, E2 and E3 that I found previously and those would be my answer? $\endgroup$
    – Hayley
    Aug 26, 2021 at 8:10
  • $\begingroup$ @Hayley Yes, and note that the inverses of elementary matrices are very easy to compute. $\endgroup$ Aug 26, 2021 at 8:15

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