To express a matrix $X = \begin{pmatrix} 2 & 1 \\ -1 & 0 \end{pmatrix}$ as a product of three elementary matrices, we need to apply row operations to transform it into an identity matrix, and then reverse these steps to write $X$ as a product of elementary matrices.

Step-by-Step Solution

Given: $X = \begin{pmatrix} 2 & 1 \\ -1 & 0 \end{pmatrix}$

Step 1: Transform $X$ into the Identity Matrix by Row Operations

To convert $X$ into the identity matrix $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$, we perform row operations and keep track of each elementary matrix used.

1. Divide Row 1 by 2 to make the first element in the first row equal to 1. $\begin{pmatrix} 2 & 1 \\ -1 & 0 \end{pmatrix} \xrightarrow{R_1 \to \frac{1}{2} R_1} \begin{pmatrix} 1 & \frac{1}{2} \\ -1 & 0 \end{pmatrix}$ The elementary matrix for this operation is: $E_1 = \begin{pmatrix} \frac{1}{2} & 0 \\ 0 & 1 \end{pmatrix}$

2. Add Row 1 to Row 2 to make the first element in the second row equal to 0. $\begin{pmatrix} 1 & \frac{1}{2} \\ -1 & 0 \end{pmatrix} \xrightarrow{R_2 \to R_2 + R_1} \begin{pmatrix} 1 & \frac{1}{2} \\ 0 & \frac{1}{2} \end{pmatrix}$ The elementary matrix for this operation is: $E_2 = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$

3. Multiply Row 2 by 2 to make the second element in the second row equal to 1. $\begin{pmatrix} 1 & \frac{1}{2} \\ 0 & \frac{1}{2} \end{pmatrix} \xrightarrow{R_2 \to 2 R_2} \begin{pmatrix} 1 & \frac{1}{2} \\ 0 & 1 \end{pmatrix}$ The elementary matrix for this operation is: $E_3 = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}$

Step 2: Write $X$ as a Product of Elementary Matrices

Now, we have $E_3 E_2 E_1 X = I$, so $X = E_1^{-1} E_2^{-1} E_3^{-1}$.

Let's compute the inverses of each elementary matrix:

1. $E_1 = \begin{pmatrix} \frac{1}{2} & 0 \\ 0 & 1 \end{pmatrix}$, so $E_1^{-1} = \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}$.
2. $E_2 = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$, so $E_2^{-1} = \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix}$.
3. $E_3 = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}$, so $E_3^{-1} = \begin{pmatrix} 1 & 0 \\ 0 & \frac{1}{2} \end{pmatrix}$.

Thus, we can express $X$ as: $X = E_1^{-1} E_2^{-1} E_3^{-1} = \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & \frac{1}{2} \end{pmatrix}$

Final Answer

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

Would you like further explanation on any of the steps?

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Here are some related questions for practice:

1. How can elementary matrices be used to find the inverse of a matrix?
2. Can every square matrix be expressed as a product of elementary matrices? Why or why not?
3. What are the properties of elementary matrices and their inverses?
4. How do row operations affect the determinant of a matrix?
5. How does matrix factorization help in solving systems of linear equations?

Tip: Remember, each elementary matrix represents a single row operation, making them a powerful tool for understanding matrix transformations.