Questions: Sea la matriz: A=[ x x-y x 2 x y -y -1 x+y 2 y ] a) Reduzca a la matriz identidad la matriz A por medio de las operaciones elementales filas.

Solution Steps

To reduce the given matrix \( A \) to the identity matrix using elementary row operations, we will perform a series of row operations to transform \( A \) into the identity matrix \( I \). This involves making the diagonal elements 1 and the off-diagonal elements 0.

We start by writing down the augmented matrix \([A | I]\), where \(I\) is the identity matrix of the same size as \(A\).

\[ \left[\begin{array}{ccc|ccc} x & x-y & x & 1 & 0 & 0 \\ 2x & y & -y & 0 & 1 & 0 \\ -1 & x+y & 2y & 0 & 0 & 1 \end{array}\right] \]

To normalize the first row, we divide the entire first row by \(x\) (assuming \(x \neq 0\)).

\[ \left[\begin{array}{ccc|ccc} 1 & 1 - \frac{y}{x} & 1 & \frac{1}{x} & 0 & 0 \\ 2x & y & -y & 0 & 1 & 0 \\ -1 & x + y & 2y & 0 & 0 & 1 \end{array}\right] \]

We need to make the elements below the pivot in the first column zero. We do this by performing row operations:

• \(R_2 \leftarrow R_2 - 2x \cdot R_1\)
• \(R_3 \leftarrow R_3 + R_1\)

\[ \left[\begin{array}{ccc|ccc} 1 & 1 - \frac{y}{x} & 1 & \frac{1}{x} & 0 & 0 \\ 0 & y - 2x(1 - \frac{y}{x}) & -y - 2x & -2 & 1 & 0 \\ 0 & x + y + 1 - \frac{y}{x} & 2y + 1 & \frac{1}{x} & 0 & 1 \end{array}\right] \]

Simplify the second row: \[ R_2: \quad 0 & y - 2x + 2y & -y - 2x & -2 & 1 & 0 \\ \Rightarrow 0 & 3y - 2x & -y - 2x & -2 & 1 & 0 \]

Simplify the third row: \[ R_3: \quad 0 & x + y + 1 - \frac{y}{x} & 2y + 1 & \frac{1}{x} & 0 & 1 \]

To normalize the second row, we divide the entire second row by \(3y - 2x\) (assuming \(3y - 2x \neq 0\)).

\[ \left[\begin{array}{ccc|ccc} 1 & 1 - \frac{y}{x} & 1 & \frac{1}{x} & 0 & 0 \\ 0 & 1 & \frac{-y - 2x}{3y - 2x} & \frac{-2}{3y - 2x} & \frac{1}{3y - 2x} & 0 \\ 0 & x + y + 1 - \frac{y}{x} & 2y + 1 & \frac{1}{x} & 0 & 1 \end{array}\right] \]

We need to make the elements above and below the pivot in the second column zero. We do this by performing row operations:

• \(R_1 \leftarrow R_1 - (1 - \frac{y}{x}) \cdot R_2\)
• \(R_3 \leftarrow R_3 - (x + y + 1 - \frac{y}{x}) \cdot R_2\)

\[ \left[\begin{array}{ccc|ccc} 1 & 0 & 1 - (1 - \frac{y}{x}) \cdot \frac{-y - 2x}{3y - 2x} & \frac{1}{x} - (1 - \frac{y}{x}) \cdot \frac{-2}{3y - 2x} & 0 - (1 - \frac{y}{x}) \cdot \frac{1}{3y - 2x} & 0 \\ 0 & 1 & \frac{-y - 2x}{3y - 2x} & \frac{-2}{3y - 2x} & \frac{1}{3y - 2x} & 0 \\ 0 & 0 & 2y + 1 - (x + y + 1 - \frac{y}{x}) \cdot \frac{-y - 2x}{3y - 2x} & \frac{1}{x} - (x + y + 1 - \frac{y}{x}) \cdot \frac{-2}{3y - 2x} & 0 - (x + y + 1 - \frac{y}{x}) \cdot \frac{1}{3y - 2x} & 1 \end{array}\right] \]

To normalize the third row, we divide the entire third row by the leading coefficient in the third row (assuming it is non-zero).

Final Answer