Question

Find the reduced row echelon form of the following matrices and then give the solution to the system that is represented by the augmented matrix. TO 4 7 0 6. a. 2 1 0 0 Lo 3 1 - 4 6. b. 54 30 71 8 6 2 -3 4 3 2 -10]

Answer 1

Answer:

a)

Reduced Row Echelon:

$\left[\begin{array}{cccc}1&1/2&0&0\\0&1&7/4&0\\0&0&1&-4\end{array}\right]$

Solution to the system:

$x_3=-4\\x_2=-\frac{7}{4}x_3=7\\x_1=-\frac{1}{2}x_2=-\frac{7}{2}$

b)

Reduced Row Echelon:

$\left[\begin{array}{cccc}4&3&0&7\\0&0&2&-17\\0&0&2&-17\end{array}\right]$

Solution to the system:  

$x_3=-\frac{17}{2}\\x_1=\frac{7-3x_2}{4}$

x_2 is a free variable, meaning that it has infinite possibilities and therefore the system has infinite number of solutions.

Step-by-step explanation:

To find the reduced row echelon form of the matrices, let's use the Gaussian-Jordan elimination process, which consists of taking the matrix and performing a series of row operations. For notation, R_i will be the transformed column, and r_i the unchanged one.

a) $\left[\begin{array}{cccc}0&4&7&0\\2&1&0&0\\0&3&1&-4\end{array}\right]$

Step by step operations:

1. Reorder the rows, interchange Row 1 with Row 2, then apply the next operations on the new rows:

$R_1=\frac{1}{2}r_1\\R_2=\frac{1}{4}r_2$

Resulting matrix:

$\left[\begin{array}{cccc}1&1/2&0&0\\0&1&7/4&0\\0&3&1&-4\end{array}\right]$

2. Set the first row to 1

$R_3=-3r_2+r_3$

Resulting matrix:

$\left[\begin{array}{cccc}1&1/2&0&0\\0&1&7/4&0\\0&0&1&-4\end{array}\right]$

3. Write the system of equations:

$x_1+\frac{1}{2}x_2=0\\x_2+\frac{7}{4}x_3=0\\x_3=-4$

Now you have the  reduced row echelon matrix and can solve the equations, bottom to top, x_1 is column 1, x_2 column 2 and x_3 column 3:

$x_3=-4\\x_2=-\frac{7}{4}x_3=7\\x_1=-\frac{1}{2}x_2=-\frac{7}{2}$

b)

$\left[\begin{array}{cccc}4&3&0&7\\8&6&2&-3\\4&3&2&-10\end{array}\right]$

1. $R_2=-2r_1+r_2\\R_3=-r_1+r_3$

Resulting matrix:

$\left[\begin{array}{cccc}4&3&0&7\\0&0&2&-17\\0&0&2&-17\end{array}\right]$

2. Write the system of equations:

$4x_1+3x_2=7\\2x_3=-17$

Now you have the reduced row echelon matrix and can solve the equations, bottom to top, x_1 is column 1, x_2 column 2 and x_3 column 3:

$x_3=-\frac{17}{2}\\x_1=\frac{7-3x_2}{4}$

x_2 is a free variable, meaning that it has infinite possibilities and therefore the system has infinite number of solutions.