Answer:
a)
Reduced Row Echelon:
[tex]\left[\begin{array}{cccc}1&1/2&0&0\\0&1&7/4&0\\0&0&1&-4\end{array}\right][/tex]
Solution to the system:
[tex]x_3=-4\\x_2=-\frac{7}{4}x_3=7\\x_1=-\frac{1}{2}x_2=-\frac{7}{2}[/tex]
b)
Reduced Row Echelon:
[tex]\left[\begin{array}{cccc}4&3&0&7\\0&0&2&-17\\0&0&2&-17\end{array}\right][/tex]
Solution to the system:
[tex]x_3=-\frac{17}{2}\\x_1=\frac{7-3x_2}{4}[/tex]
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) [tex]\left[\begin{array}{cccc}0&4&7&0\\2&1&0&0\\0&3&1&-4\end{array}\right][/tex]
Step by step operations:
1. Reorder the rows, interchange Row 1 with Row 2, then apply the next operations on the new rows:
[tex]R_1=\frac{1}{2}r_1\\R_2=\frac{1}{4}r_2[/tex]
Resulting matrix:
[tex]\left[\begin{array}{cccc}1&1/2&0&0\\0&1&7/4&0\\0&3&1&-4\end{array}\right][/tex]
2. Set the first row to 1
[tex]R_3=-3r_2+r_3[/tex]
Resulting matrix:
[tex]\left[\begin{array}{cccc}1&1/2&0&0\\0&1&7/4&0\\0&0&1&-4\end{array}\right][/tex]
3. Write the system of equations:
[tex]x_1+\frac{1}{2}x_2=0\\x_2+\frac{7}{4}x_3=0\\x_3=-4[/tex]
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:
[tex]x_3=-4\\x_2=-\frac{7}{4}x_3=7\\x_1=-\frac{1}{2}x_2=-\frac{7}{2}[/tex]
b)
[tex]\left[\begin{array}{cccc}4&3&0&7\\8&6&2&-3\\4&3&2&-10\end{array}\right][/tex]
1. [tex]R_2=-2r_1+r_2\\R_3=-r_1+r_3[/tex]
Resulting matrix:
[tex]\left[\begin{array}{cccc}4&3&0&7\\0&0&2&-17\\0&0&2&-17\end{array}\right][/tex]
2. Write the system of equations:
[tex]4x_1+3x_2=7\\2x_3=-17[/tex]
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:
[tex]x_3=-\frac{17}{2}\\x_1=\frac{7-3x_2}{4}[/tex]
x_2 is a free variable, meaning that it has infinite possibilities and therefore the system has infinite number of solutions.
