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]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%261%2F2%260%260%5C%5C0%261%267%2F4%260%5C%5C0%260%261%26-4%5Cend%7Barray%7D%5Cright%5D)
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}](https://tex.z-dn.net/?f=x_3%3D-4%5C%5Cx_2%3D-%5Cfrac%7B7%7D%7B4%7Dx_3%3D7%5C%5Cx_1%3D-%5Cfrac%7B1%7D%7B2%7Dx_2%3D-%5Cfrac%7B7%7D%7B2%7D)
b)
Reduced Row Echelon:
![\left[\begin{array}{cccc}4&3&0&7\\0&0&2&-17\\0&0&2&-17\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D4%263%260%267%5C%5C0%260%262%26-17%5C%5C0%260%262%26-17%5Cend%7Barray%7D%5Cright%5D)
Solution to the system:
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]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D0%264%267%260%5C%5C2%261%260%260%5C%5C0%263%261%26-4%5Cend%7Barray%7D%5Cright%5D)
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](https://tex.z-dn.net/?f=R_1%3D%5Cfrac%7B1%7D%7B2%7Dr_1%5C%5CR_2%3D%5Cfrac%7B1%7D%7B4%7Dr_2)
Resulting matrix:
![\left[\begin{array}{cccc}1&1/2&0&0\\0&1&7/4&0\\0&3&1&-4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%261%2F2%260%260%5C%5C0%261%267%2F4%260%5C%5C0%263%261%26-4%5Cend%7Barray%7D%5Cright%5D)
2. Set the first row to 1
![R_3=-3r_2+r_3](https://tex.z-dn.net/?f=R_3%3D-3r_2%2Br_3)
Resulting matrix:
![\left[\begin{array}{cccc}1&1/2&0&0\\0&1&7/4&0\\0&0&1&-4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%261%2F2%260%260%5C%5C0%261%267%2F4%260%5C%5C0%260%261%26-4%5Cend%7Barray%7D%5Cright%5D)
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](https://tex.z-dn.net/?f=x_1%2B%5Cfrac%7B1%7D%7B2%7Dx_2%3D0%5C%5Cx_2%2B%5Cfrac%7B7%7D%7B4%7Dx_3%3D0%5C%5Cx_3%3D-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}](https://tex.z-dn.net/?f=x_3%3D-4%5C%5Cx_2%3D-%5Cfrac%7B7%7D%7B4%7Dx_3%3D7%5C%5Cx_1%3D-%5Cfrac%7B1%7D%7B2%7Dx_2%3D-%5Cfrac%7B7%7D%7B2%7D)
b)
![\left[\begin{array}{cccc}4&3&0&7\\8&6&2&-3\\4&3&2&-10\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D4%263%260%267%5C%5C8%266%262%26-3%5C%5C4%263%262%26-10%5Cend%7Barray%7D%5Cright%5D)
1. ![R_2=-2r_1+r_2\\R_3=-r_1+r_3](https://tex.z-dn.net/?f=R_2%3D-2r_1%2Br_2%5C%5CR_3%3D-r_1%2Br_3)
Resulting matrix:
![\left[\begin{array}{cccc}4&3&0&7\\0&0&2&-17\\0&0&2&-17\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D4%263%260%267%5C%5C0%260%262%26-17%5C%5C0%260%262%26-17%5Cend%7Barray%7D%5Cright%5D)
2. Write the system of equations:
![4x_1+3x_2=7\\2x_3=-17](https://tex.z-dn.net/?f=4x_1%2B3x_2%3D7%5C%5C2x_3%3D-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_2 is a free variable, meaning that it has infinite possibilities and therefore the system has infinite number of solutions.