Question

A = ( −2 −1 2 −2 2 3 −4 1 3 ) b = ( −1 −1 4 ) x = ( x1 x2 x3 ) (a) (2 pts) Write down the augmented matrix (A|b). (b) (4 pts) Use Gauss-Jordan elimination to find the Reduced Row Echelon Form (RREF) of the augmented matrix. (c) (2 pts) What is the rank of A? What is the rank of (A|b) (d) (2 pts) State whether the system is consistent or inconsistent. State how many solutions the system has; if there is/are a solution/s, write it/them down.

Answer 1

The augmented matrix is
`\left[\begin{array}c-2&-1&2&-1\\-2&2&3&-1\\-4&1&3&4\end{array}\right]`

The Reduced Row Echelon Form of the augmented matrix is
`\left[\begin{array}{cccc}1&0&0&-3\\0&1&0&1\\0&0&1&-3\end{array}\right]`

The rank of matrix (A|B) is 3

The system is consistent and the solutions are
`x_(1)= -3, x_(2) = 1, x_(3)= -3`

Explanation:

We have the following information:

`A=\left[\begin{array}{ccc}-2&-1&2\\-2&2&3\\-4&1&3\end{array}\right], X=\left[\begin{array}{c}x_(1)&x_(2)&x_(3)\end{array}\right] and \:B=\left[\begin{array}{c}-1&-1&4\end{array}\right]`

1. The augmented matrix is

We take the matrix A and we add the matrix B we use a vertical line to separate the coefficient entries from the constants.

`\left[\begin{array}c-2&-1&2&-1\\-2&2&3&-1\\-4&1&3&4\end{array}\right]`

2. To transform the augmented matrix to the Reduced Row Echelon Form (RREF) you need to follow these steps:

• Row operation 1: multiply the 1st row by -1/2

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

• Row Operation 2: add 2 times the 1st row to the 2nd row

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

• Row Operation 3: add 4 times the 1st row to the 3rd row

`\left[\begin{array}{cccc}1&1/2&-1&1/2\\0&3&1&0\\0&3&-1&6\end{array}\right]`

• Row Operation 4: multiply the 2nd row by 1/3

`\left[\begin{array}{cccc}1&1/2&-1&1/2\\0&1&1/3&0\\0&3&-1&6\end{array}\right]`

• Row Operation 5: add -3 times the 2nd row to the 3rd row

`\left[\begin{array}{cccc}1&1/2&-1&1/2\\0&1&1/3&0\\0&0&-2&6\end{array}\right]`

• Row Operation 6: multiply the 3rd row by -1/2

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

• Row Operation 7: add -1/3 times the 3rd row to the 2nd row

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

• Row Operation 8: add 1 times the 3rd row to the 1st row

`\left[\begin{array}{cccc}1&1/2&0&-5/2\\0&1&0&1\\0&0&1&-3\end{array}\right]`

• Row Operation 9: add -1/2 times the 2nd row to the 1st row

`\left[\begin{array}{cccc}1&0&0&-3\\0&1&0&1\\0&0&1&-3\end{array}\right]`

3. What is the rank of (A|B)

To find the rank of a matrix, we simply transform the matrix to its row echelon form and count the number of non-zero rows.

Because the row echelon form of the augmented matrix has three non-zero rows the rank of matrix (A|B) is 3

4. Solutions of the system

This definition is very important: "A system of linear equations is called inconsistent if it has no solutions. A system which has a solution is called consistent"

This system is consistent because from the row echelon form of the augmented matrix we find that the solutions are (the last column of a row echelon form matrix always give you the solution of the system)

`x_(1)= -3, x_(2) = 1, x_(3)= -3`