Question

Given the following system of equations:

6X1 - 6x2 - 4x3 = 0
X1 - 7x2 - 6x3 = 2
X1 +5x2 + nx3 = -2

Rewrite the system in Ax = b format and determine the following:

a. By reduction of the augmented matrix [A|b] to ref, find a value for n such that the system is consistent with an infinite number of solutions.
b. Based on your solution in part A, identify the rank of matrix A and rank of the augmented matrix [A|b].
c. Based on the value of the rank, how many equations (the row vectors of the augmented matrix [Ab]) are linearly independent?
d. Using your solution in part A, solve the system of equations using Gauss-jordan elimination.

Answer 1

Answer:

Step-by-step explanation:

Given:-

- The following system of equations is given:

                       $6x_1 - 6x_2 -4x_3 = 0\\\\x_1 - 7x_2 -6x_3 = 0\\\\x_1 - 5x_2 -nx_3 = 0$

Solution:-

- The matrix equation consists of coefficient matrix "A" and a variable matrix " x ". These two matrices undergo multiplication to yield a solution column vector "b".

- The matrix A, is a symmetrical square matrix with its elements representing the coefficients of each variable as follows:

            $A = \left[\begin{array}{ccc}a_1_1&a_1_2&a_1_3\\a_2_1&a_2_2&a_2_3\\a_3_1&a_3_2&a_3_3\end{array}\right]$

- Where the elements first subscript denotes the equation number and second subscript denotes the variable number.

             $A = \left[\begin{array}{ccc}6&-6&-4\\1&-7&-6\\1&5&n\end{array}\right]$

- Similarly, the variable matrix " X " is a column vector that lists all the variables in the the system of equations in a ascending order.

             $X = \left[\begin{array}{c}x_1&x_2&x_3\end{array}\right]$

- The solution vector " b " is the corresponding solution or any number written on the right hand side of the equals to sign " = " :

              $b = \left[\begin{array}{c}0&2&-2\end{array}\right]$

- Now, we can express the given system in the asked format:

            $A*X = b\\\\\left[\begin{array}{ccc}6&-6&-4\\1&-7&-6\\1&5&n\end{array}\right]*\left[\begin{array}{c}x_1&x_2&x_3\end{array}\right] = \left[\begin{array}{c}0&2&-2\end{array}\right]$

- The augmented matrix is a matrix that combines the coefficient matrix " A " and the solution vector " b ". A solution vector "b" as an extra column to the coefficient matrix:

            $[ A | b ]\\\\ \left[\begin{array}{ccccc}6&-6&-4&|&0\\1&-7&-6&|&2\\1&5&n&|&-2\end{array}\right]$

- Now we will perform row reduction operation such that the system is consistent and has infinite number of solution.

- Row operation: R3 - R2 & R1/6

            $\left[\begin{array}{ccccc}1&-1&-\frac{2}{3} &|&0\\1&-7&-6&|&2\\0&12&n+6&|&-4\end{array}\right]$  

- Row operation: R2 - R1 & R3 / 12

            $\left[\begin{array}{ccccc}1&-1&-\frac{2}{3} &|&0\\0&-6&-\frac{16}{3} &|&2\\0&1&\frac{n+6}{12} &|&-\frac{1}{3}\end{array}\right]$

- Row operation: R2 / 6

           $\left[\begin{array}{ccccc}1&-1&-\frac{2}{3} &|&0\\0&-1&-\frac{8}{9} &|&\frac{1}{3} \\0&1&\frac{n+6}{12} &|&-\frac{1}{3}\end{array}\right]$

For the above system to be consistent and have infinite many solution then the coefficient of " x3 " for the 2nd and 3rd row must be equal:

           $-x_2 - ( \frac{n+6}{12})*x_3 = \frac{1}{3}$

           $-x_2 - ( \frac{8}{9})*x_3 = \frac{1}{3}$

           

The coefficient of " x_3 " must be equal:

           $( \frac{n+6}{12}) = \frac{8}{9} \\\\\\\n = \frac{14}{3}$

- The augmented matrix in reduced form becomes:

           $\left[\begin{array}{ccccc}1&-1&-\frac{2}{3} &|&0\\0&1&\frac{8}{9} &|&-\frac{1}{3} \\0&0&0 &|&0\end{array}\right]$

Answer: Rank = Number of non-zero rows = 2

- The number of linearly independent rows are equal to the rank of the augmented matrix.

Hence,

Answer: Number of linearly independent rows = 2

Row operation: R1 + R2

             $\left[\begin{array}{ccccc}1&0&\frac{2}{9} &|&-\frac{1}{3} \\0&1&\frac{8}{9} &|&-\frac{1}{3} \\0&0&0 &|&0\end{array}\right]$

- The variable "x_3" will take any arbitrary value for which the solution holds infinitely many solutions.

           $x_2 + \frac{8}{9}*x_3 = -\frac{1}{3} \\\\x_2 = - ( \frac{8}{9}*x_3 + \frac{1}{3} )\\\\x_1 + \frac{2}{9}*x_3 = -\frac{1}{3} \\\\x_1 = - ( \frac{2}{9}*x_3 + \frac{1}{3} )$

- Taking x_3 = α:

Answers:

           $x_1 = -\frac{1}{3} + \frac{2}{9} \alpha \\\\x_2 = -\frac{1}{3} + \frac{8}{9} \alpha$