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

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&-(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&-(2)/(3) &|&0\\0&-6&-(16)/(3) &|&2\\0&1&(n+6)/(12) &|&-(1)/(3)\end{array}\right]`

- Row operation: R2 / 6

`\left[\begin{array}{ccccc}1&-1&-(2)/(3) &|&0\\0&-1&-(8)/(9) &|&(1)/(3) \\0&1&(n+6)/(12) &|&-(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 - ( (n+6)/(12))*x_3 = (1)/(3)`

`-x_2 - ( (8)/(9))*x_3 = (1)/(3)`

The coefficient of " x_3 " must be equal:

`( (n+6)/(12)) = (8)/(9) \\\\\\\\ = (14)/(3)`

- The augmented matrix in reduced form becomes:

`\left[\begin{array}{ccccc}1&-1&-(2)/(3) &|&0\\0&1&(8)/(9) &|&-(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&(2)/(9) &|&-(1)/(3) \\0&1&(8)/(9) &|&-(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 + (8)/(9)*x_3 = -(1)/(3) \\\\x_2 = - ( (8)/(9)*x_3 + (1)/(3) )\\\\x_1 + (2)/(9)*x_3 = -(1)/(3) \\\\x_1 = - ( (2)/(9)*x_3 + (1)/(3) )\\`

- Taking x_3 = α:

Answers:

`x_1 = -(1)/(3) + (2)/(9) \alpha \\\\x_2 = -(1)/(3) + (8)/(9) \alpha`