Answer to Question #294618 in Linear Algebra for Da3

Question

consider the following system of equations:

3x + 4y +5z = 66

7x + 4y +3z = 74

8x + 8y +9z = 136

a. write down the associated augmented matrix for this system of equations and the coefficient matrix A.

b. by performing elementary row operations of the augmented matrix, solve the system of equations or show that no solution exists. In case there exist infinitely many solutions, then the solution to the system must be written in parametric form.

c. Based on your answer in b, what is the rank of the coefficient matrix A?

Answer 1

Given, the system of linear equations

$\begin{aligned} 3x + 4y +5z &= 66\\ 7x + 4y +3z &= 74\\ 8x + 8y +9z &= 136 \end{aligned}$

a) The augmented matrix of the system is

$[A \mid B] = \begin{bmatrix} 3& 4 & 5 & \mid &66\\ 7& 4 & 3 &\mid &74\\ 8 & 8 & 9 & \mid &136\\\end{bmatrix}$ and the coefficient matrix is

$A = \begin{bmatrix} 3& 4 & 5 \\ 7& 4 & 3 \\ 8 & 8 & 9 \end{bmatrix}$

b) Using elementary row operations of the augmented matrix,

$\begin{aligned} [A \mid B] &= \begin{bmatrix} 3& 4 & 5 & \mid &66\\ 7& 4 & 3 &\mid &74\\ 8 & 8 & 9 & \mid &136\\\end{bmatrix}\\ &= \begin{bmatrix} 3& 4 & 5 & \mid &66\\ 0 & \frac{-16}{3} & \frac{-26}{3} &\mid & -80\\ 0 & \frac{-8}{3} & \frac{-13}{3} &\mid & -40\\\end{bmatrix}~~ \begin{aligned}&R_{2} \rightarrow R_{2} - \frac{7R_1}{3}\\&R_{3} \rightarrow R_{3} - \frac{8R_1}{3} \end{aligned}\\ &= \begin{bmatrix} 3 & 4 & 5 & \mid & 66\\ 0 & \frac{-16}{3} & \frac{-26}{3} &\mid & -80\\ 0 & 0 & 0 &\mid & 0\\ \end{bmatrix}~~ R_{3} \rightarrow R_{3} - \frac{R_2}{2}\\ \end{aligned}$

The system has infinitely many solutions. Taking, $z = t$ and back substituting

$\begin{aligned} -\frac{16y}{3}-\frac{26z}{3} &= -80\\ 16y &= 240-26t\\ y &= 15 - \frac{13t}{8} \end{aligned}\\ \begin{aligned} 3x+4y+5z &= 66\\ 3x &= 66 - 4y - 5z\\ x &= \frac{1}{3}\left[66 - 4\left(15 - \frac{13t}{8}\right) - 5t\right]\\ & = \frac{1}{3}\left[66 - 60 + \frac{13t}{2}- 5t\right]\\ &= \frac{1}{3}\left[6 + \frac{3t}{2}\right]\\ &=2 + \frac{t}{2}\\ \therefore x&= \frac{t}{2} + 2 \end{aligned}$

c) Based on the solution in (b), we find that the rank of the coefficient matrix A is 2.