Given the following system of linear equations: -4x-1 + 2x_2 = 6x-3 = 4 \\ \ \ \ \ \ \ \ \ \...

Question:

Given the following system of linear equations:

$$-4x-1 + 2x_2 = 6x-3 = 4 \\ \ \ \ \ \ \ \ \ \ \ \ 12x_1+ 16x_3 = 9\\ 8x-1 - x_2 + 5x_3 = 1 $$

(a) Write it in the form of AX = b

(b) Find all solutions to the system using Gauss elimination method

(c) Find the rank of A

(d) Use cramer's method to solve for {eq}x_3 {/eq} if possible.

Solving Systems of Linear Equations with matrices:

Let's consider the following system of {eq}m {/eq} linear equations

{eq}\begin{aligned} &a_{11} x_{1}+a_{12} x_{2}+\ldots+a_{1 n} x_{n}=b_{1}\\ &a_{21} x_{1}+a_{22} x_{2}+\ldots+a_{2 n} x_{n}=b_{2}\\ &\vdots\\ &a_{m 1} x_{1}+a_{m 2} x_{2}+\ldots+a_{m n} x_{n}=b_{n} \end{aligned} {/eq}

This system, can be written in a matrix form {eq}AX=B {/eq}, where

{eq}A=\left[\begin{array}{cccc} a_{11}&a_{12}&\ldots&a_{1 n} \\ a_{21}&a_{22}&\ldots&a_{2 n} \\ \vdots&\vdots&\vdots&\vdots\\ a_{m1}&a_{m2}&\ldots&a_{m n} \\ \end{array}\right] {/eq}

is the coefficient matrix,

{eq}X=\left[\begin{array}{c} x_{1} \\ x_{2} \\ \vdots\\ x_{m} \\ \end{array}\right] {/eq} is the variable vector and

{eq}B=\left[\begin{array}{c} b_{1} \\ b_{2} \\ \vdots\\ b_{m} \\ \end{array}\right] {/eq} is the constant vector

Answer and Explanation:

It is given the system of linear equations

{eq}\begin{aligned} -4 x-1+2 x_{2}&=6 x-3=4 \\ 12 x_{1}+16 x_{3}&=9 \\ 8 x-1-x_{2}+5 x_{3}&=1 \end{aligned} {/eq}

a)This system has four equations and four variable {eq}x_1,x_2,x_3,x {/eq} and it can be rewritten as:

{eq}\left\{\begin{aligned} 12 x_{1} +16 x_{3} &=9 & \text { (second) } \\ -x_{2}+5 x_{3}+8 x &=2 & \text { (third) } \\ 2 x_{2} -4 x &=5 & \text { (first) } \\ 6 x &=7 &\text { (first) } \end{aligned}\right. {/eq}

The coefficient Matrix is:

{eq}A=\left[\begin{array}{cccc} 12 & 0 & 16 & 0 \\ 0 & -1 & 5 & 8 \\ 0 & 2 & 0 & -4 \\ 0 & 0 & 0 & 6 \end{array}\right] {/eq} The variable vector is:

{eq}x=\left[\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ x \end{array}\right] {/eq}

and the constant column vector is:

{eq}B=\left[\begin{array}{l} 9 \\ 2 \\ 5 \\ 7 \end{array}\right] {/eq}

Therefore, the given system can be rewritten as:

{eq}A X=B \Rightarrow\left[\begin{array}{cccc} 12 & 0 & 16 & 0 \\ 0 & -1 & 5 & 8 \\ 0 & 2 & 0 & -4 \\ 0 & 0 & 0 & 6 \end{array}\right]\left[\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ x \end{array}\right]=\left[\begin{array}{c} 9 \\ 2 \\ 5 \\ 7 \end{array}\right] {/eq}

b)

The augmented matrix is:

{eq}[A | B]=\left[\begin{array}{cccc} 12 & 0 & 16 & 0 \quad \vdots \quad 9 \\ 0 & -1 & 5 & 8 \quad \vdots \quad 2 \\ 0 & 2 & 0 & -4 \quad \vdots \quad 5\\ 0 & 0 & 0 & 6 \quad \vdots \quad 7 \end{array}\right] {/eq}

By using elementary row operations:

{eq}[A | B]=\left[\begin{array}{ccccc} 12 & 0 & 16 & 0 \quad \vdots \quad 9 \\ 0 & -1 & 5 & 8 \quad \vdots \quad 2 \\ 0 & 0 & 10 & 12 \quad \vdots \quad 9 \\ 0 & 0 & 0 & 6 \quad \vdots \quad 7 \end{array}\right] \quad\left(r_{3}+2 r_{2} \rightarrow r_{3}\right) {/eq}

Now, we use Backward substitution.

{eq}\left\{\begin{array}{l} 12 x_{1}+16 x_{2}=9 \\ -x_{2}+5 x_{3}+8 x=-2 \\ 10 x_{3}+12 x=9 \\ 6 x=7 \end{array} \Rightarrow\left\{\begin{array}{l} x_{1}=\left(9-16 x_{2}\right) / 12 \\ x_{2}=5 x_{3}+8 x-2 \\ x_{3}=(9-12 x) / 10 \\ x = 7 / 6 \end{array}\right.\right. {/eq}

For {eq}x=\frac{7}{6} {/eq},

{eq}\begin{aligned} x_{3} &=(9-12 x) / 10 \\ &=9-12(7 / 6)] \frac{1}{10}=-\frac{1}{2} \end{aligned} {/eq}

For {eq}x=\frac{7}{6}\ \text{and} \ x_3=-\frac{1}{2} {/eq},

{eq}\begin{aligned} x_{2} &=5\left(-\frac{1}{2}\right)+8\left(\frac{7}{6}\right)-2 \\ &=\frac{29}{6} \end{aligned} {/eq}

For {eq}x=\frac{7}{6} {/eq}, {eq}x_3=-\frac{1}{2} {/eq} and {eq}\frac{29}{6} {/eq},

{eq}x_{1}=\left[9-16\left(\frac{29}{6}\right)\right] \frac{1}{12}=\frac{17}{12} {/eq}

Finally, we have found that:

{eq}x_{1}=\frac{17}{12}, \quad x_{2}=\frac{29}{6}, \quad x_{3}=-\frac{1}{2} \quad \text { and } \quad x=7 / 6 {/eq}

c)

Since the given system has only one solution we have that:

{eq}\operatorname{rank}=n=4 \rightarrow \text { number of variables } {/eq}

d)

First, we evaluate {eq}\text{det}(A) {/eq}

{eq}\begin{aligned} |A| &=\left|\begin{array}{cccc} 12 & 0 & 16 & 0 \\ 0 & -1 & 5 & 8 \\ 0 & 2 & 0 & -4 \\ 0 & 0 & 0 & 6 \end{array}\right|=12\left|\begin{array}{ccc} -1 & 5 & 8 \\ 2 & 0 & -4 \\ 0 & 0 & 6 \end{array}\right| \\ &=12(6)\left|\begin{array}{cc} -1 & 5 \\ 2 & 0 \end{array}\right|=12(6)(-10)=-720 \end{aligned} {/eq}

where {eq}A {/eq} is a {eq}4\times 4 {/eq} matrix in which we have replaced the third column vector by {eq}B {/eq}. Thus,

{eq}\begin{aligned} \left|A_{3}\right| &=\left|\begin{array}{cccc} 12 & 0 & 9 & 0 \\ 0 & -1 & 2 & 8 \\ 0 & 2 & 5 & -4 \\ 0 & 0 & 7 & 6 \end{array}\right|=2\left|\begin{array}{ccc} -1 & 2 & 8 \\ 2 & 5 & -4 \\ 0 & 7 & 6 \end{array}\right| \\ &=12(-1)\left|\begin{array}{cc} 5 & -4 \\ 7 & 6 \end{array}\right|+12(-2)\left|\begin{array}{cc} 2 & 8 \\ 7 & 6 \end{array}\right| \\ &=-12(30+28)-24(16-56)=360 \end{aligned} {/eq}

Finally, {eq}x_{3}=\frac{\left|A_{3}\right|}{|A|}=\frac{360}{-720}=-\frac{1}{2} {/eq}


Learn more about this topic:

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Inconsistent and Dependent Systems: Using Gaussian Elimination

from Algebra II Textbook

Chapter 10 / Lesson 8
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