Introduction to System of Linear Equations_Exercises

Exercise 1.1

In each case verify that the following is the solution for all values of s and t.

$$\begin{align} & x=19t-35, \\ & y=25-13t, \\ & z=t, \\ \end{align}$$

Is the solution of the following linear system:

$$\begin{align} & 2x+3y+z=5, \\ & 5x+7y-4z=0 \\ \end{align}$$

Solution: To prove that x,y and z are the solutios of the linear system we can simply substitute them into linear system. With this substitution into the first equation we get:

$$\begin{align} & 2x+3y+z=5 \\ & 2\left( 19t-35 \right)+3\left( 25-13t \right)+t=5 \\ & 38t-70+75-39t+t=5 \\ & 5=5, \\ \end{align}$$

and with substitution into the second equation of the system we get:

$$\begin{align} & 5x+7y-4z=0 \\ & 5\left( 19t-35 \right)+7\left( 25-13t \right)-4t=0 \\ & 95t-175+175-91t-4t=0 \\ & 0=0 \\ \end{align}$$

With this substitution we have proved that x,y and z are the solution of this linear system and t can be any number.

Exercise 1.2

In each case verify that the following is the solution for all values of s and t.

$$\begin{align} & {{x}_{1}}=2s+12t+13, \\ & {{x}_{2}}=s, \\ & {{x}_{3}}=-s-3t-3, \\ & {{x}_{4}}=t \\ \end{align}$$   

Is the solution of the following linear system:

$$\begin{align} & 2{{x}_{1}}+5{{x}_{2}}+9{{x}_{3}}+3{{x}_{4}}=-1, \\ & {{x}_{1}}+2{{x}_{2}}+4{{x}_{3}}=1 \\ \end{align}$$

Solution: To prove that $${{x}_{1}},{{x}_{2}},{{x}_{3}}\text{ and }{{x}_{4}}$$ are the solutions of the linear system we can simply substitute them into linear system. With this substitution into the first equation we get:

$$\begin{align} & 2\left( 2s+12t+13 \right)+5\left( s \right)+9\left( -s-3t-3 \right)+3\left( t \right)=-1, \\ & 4s+24t+26+5s-9s-27t-27+3t=-1, \\ & -1=-1 \\ \end{align}$$

and with substitution into the second equation of the system we get:

$$\begin{align} & 2s+12t+13+2\left( s \right)+4\left( -s-3t-3 \right)=1 \\ & 2s+12t+13+2s-4s-12t-12=1 \\ & 1=1 \\ \end{align}$$

Exercise 1.3

Find all solutions of the following equation in parametric form:  
a)
  $$3x+y=2$$

The solution for the previous equation can be written in the following form:

$$y=3x-2$$

If we say that x=t than we can write the solution in the parametric form.

$$\begin{align} & x=t, \\ & y=3t-2, \\ \end{align}$$

where t is arbitrary. Note that there are infinitely many distinct solutions, one for each choice of the parameter t.

The second parametric solutions we get if we write the first equation in the following form:

$$x=\frac{1}{3}\left( 2-y \right)$$

If we write y=s where s is a parameter we get the solution in the parametric form:

$$\begin{align} & x=\frac{1}{3}\left( s+4 \right) \\ & y=s \\ \end{align}$$ 

b)

$$2x+3y=1$$

The solution of the previous equation can be written in the following form. 

$$y=\frac{1}{3}\left( 1-2x \right)$$  

If we write x=t where t is parameter we will get the solution in the parametric form: 

$$\begin{align} & x=t \\ & y=\frac{1}{3}\left( 1-2t \right) \\ \end{align}$$  


We can also write a main equation in the following form: 

$$x=\frac{1}{2}\left( 1-3y \right)$$

If we write y=s where s is parameter we will get the solution in the parametric form: 

$$\begin{align} & x=\frac{1}{2}\left( 1-3s \right) \\ & y=s \\ \end{align}$$

c) 

 

$$3x-y+2z=5$$

First let’s start by moving the variables x and z on the right side of the equation. Then we get: 

$$y=3x+2z-5$$

Now let’s say that x=t and z=s then we will get the following solution: 

$$\begin{align} & x=t \\ & y=3t+2s-5 \\ & z=s \\ \end{align}$$

We could solve the equation for x in terms of y and z so let’s say that y=q and z=p where q and p are the parameters. The parametric solution can be written in the following form: 

$$\begin{align} & x=\frac{1}{3}\left( 5+p+2q \right) \\ & y=p \\ & z=q \\ \end{align}$$

Exercise 1.4 Find the solution of each of the following systems of linear equations using augmented matrices: 
$$\begin{align} & x-3y=1 \\ & 2x-7y=3 \\ \end{align}$$ 

So as you can see this is a very simple system but before solving it we need to convert it into augmented matrix.  
$$\left[ \left. \begin{matrix} 1 & -3 \\ 2 & -7 \\ \end{matrix} \right|\begin{matrix} 1 \\ 3 \\ \end{matrix} \right]$$

And now we need to solve it.  
$$\left[ \left. \begin{matrix} 1 & -3 \\ 2 & -7 \\ \end{matrix} \right|\begin{matrix} 1 \\ 3 \\ \end{matrix} \right]\begin{matrix} {} \\ {{R}_{2}}-2{{R}_{2}} \\ \end{matrix}\sim \left[ \left. \begin{matrix} 1 & -3 \\ 0 & 1 \\ \end{matrix} \right|\begin{matrix} 1 \\ 1 \\ \end{matrix} \right]$$

Now we can write the solution in standard form. 

$$\begin{align} & x-3y=1 \\ & y=1 \\ \end{align}$$   

The solution of the first equation is: 

$$x=1+3 \Rightarrow x=4$$

To check if solution is correct simply substitute these values into one of the equations of linear equation system. Let’s substitute x and y into the first equation of the system.  
$$\begin{align} & x-3y=1 \\ & 4-3(1)=1 \\ & 1=1 \\ \end{align}$$   

As you can see the solution is correct.

Exercise 1.5

Find the solution of each of the following systems of linear equations using augmented matrices. 
$$\begin{align} & x+y+2z=-1 \\ & 2x+y+3z=0 \\ & -2y+z=2 \\ \end{align}$$  

Augmented matrix form of the system is:  
$$\left[ \left. \begin{matrix} 1 & 1 & 2 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \\ \end{matrix} \right|\begin{matrix} -1 \\ 0 \\ 2 \\ \end{matrix} \right]$$

Now let’s solve the augmented matrix.  
$$\begin{align} & \left[ \left. \begin{matrix} 1 & 1 & 2 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \\ \end{matrix} \right|\begin{matrix} -1 \\ 0 \\ 2 \\ \end{matrix} \right]\begin{matrix} {} \\ {{R}_{2}}-2{{R}_{1}} \\ {} \\ \end{matrix}\sim \left[ \left. \begin{matrix} 1 & 1 & 2 \\ 0 & -1 & -1 \\ 0 & -2 & 1 \\ \end{matrix} \right|\begin{matrix} -1 \\ 2 \\ 2 \\ \end{matrix} \right]\begin{matrix} {} \\ (-1){{R}_{2}} \\ {} \\ \end{matrix} \\ & \left[ \left. \begin{matrix} 1 & 1 & 2 \\ 0 & 1 & 1 \\ 0 & -2 & 1 \\ \end{matrix} \right|\begin{matrix} -1 \\ -2 \\ 2 \\ \end{matrix} \right]\begin{matrix} {} \\ {} \\ {{R}_{3}}+2{{R}_{2}} \\ \end{matrix}\sim \left[ \left. \begin{matrix} 1 & 1 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 3 \\ \end{matrix} \right|\begin{matrix} -1 \\ -2 \\ -2 \\ \end{matrix} \right] \\ \end{align}$$

Solving this system we’ve got the following system written in standard form: 
$$\begin{align} & \left[ \left. \begin{matrix} 1 & 1 & 2 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \\ \end{matrix} \right|\begin{matrix} -1 \\ 0 \\ 2 \\ \end{matrix} \right]\begin{matrix} {} \\ {{R}_{2}}-2{{R}_{1}} \\ {} \\ \end{matrix}\sim \left[ \left. \begin{matrix} 1 & 1 & 2 \\ 0 & -1 & -1 \\ 0 & -2 & 1 \\ \end{matrix} \right|\begin{matrix} -1 \\ 2 \\ 2 \\ \end{matrix} \right]\begin{matrix} {} \\ (-1){{R}_{2}} \\ {} \\ \end{matrix} \\ & \left[ \left. \begin{matrix} 1 & 1 & 2 \\ 0 & 1 & 1 \\ 0 & -2 & 1 \\ \end{matrix} \right|\begin{matrix} -1 \\ -2 \\ 2 \\ \end{matrix} \right]\begin{matrix} {} \\ {} \\ 
{{R}_{3}}+2{{R}_{2}} \\ \end{matrix}\sim \left[ \left. \begin{matrix} 1 & 1 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 3 \\ \end{matrix} \right|\begin{matrix} -1 \\ -2 \\ -2 \\ \end{matrix} \right] \\ \end{align}$$ 



 So the solutions of variables are:

$$x=\frac{5}{3},y=-\frac{4}{3},z=-\frac{2}{3}$$

Checking of solution
 $$\begin{align} & x+y+2z=-1 \\ & \frac{5}{3}-\frac{4}{3}-\frac{4}{3}=-1 \\ & \frac{5-4-4}{3}=-1 \\ & -\frac{3}{3}=-1 \\ & -1=-1 \\ \end{align}$$




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