Homework 1.1, 1.2 & Quiz 1 Solutions Quizlet

Solve the following linear system x1 + 2x2 - 3x3 = 2 -x1 - x2 + 4x3 = 1 -2x1 - 4x2 + 7x3 = -3 -3x1 + 2x2 + 3x3 = 4

(x1, x2, x3) = (1, 2, 1)

Balance the following reaction x1C3H8 + x2O2 ---> x3H20 + x4CO2

(x1, x2, x3, x4)= ( t/3, ((4t/3)+2t)/2, 4t/3, t) When t=3 x1, x2, x3, x4= (1, 5, 4, 3)

What are the 2 conditions for a matrix to be in Reduced Row-Echelon form?

1) The matrix must be a row-echelon matrix 2) The first non-zero entry in each row is the only non zero entry in each column (if there is a column with a free variable, this doesn't apply for that column)

What are the 3 conditions for a matrix to be in Row-Echelon form?

1) the first non-zero entry in each row is 1 2) If a row does not consist entirely of zeroes, the first non zero element lies strictly to the right of the first non-zero element in the preceding row (looks like stairs- the stairs can be 2 zeroes wide, but not 2 zeroes long) 3) if there are rows entirely of all zero coefficients, they are below the rows having non-zero elements

Solve the system of equations 2x1 + x2 =8 4x1 - 3x2 = 6

1. Eliminate x2 by multiplying the top row by 3 and adding the rows together. This gives us 10x1 ​= 30 2. Solve for x1 -> x1 = 3 3. Substitute to find x2 -> 2(3) + x2 =8 -> x2 = 2 Solution: (x1, x2) = (3, 2)

Use back substitution to solve the system of linear equations x1 ​− 3x2 ​= 2 2x2 ​= 6​

1. solve for x2 -> x2=3 2. substitute x1-3(3)=2 ->x1=11 Solution: (x1, x2)= (11,3)

Use back substitution to solve the system of linear equations x1​ + 2x2​ + 2x3​ + x4​ = 5 3x2 ​+ x3​ − 2x4​ = 1 −x3​ + 2x4 ​= −1 4x4​ = 4​

1. solve for x4 in row 4 -> x4 =1 2. substitute into row 3 to find x3 -x3 +2(1) = -1 -> x3 = 3 3. substitute x3 and x4 into row 2 to find x2 3x2 + (3) -2(1) =1 -> x2 = 0 4. repeat for x1 in row 1 x1 +2(0) + 2(3) + (1) = 5 -> x1= -2 Solution: (x1, x2, x3, x4) = (-2, 0, 3, 1)

Determine the value of x1, x2, x3 and x4 to balance the equation x1 C6H6 + x2 O2 --> x3 C + x4 H20

Create 3 equations to balance the carbon hydrogen and oxygen 6x1 = x3 -> 6x1 - x3 =0 (for carbon) 6x1= 2x4 -> 3x1 -x4 =0 (for hydrogen) 2x2=x4 -> 2x2 - x4 =0 (for oxygen) Since there are 3 equations and 4 variables, there will be a free variable. Therefore there will not be a unique solution (there are lots of solutions that will work). Let the free variable x4 = t, then solve. Plug in any number for t (preferably one that will make everything a whole number) Lets say t=6. Solution: (x1, x2, x3, x4) = (2, 3, 12, 6)

Use Gaussian Elimination to solve x1 + x2 + x3 + x4 = 0 2x1 +3x2 -x3 -x4 = 2 3x1 +2x2 +x3 +x4 = 5 3x1 +6x2 -x3 -x4 = 4

From the last row of the matrix follows that 0x1 + 0x2 + 0x3 + 0 x4 = 5, which is not satisfied for any values of these variables. Therefore, the system is inconsistent.

Use Gaussian Elimination to solve the system of equations. When will it have infinitely many solutions? 1 2 1 (0) 2 5 3 (0) -1 1 B (0)

From the last row of the row echelon form, we have that (β−2)x3​ = 0. This equation will have infinitely many solutions only if β−2=0 (in that case x3​ would be a free variable), which implies that β=2. In every other case (when β does not equal 2), we would have a unique solution using back substitution. Solution: The system has infinitely many solutions only if β=2​.

Is it possible for the following system of equations to be inconsistent? 1 2 1 (0) 2 5 3 (0) -1 1 B (0)

Given coefficient matrix tells us that the corresponding system of equations is homogenous. Since homogenous system of equations always has a trivial solution, it can never be inconsistent. In other words, (x1​,x2​,x3​)=(0,0,0) is always a solution, regardless of the value of parameter β.

Is the Matrix in Row-echelon form? Is it reduced? 1 4 6 0 0 1 0 1 3

Not in Row- Echelon form. If R2 and R3 were switched it would be (but it wouldn't be reduced)

Is the matrix in row-echelon form. Is it reduced? 1 1 1 0 1 2 0 0 3

Not in Row-Echelon form

Graph the lines and determine geometrically the number of solutions x1 + x2 =4 x1 - x2 =2

One way to solve: think of x1 as x and x2 as y. Solve for x2 (aka y) to get an equation in the form y=mx+b For the first equation, the slope will be -1 and the y intercept will be 4 For the second equation the y intercept will be 1 and y intercept will be - 2. Sketching these lines we can determine they will intersect at 1 point

Consider a linear system whose augmented matrix is in the form 1 2 1 (1) -1 4 3 (2) 2 -2 a (3) For what values of a will the system have a unique solution?

Since parameter a is present only in the third row of the row echelon form, equation (a+2)x3 = 4 determines whether the system has a solution or not. The only scenario in which this equation cannot be solved is if a+2=0, which implies that a = -2. For any other value of a, we can use back substitution the obtain the unique solution of the given system. Thus, the system will have a unique solution when a does not equal 2

Use Gaussian Elimination to solve -1x1 + 2x2 - x3 = 2 -2x1 +2x2 + x3 = 4 3x1 + 2x2 + 2x3 =5 -3x1 + 8x2 + 5x3 = 17

Since there are no free variables, there is a unique solution to the given system. Using back substitution, we get the following 1. x3 = 1 2. x2- (3/2)x3 = 0 -> x2 -(3/2)= 0 -> x2 = 3/2 3. x1 - 2x2 + x3 = -2 -> x1 -2(3/2) +1 = -2 -> x1 = 0 solution: (x1, x2, x3) = (0, 3/2 , 1)

Consider a linear system whose augmented matrix is in the form 1 1 3 (2) 1 2 4 (3) 1 3 a (b) a) For what values of a & b will the system have infinitely many solutions? b) For what values will the system have no solution?

Solution: The system will have infinitely many solutions when a=5 and b =4 The system will have no solution when a = 5 and b does not equal 4

Solve the system of equations 2x1 + x2 + 3x3 = 1 4x1 + 3x2 +5x3 = 1 6x1 + 5x2 +5x3 = -3

Solution: (x1, x2, x3) = (-3, 1, 2)

Solve the system of equations x1 + 2x2 -x3 =1 2x1 -x2 + x3 =3 -x1 +2x2 +3x3 =7

Solution: (x1, x2, x3)= (1, 1, 2)

Use the diagram for question 15 from Ch 1.2 to determine the values of x1, x2, x3, x4

Solution: (x1, x2, x3, x4) = (280, 230, 350, 590)

Solve the system of equations 0 + x2 + x3 + x4 = 0 3x1+ 0 +3x3 -4x4 = 7 x1 + x2 + x3 + 2x4 =6 2x1+3x2+ x3 + 3x4 = 6

Solution: (x1, x2, x3, x4) = (4, -3, 1, 2)

How would you go about solving a problem that involves a traffic flow diagram with 4 intersections?

Step 1: Label each intersection, A, B, C and D Step 2: Note that the inflow of traffic must be equal to the outflow of traffic. Step 3: Create an equation for the 4 intercepts. At each intersection (A, B, C and D) look where the arrows are pointing in and where they are pointing out. Set those numbers (or variables) equal to each other (arrows pointing in = arrows pointing out). For each of the four intersections you will be left with a system of 4 linear equations. Step 4: Solve using matrixes

Use the diagram from 22a from Ch.1.2 to Determine the amount of each current (i) flowing through the network.

Step 1: create a system of equations (remember V=iR)(remember current connecting potion= current of upper part + current of lower part) equation 1. i2 = i1 + i3 -> i1 - i2 +i3 = 0 equation 2. Voltage of upper loop: 2i1 + 2i2 = 16 equation 3. Voltage of lower loop: 3i3 + 2i2 = 0 Step 2: Create a matrix and solve Solution: (i1, i2, i3) = (5, 3, 2) Amps

Is this in row echelon form? Is it reduced? 0 1 0 0 0 0

This is in Reduced Row echelon form (and also r/e form)

Is the matrix in row-echelon form? Reduced Row echelon form? 1 2 3 4 0 0 1 2

This is in Row- Echelon Form ( remember "steps" can be long but not tall). It does not meet the criteria for reduced row-echelon form.

Is the matrix in Row-Echelon form? Is it reduced? 1 0 0 1 2 0 1 0 2 4 0 0 1 3 6

This is in reduced row-echelon form (and therefore also in row-echelon form)

Is this in Row-echelon form. Is it reduced? 0 1 3 4 0 0 1 3 0 0 0 0

This is in row-echelon form but it is not reduced

Is this in row-echelon form? Is it reduced? 1 3 0 0 0 1 0 0 0

This is in row-echelon form. It is also in reduced row-echelon form. Note that the second column does not have a leading variable and the 3 is therefore a free variable, so it does not need to be reduced to 0

Is the matrix in row-echelon form? reduced? 1 0 0 0 0 0 0 0 1

This is not in Row-echelon form (and therefore not reduced). If you switched rows 2 & 3 the matrix would be both in r/e and Rr/e form

Consider the following linear system ( 1 1 2 (1) (2 1 2 (2) (1 2 a (3) Find "a" such that the system has a unique solution Find "a" such that the system has infinitely many solutions

Unique solution when a does not equal 4 There is no such a that will cause this system to have infinitely many solutions

What is Ohms Law and how would you use it so help solve problems 22 a & b

Voltage = Current x Resistance (V=IR) Use this problem to relate current (i) to voltage. This will give you an equation for each "loop". You can put these together to relate a system of equations. Note: -if a loop does not have a power source, then its voltage is zero -if there are 2 resisters in a loop, the current is multiplied by each resistance and added (in class example for i3) -i2 is the ONLY current that will "experience" a resister that is placed in the center wire.

Graph the lines and determine geometrically the number of solutions x1​ + 2x2​ = 4 −2x1​ − 4x2 ​​= 4

When both of these equations are put into slope- intercept form, you'll notice that both the slopes are the same, but they have different y- intercepts. Therefore, these lines must be parallel & there will be no solutions.

Graph the lines and determine geometrically the number of solutions 2x1 ​− x2​ = 3 −4x1​ + 2x2​​ = −6​

When these equations are put into slope-intercept form, you'll notice their equations are exactly the same. Therefore they are the same line and there will be infinitely many solutions

Use the diagram from 22b (Ch 1.2) to Determine the amount of each current (i) flowing through the network.

i2 = i1 + i3 -> i1 - i2 +i3 = 0 Voltage of upper loop: 2i1 + 4i2 = 20 Voltage of lower loop: 4i2 + 2i3 =20 (i1, i2, i3) = (2, 4, 2) Amps

There are 3 currents; i1, i2, i3. i2 is in the center to create 2 "loops" with i1 above and i3 below. How are these currents related?

i2 = i1 + i3 -> i1 - i2 +i3 = 0 Use this along with ohms law to create a system of equations

Graph the lines and determine geometrically the number of solutions x1 + x2 =1 x1 - x2 =1 -x1 + 3x2 = 3

solve for x2 x2= 1 - x1 x2 = x1-1 x2 = 1+1/3x1 lets set rows 1 & 2 equal to each other to find x1 x1-1 = 1-x1 x1=1 substitute into R1 -1 + x2 =1 x2 =0 See if these work in R3 -(1) + 3(0) does not equal 3 Therefore, there is no common solution

Use Gaussian Elimination to solve x1 + 2x2 - 3x3 + x4 =1 -x1 -x2 +4x3 -x4 =6 -2x1 -4x2 +7x3 -x4 =1

x4 is a free variable, so set it equal to t (parameter), then use back substitution to solve. Solution: (x1, x2, x3, x4) = (2-6t, 4+t, 3-t, t)