course mth 164 i understand the theory of this better than most but am still going to try to finish all test three things up and then take it asap. I will try to get test four done by the first of the year. z|z{Passignment #014
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18:04:26 `q001. The determinant of a matrix [ [ a b ] , [ c d ] ] is ad - bc. Figure 59 shows how the determinant of a 2 x 2, i.e., a matrix with 2 rows and 2 columns, can be visualized. Find the determinant of a Sane-Demented transition in which 10% of the sane become demented and 20% of the demented become sane.
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RESPONSE --> the determinant is .1xaxd-.2xbxc confidence assessment: 2
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18:12:56 The transition matrix is [ [ .90 .20 ], [ .10, .80] ], with a = .90, b = .20, c = .10 and d = .80. The determinant is therefore ad - bc = .90 * .80 - .20 * .10 = .70. This calculation is represented in Figure 20.
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RESPONSE --> ok self critique assessment: 1
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18:21:23 `q002. For a 3 x 3 matrix, i.e., a matrix with 3 rows and 3 columns, the determinant is calculated by a slightly more complex process. The 1, 1 minor of the matrix is the 2 x 2 matrix left when the first row and first column are eliminated. The 1, 2 minor is the 2 x 2 matrix left when the first row and second column are eliminated. The 1, 3 minor is the 2 x 2 matrix left when the first row and third column are eliminated. Give the 1, 1 minor, the 2, 2 minor and the 3, 3 minor of the matrix T = [ [ .75 .15 .10 ], [ .15 .60 .25 ], [ .20 .15 .65 ] ] representing the sane-borderline-demented problem of the preceding section.
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RESPONSE --> 1,1=.6,.25 .15,.65 2,2=.15,.25 .2, .65 3,3=.15,.6 .2, .15 confidence assessment: 2
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18:23:59 Figure 26 depicts the matrix and its minors. As shown in the figure the 1,1 minor is obtained by eliminating the first row and the first column to obtain [ [.6 .15], [.25 .65] ], which incidentally has determinant .6 * .65 - .15 * .25. The 2, 2 and 3, 3 minors and the expression for the determinant of each are also shown in the Figure.
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RESPONSE --> ok used the wrong directions but figured out that the system for the using the elimanating the and coordinating the system self critique assessment: 3
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18:34:53 `q003. Letting T(1,1), T(1, 2) and T(1, 3) represent the elements in row 1 column 1, row 1 column 2 and row 1 column 3, and letting M(1, 1), M(1, 2) and M(1, 3) represent the 1, 1 minor, the 1, 2 minor and the 1, 3 minor of the matrix T = [ [ .75 .15 .10 ], [ .15 .60 .25 ], [ .20 .15 .65 ] ] of the preceding problem, the determinant of T is T(1, 1) * det M(1, 1) - T(1, 2) * det M(1, 2) + T(1, 3) * det M(1, 3). What therefore is the determinant of the matrix T?
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RESPONSE --> t=0 and this is due to the 0 that is multiplied by the 1,1 and the 1,1 confidence assessment: 2
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18:39:54 Figure 24 shows how the determinant is expanded. T(1, 1) = .75, T(1, 2) = .15 and T(1, 3) = .2. So the determinant is T(1, 1) * det M(1, 1) - T(1, 2) * det M(1, 2) + T(1, 3) * det M(1, 3) = .75 * det [ [.15 .6], [.1 .25] ] - .15 * det [ [.75 .2], [ .1, .65] ] + .2 * det [ [ .75 .15 ], [ .15 .6 ] ] = .75 * (.15 * .25 - .6 * .1) - .15 * (.75 * .65 - .2 * .1) + .2 * (.75 * .6 - .15 * .15) = .2475.
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RESPONSE --> misunderstood the problem and used the det numbers instead of the figure 24 numbers. self critique assessment: 2
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18:45:05 `q004. Speculate on how we would find the determinant of a 4 x 4 matrix.
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RESPONSE --> we would use a formula that multiplied all the factors confidence assessment: 2
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20:34:50 A 4 x 4 matrix would have four numbers in its first row, which we could call T(1, 1), T(1, 2), T(1, 3) and T(1, 4). We could form the corresponding minors M(1, 1), M(1, 2), M(1, 3) and M(1, 4), forming each by eliminating the appropriate row and column (e.g,. eliminating row 1 and column 3 to get M(1, 3) ). Each minor would be a 3 x 3 matrix, whose determinant could in turn be calculated by the means of the preceding problem. So we could perform the calculation T(1, 1) * M(1, 1) - T(1, 2) * M(1, 2) + T(1, 3) * M(1, 3) - T(1, 4) * M(1, 4). Note that as for the 3 x 3 matrix we alternate the + and - as we 'march' across the first row.
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RESPONSE --> ok self critique assessment: 2
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20:42:58 `q005. Solve by elimination the system of equations 2x + 3 y = 9, 4 x + 5 y = 5.
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RESPONSE --> x=4.5-1.5y y=3-2/3x x=1.25-1.25y y=-4x confidence assessment: 1
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20:45:05 We start by writing the system in the form 2x + 3 y = 9 4 x + 5 y = 5. We then multiply the first equation by -2, leaving the second equation alone, to obtain -4x -6 y = -18 4 x + 5 y = 5. We add the two equations to obtain 0 x - 1 y = -13, which we multiply by -1 to get y = 13. We then substitute y = 13 back into the first original equation to get 2x + 3 * 13 = 9, which becomes 2x + 39 = 9. Adding -39 to both sides we get 2x = -30, the dividing by 2 we finally obtain {}x = -15. We check our solution x = -15, y = 13 in the second equation: 4 * -15 + 5 * 13 = -60 + 65 = 5, which agrees with the equation and confirms our result.
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RESPONSE --> ok didnt understand the complete formula but now have concept self critique assessment: 2
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20:57:02 `q006. Solve by elimination the system of equations 2x + 3 y - z = 7, 4 x - y + z = 3, 5 x + y - 2z = 5.
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RESPONSE --> -x+6y=6+z confidence assessment: 2
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20:58:27 We start by writing the system in the form 2x + 3 y - z = 7 4 x - y + z = 3 5 x + y - 2z = 5. We multiply the first equation by -2 to get -4x - 6y + 2z = -14, which we add to the second equation 4x - y + z = 3, resulting in solution -7y + 3z = -11. Then we multiply the first equation by -5/2 to get -5 x - 15/2 y + 5/2 z = -35/2, which we add to the third equation 5x + y - 2z = 5 to obtain -13/2 y + 1/2 z = -25/2, which we multiply by the common denominator 2 to get -13 y + z = -25. We have therefore eliminated x from the second and third equations. We form a system consisting of these two equations: -7y + 3z = -11 -13y + z = -25. We can choose to eliminate z from these equations, multiplying the second by -3 to getthe system -7y + 3z = -11 39y - 3z = 75. Adding these equations we obtain 32 y = 64, which has solution y = 2. Substituting y = 2 into either of the two equations in this system, choosing the first we get -7 * 2 + 3 z = -11 which simplifies to -14 + 3z = -11 then to 3z = 3 and z = 1. Substituting z = 1 and y = 2 back into any equation of the original system, choosing the second we get 4 x - 2 + 1 = 3 so that 4x - 1 = 3 and 4x = 4 with solution x = 1 we have a complete solution to the original system: x=1 y=2 z=1. We can check this solution by substituting these values into each of the three equations; the result will verify the solution.
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RESPONSE --> ok multiplied the wrong equation which created an error self critique assessment: 2
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21:13:59 `q007. Find the matrix product [ [ 2 3], [ 4 5 ] ] * [ x, y ]` and use it to simplify the equation [ [ 2 3], [ 4 5 ] ] * [ x, y ]` = [ 9, 5 ]`. Explain how the system 2x + 3 y = 8, 4 x - y = 5 can be represented by the matrix equation A * X = B, where A = [ [ 2 3], [ 4 5 ] ], X = [ x, y ]` and B = [ 9, 5 ]`.
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RESPONSE --> x=1.125 y=1/3 confidence assessment: 2
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21:15:20 As seen in Figure 80 the product is [ [ 2 3], [ 4 5 ] ] * [ x, y ]` = [ 2x + 3y, 4x + 5y]. Setting [ [ 2 3], [ 4 5 ] ] * [ x, y ]` = [ 9, 5 ]` we therefore have [ 2x + 3y, 4x + 5y ] = [ 9, 5 ]`, as also depicted in Figure 80. The first row of the left-hand side is equal to the first row of the right-hand side so 2x + 3y = 9; and The second row of the left-hand side is equal to the second row of the right-hand side so 4x + 5y = 5. This gives us the system 2x + 3y = 9 4x + 5y = 5. So the original matrix product represents a system of two equations in two unknowns. Figure 81 shows how these matrices can be identified as A, X and B in the product A * X = B.
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RESPONSE --> i see how the 234and5 are multiplied but are the x and y ever defined confidence assessment: 2
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21:17:32 `q008. If the system 2x + 3 y - z = 7, 4 x - y + z = 3, 5 x + y - 2z = 5 is represented by the matrix equation A * X = B, then what are A, X and B? Verify your conclusion by multiplying your matrices.
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RESPONSE --> 2 3 -1 x y z 4 -1 1 x y z 5 1 -2 x y z confidence assessment: 2
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21:19:22 Figure 12 illustrates the matrix product. A is the matrix [ [ 2 3 -1], [ 4, -1, 1 ], [ 5, 1, -2 ] ], X is the matrix [ x y z ]` and B is the matrix [ 7 3 5 ].
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RESPONSE --> ok i see that the x y and z are all combined and also that the results to the problems are all combined self critique assessment: 2
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21:20:37 `q009. We can represent the system 2x + 3 y = 9, 4 x + 5 y = 5 by the 'augmented matrix' [ [ 2 3 | 9 ], [ 4 5 | 5 ] ]. Perform the following sequence of operations on the matrix, showing your work at each step: Keep the first row the same while you replace the second row by the row you get when you subtract double the first row from the second. This operation will result in a matrix with 0 in the second row, second column. Now replace the second row with -1 times the 'new' second row.{} Add -3 times the second row first row to the first row and replace the first row with this row, leaving the second row the same as it was in the preceding step, noting that this gives you 0 in the first row, second column. Replace the first row with the row you get when you multiply the 'new' first row by 1/2, leaving the second row the same. What matrix do you end up with? What does this procedure have to do with the solution of the system in question 5?
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RESPONSE --> it is how to solve the equation given previously using the matrices confidence assessment: 2
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21:21:04 Figure 36 illustrates the augmented matrix and the process specified in the problem. We start with [ [ 2 3 | 9 ], [ 4 5 | 5 ] ]. Double the first row is [ 4 6 | 18 ]; subtracting this from the second row [ 4 5 | 5 ] we get [ 4 - 4 5 - 6 5 - 18 ] = [ 0 -1 -13 ], which we write as the new second row. Multiplying this row by -1 we get [ 0 1 13 ], which in the third step replaces the second row. Subtracting 3 * 2d row (using the present 2d row) from 1st row we have [ 2 3 | 9 } - [ [ 0 3 | 39 ] = [ 2 0 -30 ], as in the fourth step in the figure. Finally we take half of the first row to get [ 1 0 | -15 ], which we use as the first row in the final matrix.
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RESPONSE --> ok self critique assessment: 2
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21:23:02 `q010. The system you ended up with was [ [1 0 | -15], [ 0 1 | 13 ] ]. If this is the augmented system for a set of two equation, what are the equations? What therefore are the solutions x and y?
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RESPONSE --> x=1 y=15/13 confidence assessment: 1
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21:23:48 The final matrix stands for the equations 1 x + 0 y = -15 0 x + 1 y = 13. These equations therefore give us the solution x = -15, y = 13 of the system. This is illustrated in Figure 89.
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RESPONSE --> i see what is right now the augmented portion is what is to the right or what is the result of the equation self critique assessment: 2
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21:28:21 `q011. The augmented matrix for the system 3 x - 4 y = -8, 2 x + 3 y = 23 is [ [ 3 4 | -8 ], [ 2 3 | 23 ] ]. Using a procedure similar to that used previously reduce this system by the following steps: Figure out a multiple of the first row which, added to the second row, leaves 0 in the first position. Note that one of the multipliers will be a negative number, the other a positive. Replace the second row by this new row. Divide the 'new' second row by a number which will make the second number in the row 1; note that the first number will remain 0. Add to the first row a multiple of the second row that will leave 0 in the second position, and replace the first row with this new first row. Divide the 'new' first row by a number that leaves 1 in the first position. Figure out what system the new augmented matrix represents and determine from this the values of x and y that solve the equations. Verify that your solution is indeed a solution to the equations.
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RESPONSE --> x=2.667 y=23 confidence assessment: 2
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21:29:16 The process is illustrated in Figure 39. The final matrix is interpreted as the system 1 x + 0 y = -116, 0 x + 1 y = 85. Thus the solution of the system is x = -116, y = 85. This is verified by substituting these values into the two equations. The first gives us 3 * (-116) - 4 * 85 = -8 which simplifies to the identity -8 = -8; the second gives us 2 * (-116) + 3 * 85 = 23, which simplifies to the identity 23 = 23. The solution is therefore verified.
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RESPONSE --> misunderstood the directions and go them to cancel eachother self critique assessment: 2
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21:30:33 `q012. Write the matrix product A * X = B representing the system of equations 2x + 3 y - z = 7, 4 x - y + z = 3, 5 x + y - 2z = 5, where X = [ x y z ]`.
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RESPONSE --> 2 3 -1 7 x y z 4 -1 1 3 5 1 -2 5 confidence assessment: 3
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21:30:56 Figure 12 illustrates the matrix product, with A = [ [ 2 3 -1 ], [ 4 -1 1 ], [ 5 1 -2 ] ] and B = [ 7 3 5 ].
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RESPONSE --> ok correct self critique assessment: 3
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21:32:12 `q013. Write the augmented matrix representing the system of equations 2x + 3 y - z = 7, 4 x - y + z = 3, 5 x + y - 2z = 5.
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RESPONSE --> 2 3 -1 7 4 -1 1 3 5 1 -2 5 confidence assessment: 3
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21:32:21 The augmented matrix [ [ 2 3 -1 | 7 ], [ 4 -1 1 | 3 ], [ 5 1 -2 | 5 ] ] is illustrated in Figure 16.
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RESPONSE --> ok correct self critique assessment: 3
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21:35:18 `q014. The augmented matrix representing the system of equations 2x + 3 y - z = 7, 4 x - y + z = 3, 5 x + y - 2z = 5 is [ ]. By replacing rows with sums of multiples of rows and by multiplying or dividing rows by appropriate numbers, change the matrix to the form [ [ 1 0 0 | * ], [ 0 1 0 | * ], [ 0 0 1 | * ] ], where the *'s represent the numbers that end up to the right of the | mark.
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RESPONSE --> [2 3 -1/ 7 ]*[4 -1 1/ 3]*[5 1 -2/ 5] confidence assessment: 2
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21:36:41 Figure 18 then Figure 42 illustrate the process. The steps of the matrix reduction are illustrated to the left, the corresponding systems of equations to the right. We begin with the goal of obtaining zeros in the second and third rows of the first column. We 'match up' and 'cancel' the 4 and 5 in these positions by subtracting 2 * the first row from the second and 2.5 times the first row from the third, obtaining the result in the second step. We then obtain the 1's in the desired positions in the first and second rows by taking 1/2 of the first row and -1/7 of the second, as indicated in the third step. We proceed to obtain the 0 required in the second position of the third row. In the final step of Figure 18 we add 13/2 of the second row to obtain the third row. Moving on to Figure 42 we easily obtain the required 1 in the third row, multiplying the third row by -7/16. We can now easily use the third row to obtain the zeros required in the third positions of the first two rows. This is illustrated in the second step of Figure 42. We finally use the second row to obtain the last required zero. As illustrated in the system accompanying the finat step, the matrix [[ 1 0 0 | 1 ], [ 0 1 0 | 2 ], [ 0 0 1 | 1 ] ] represents the solution x = 1, y = 2, z = 1 of the system.
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RESPONSE --> misunderstood the quiestion and thought that the desired form was linear not to reduce the matrices self critique assessment: 2
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21:38:44 `q015. Write the matrix equation A * X = C which represents the matrix [ [ 1 0 0 | * ], [ 0 1 0 | * ], [ 0 0 1 | * ] ] obtained in the preceding exercise. Multiply these matrices and explain what the product means.
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RESPONSE --> the product is the reduced form of all equations 000 confidence assessment: 2
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21:39:24 The matrix equation is [ [1 0 0], [0 1 1], [0 0 1] ] * [ x y z ] ` = [ 1 2 1 ] `. The product of the matrices on the left-hand side is [ x y z ] ` so we have [ x y z ] = [ 1 2 1]. The corresponding system of equations is x = 1, y = 2, z = 1 (see Figure 61).
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RESPONSE --> 011 was not given earlier 010 was self critique assessment: 2
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zz~v굯 assignment #014 014. Determinants; Systems of Equations and their Geometrical Interpretation Precalculus II 12-17-2008"