y y = 4x + 1 y = 3x + 1 in Software

Maker QR Code 2d barcode in Software y y = 4x + 1 y = 3x + 1

Check! Finally the fourth: y = 4x + 1 1=4 0+1 1=0+1 1=1 We can now be confident that the solution to the system is indeed x = 0 and y = 1 10 Figure B-10 shows graphs of all four equations on the Cartesian plane It s visually apparent that if we choose any pair or triplet of these lines, they intersect at the point (0, 1) and nowhere else Therefore, any pair or triplet of the equations, taken as a twoby-two or three-by-two linear system, has the unique solution x = 0 and y = 1

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1 Here are the original three equations, stated again for convenience, followed by the step-by-step processes that get the equations into form for conversion to a matrix: x=y z 7 y = 2x + 2z + 2 z = 3x 5y + 4

We morph the first equation as follows: x=y z 7 x y = z 7 x y + z = 7 Next, the second equation: y = 2x + 2z + 2 2x + y = 2z + 2 2x + y 2z = 2 Finally, the third: z = 3x 5y + 4 3x + z = 5y + 4 3x + 5y + z = 4 Now we have this set of equations representing our three-by-three linear system: x y + z = 7 2x + y 2z = 2 3x + 5y + z = 4 2 Before we write down the matrix, we must be sure we have the correct signs for the coefficients Subtraction of a positive is equivalent to addition of a negative With that in mind, we can pigeonhole the coefficients into the matrix form:

1 2 3 1 1 5 1 2 1 7 2 4

3 It s easy to convert a matrix into a set of linear equations, but we must pay attention to the signs Here s the matrix:

0 5 1 4 3/2 1 1 8 1 2 1 1

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Here are the equations, derived directly from the coefficients in the matrix: 0x + 4y + ( z) = 2 5x + ( 3/2)y + 8z = 1 x+y+z=1 Here are the equations with the negative additions converted to subtractions, and the term 0x eliminated from the first equation for simplicity: 4y z = 2 5x (3/2)y + 8z = 1 x+y+z=1 4 Here s the matrix again, for reference:

0 5 1 4 3/2 1 1 8 1 2 1 1

5 0 1 3/2 4 1 8 1 1 1 2 1

10 0 1 3 4 1 16 1 1 2 2 1

10 0 10 3 4 10 16 1 10 2 2 10

10 0 0 3 4 13 16 1 6 2 2 8

10 0 0 3 52 52 16 13 24 2 26 32

10 0 0 3 52 0 16 13 11 2 26 58

This matrix is in echelon form 5 We want to put the matrix from solution to Prob 4 into diagonal form We can multiply the second row by 11 and the third row by 13 to obtain

10 0 0 3 572 0 16 143 143 2 286 754

10 0 0 3 572 0 16 0 143 2 1,040 754

5,720 0 0 1,716 1,716 0 9,152 0 143 1,144 3,120 754

5,720 0 0 0 1,716 0 9,152 0 143 1,976 3,120 754

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We need to turn the 9,152 in the first row into 0 We ll have to work with the third row to make it happen Suppose 9,152 divides cleanly by 143 Let s give it a try A calculator tells us that 9,152 / 143 = 64 Let s multiply the third row by 64 to get the matrix

5,720 0 0 0 1,716 0 9,152 0 9,152 1,976 3,120 48,256