104.4.1 Formal process (Handout 3) a'nna12a13a1na'2na'n2a'23a'32Create zeros
114.4.1 Formal process (Handout 3) a'nna12a13a1na'2na'n2a'23Create zeros
124.4.1 Formal process (Handout 3) a''nna12a13a1na'2na'23Create zeros
134.4.1 Formal process (Handout 3) a''nna12a13a1na'2na'23Create zeros
144.4.2 Augmented Matrix notation We perform row operations on the matrix and the opposite row:12-1xyz65=Combine both of these into one matrix called the augmented matrix:12-165
154.4.3 Row sums A way to check calculations: Do a row operation: e.g. 1 Add up rows2. Write totals on right12-165937Do a row operation: e.g.1-126597r2 r2 - 2r13-2x9
164.4.3 Row sums A way to check calculations: Do a row operation: e.g. Add up rows2. Write totals on right12-165937Do a row operation: e.g.1-12659r2 r2 - 2r1-1-3-11-157Check row sums: e.g. 0 + (-1) – 3 – 11 = -15
174.5 Examples 1 4 3 2 8 6 -2 x y z 9 18 = Matrix-vector system Write in augmented form143286-2xyz918=
184.5 Examples – EXAMPLE 1Matrix-vector systemWrite in augmented formWrite on row sums-214328691815366Use the top left entry to create zeros below itFirst get a zero in the second row:-21329415r2 r2 - 4r136 – 4x156
194.5 Examples – EXAMPLE 1Matrix-vector systemWrite in augmented formWrite on row sums-214328691815366Use the top left entry to create zeros below itFirst get a zero in the second row:-21329415-246r2 r2 - 4r1
204.5 Examples – EXAMPLE 1Matrix-vector systemWrite in augmented formWrite on row sums-214328691815366Use the top left entry to create zeros below itFirst get a zero in the second row:-21329415-246r2 r2 - 4r1-6-18
214.5 Examples – EXAMPLE 1 Check row sums before continuing... -21432869181536-2132-69-18415-246r2 r2 - 4r1Check row sums before continuing...– 18 = OK!= OK!= OK!
224.5 Examples – EXAMPLE 1 Now get a zero in the third row: -21432869181536-2132-69-18415-246r2 r2 - 4r1Now get a zero in the third row:-11123-6-59-18-2315-24-39r3 r3 - 3r1Want upper triangular form so swap rows 2 and 3123-69-1815-24-11-5-23-39r r3
234.5 Examples – EXAMPLE 1 Now solve by backwards substitution: 23-69-1815-24-5-11-23-39Now solve by backwards substitution:r3 : -6z = z = 3r2 : -5y – 11z = -23-5y -33 =-23y = -2r1 : x + 2y +3z = 9x = 9x = 4Hence: x = 4, y = -2, z = 3 is the unique solution.
244.5 Examples 1 -1 x y z 6 = Matrix-vector system Write in augmented form1-1xyz6=
254.5 Examples – EXAMPLE 2Matrix-vector systemWrite in augmented formWrite on row sums1-1616Use the top left entry to create zeros below itFirst get a zero in the third row:1-16-2r3 r3 - r1
264.5 Examples – EXAMPLE 2r3 r3 - r11-16-2Use second row to get a zero in the third row:r3 r3 – r221-16-8-6
274.5 Examples – EXAMPLE 2 Solve by backwards substitution: 2 1 -1 6 -8 -16-8-6Solve by backwards substitution:r3 : 2z = -8z = -4UNIQUE SOLUTIONr2 : y - z = 6y + 4 = 6y = 2r1 : x - y = 1x - 2 = 1x = 3
284.6 DeterminantsQuestion: During the elimination process, what has changed about the determinant of the matrix?Swapping rows multiplies the determinant by (-1)Adding or subtracting multiples of rows does not change the determinant
294.6 Determinants In EXAMPLE 1 we used one swap operation to get from 23-6-11-5486-2= BA =Calculating the determinant:|A|= (-1) |B|= (-1)x(1)(-5)(-6) = -30Non-zero, so we got a unique solution
304.6 Determinants In EXAMPLE 2 we used no swaps to get from = B A = 1-12= BA =Calculating the determinant:|A|= |B|= (1)(1)(2) = 2Non-zero, so got a unique solution
314.6 Non-Standard Gaussian Elimination In standard Gaussian Elimination the following operation were allowed:Swap two rows;Add or Subtract a multiple of a row from another row.In Non-Standard Gaussian Elimination we are also allowed to do the following:Multiply a row by a constant. E.g.r r3
324.6 Non-Standard Gaussian Elimination Quick Example: In Standard G.E.315-14831-8/348-16/3r r2 - 5r1/3This is a bit messy with the fractions. However, in Non-Standard G.E.315-14831-848-16r r2 - 5r1However, in doing this we have multiplied the determinant by 3.
334.7 Backwards substitution: more general case Two cases after elimination process:All diagonal entries non-zero, then determinant is non-zero. Hence, get answer by backwards substitution.Is a zero on the diagonal, then determinant is zero. Either get infinite solutions or no solutions.
344.7.1 Case of No SolutionsSuppose we followed the elimination process and got to:312-1xyz11-39=312-111-39Zeros on the diagonal, so determinant is zero.ROW 3 gives the equation0x + 0y + 0z = 9This is impossible. Hence there are no solutions.
354.7.1 Case of Infinite solutions Suppose instead that312-1xyz11-3=312-111-3ROW 3 now OK: x + 0y + 0z = 0Have two equations in three unknownsGet infinitely many solutions
364.7.1 Case of Infinite solutions Three steps:In the final (echelon-form) of the matrix, circle the first non-zero entry in each rowFind the columns that have no circles in. Each column corresponds to a variable.Assign a new name to each of the chosen variables, then use back substitution on the non-zero rows.
374.7.1 Case of Infinite solutions 312-1xyz11-3=312-111-3Circle first non-zero row entriesFind column with no circles inColumn 2 corresponds to the y variable3. Assign a name to y: let y = α
384.7.1 Case of Infinite solutions 312-1xyz11-3=312-111-3Solve by back substitution:r3 : Tells us nothingr2 : -z = z = 3r1 : 3x + y + 2z = 113x + α + 6 = 11x = (5 – α)/3So, solution is x = (5 – α)/3, y = α, z = 3 for any α