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2017-09-06

-- Sep 6 In-Class Exercise Thread
Row1 = [1,2,3] Row2 = [4,5,6] Row3 = [7,8,9]
Row2' = Row2- Row1 Row3' = Row3 - Row2
This gives Row2' = Row3' Therefore det(A) = 0
Row1 = [1,2,3] Row2 = [4,5,6] Row3 = [7,8,9] Row2' = Row2- Row1 Row3' = Row3 - Row2 This gives Row2' = Row3' Therefore det(A) = 0

-- Sep 6 In-Class Exercise Thread
Matrix = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
Row 1 = Row 1 + Row 3
Matrix = [[8, 10, 12], [4, 5, 6], [7, 8, 9]]
Row 2 = 2 * Row 2
Matrix = [[8, 10, 12], [8, 10, 12], [7, 8, 9]]
Since two rows are zero, the determinant is 0.
Matrix = [[1, 2, 3], [4, 5, 6], [7, 8, 9]] Row 1 = Row 1 + Row 3 Matrix = [[8, 10, 12], [4, 5, 6], [7, 8, 9]] Row 2 = 2 * Row 2 Matrix = [[8, 10, 12], [8, 10, 12], [7, 8, 9]] Since two rows are zero, the determinant is 0.

-- Sep 6 In-Class Exercise Thread
R2=R3-R2 R3=R3-2R2 R3=R3-R1 Matrix=[[1,2,3][3,3,3][0,0,0]][[1,2,3][3,3,3][0,0,0]] Hence Determinant=0
R2=R3-R2 R3=R3-2R2 R3=R3-R1 Matrix=[[1,2,3][3,3,3][0,0,0]][[1,2,3][3,3,3][0,0,0]] Hence Determinant=0

-- Sep 6 In-Class Exercise Thread
Given Matrix: '[1,2,3],[4,5,6],[7,8,9]'
R2-4R1 and R3-7R1 gives: '[1,2,3],[0,-3,-6],[0,-6,-12]'
R3-2R2 gives: '[1,2,3],[0,-3,-6],[0,0,0]'
Determinant = 1*-3*0 =0
Given Matrix: '[1,2,3],[4,5,6],[7,8,9]' R2-4R1 and R3-7R1 gives: '[1,2,3],[0,-3,-6],[0,-6,-12]' R3-2R2 gives: '[1,2,3],[0,-3,-6],[0,0,0]' Determinant = 1*-3*0 =0

-- Sep 6 In-Class Exercise Thread
`[1,2,3, 4,5,6, 7,8,9] `
R1=R1+R3 then R2=2*R2 R1=R2 hence by property det(A)=0
(Edited: 2017-09-06)
@BT@[1,2,3, 4,5,6, 7,8,9] @BT@ R1=R1+R3 then R2=2*R2 R1=R2 hence by property det(A)=0

-- Sep 6 In-Class Exercise Thread
`[[1,2,3],[4,5,6],[7,8,9]]`
row 2 - (4 * row 1)
`[[1,2,3],[0,-3,-6],[7,8,9]]`
row 3 - (7 * row 1)
`[[1,2,3],[0,-3,-6],[0,-6,-12]]`
row 3 - (2 * row 2)
`[[1,2,3],[0,-3,-6],[0,0,0]]`
multiply diagonal to get Determinant = `1 * -3 * 0 =`0
@BT@[[1,2,3],[4,5,6],[7,8,9]]@BT@ row 2 - (4 * row 1) @BT@[[1,2,3],[0,-3,-6],[7,8,9]]@BT@ row 3 - (7 * row 1) @BT@[[1,2,3],[0,-3,-6],[0,-6,-12]]@BT@ row 3 - (2 * row 2) @BT@[[1,2,3],[0,-3,-6],[0,0,0]]@BT@ multiply diagonal to get Determinant = @BT@1 * -3 * 0 =@BT@'''0'''

-- Sep 6 In-Class Exercise Thread
R2=R2-4*R1 , R3=R3-7*R1 `[[1,2,3],[0,-3,-6],[0,-6,-12]]` R3=R3-2*R1
`[[1,2,3],[0,-3,-6],[0,0,0]]` since this is a upper tringular matrix with 0 in diagonal : determinant = 0
(Edited: 2017-09-06)
R2=R2-4*R1 , R3=R3-7*R1 @BT@[[1,2,3],[0,-3,-6],[0,-6,-12]]@BT@ R3=R3-2*R1 @BT@[[1,2,3],[0,-3,-6],[0,0,0]]@BT@ since this is a upper tringular matrix with 0 in diagonal : determinant = 0

-- Sep 6 In-Class Exercise Thread
 Original Matrix: `[1,2,3],[4,5,6],[7,8,9]`
 Applying rules of multilinearity:
 Row1 = Row1 + Row3
 
 `[8,10,12],[4,5,6],[7,8,9]`
 Row1 = 2 * Row2
 2 * `[4,5,6],[4,5,6],[7,8,9]`
 Since the first and second rows are the same the determinant value is 0.
 2*0 = 0
(Edited: 2017-09-06)
Original Matrix: @BT@[1,2,3],[4,5,6],[7,8,9]@BT@ Applying rules of multilinearity: Row1 = Row1 + Row3 @BT@[8,10,12],[4,5,6],[7,8,9]@BT@ Row1 = 2 * Row2 2 * @BT@[4,5,6],[4,5,6],[7,8,9]@BT@ Since the first and second rows are the same the determinant value is 0. 2*0 = 0

-- Sep 6 In-Class Exercise Thread
A = [1,2,3; 4,5,6; 7,8,9]
Since A is multilinear, the following can be applied:
Det(A) = |1,2,3; 4,5,6; 7, 8, 9|
 = |1,2,3; 4,5,6; 7, 8, 9| - 4*|0, 0, 0; 1, 2, 3; 0, 0, 0|
 = |1, 2, 3; 0, -3, -6; 7, 8, 9| - 7 |0, 0, 0; 0, 0, 0; 1, 2, 3|
 = |1, 2, 3; 0, -3, -6; 0, -6, -12| + 2 |0, 0, 0; 0, 0, 0; 0, -3, -12|
 = |1, 2, 3; 0, -3, -6; 0, 0, 0|
 = 0
(Edited: 2017-09-06)
A = [1,2,3; 4,5,6; 7,8,9] Since A is multilinear, the following can be applied: Det(A) = |1,2,3; 4,5,6; 7, 8, 9| = |1,2,3; 4,5,6; 7, 8, 9| - 4*|0, 0, 0; 1, 2, 3; 0, 0, 0| = |1, 2, 3; 0, -3, -6; 7, 8, 9| - 7 |0, 0, 0; 0, 0, 0; 1, 2, 3| = |1, 2, 3; 0, -3, -6; 0, -6, -12| + 2 |0, 0, 0; 0, 0, 0; 0, -3, -12| = |1, 2, 3; 0, -3, -6; 0, 0, 0| = 0
2017-09-13

-- Sep 6 In-Class Exercise Thread
1(45-48)-2(36-63)+3(32-35)=0
1(45-48)-2(36-63)+3(32-35)=0
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