-- Sep 6 In-Class Exercise Thread

1 * (5*9 - 8*6) - 2 * (4*9 - 6*7) + 3 * (4 * 8 - 5*7) 1* (-3) -2(-6)+3(-3) -3+12-9=0

-- Sep 6 In-Class Exercise Thread

A = [[1, 2, 3], [4,5,6], [7,8,9]] R2 -4 * R1; R3 - 7*R1 A = [[1, 2, 3], [0,-3,-6], [0,-6,-12]] R3 - 2 * R2 A = [[1, 2, 3], [0, -3, -6], [0,0,0]] Now A becomes a upper triangle matrix => det(A) = 1*-3*0 = 0

-- Sep 6 In-Class Exercise Thread

Original matrix is : @BT@[ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ]@BT@ -> R2 - 4 * R1 @BT@[ [ 1, 2, 3 ], [ 0, -3, -6 ], [ 7, 8, 9 ] ]@BT@ -> R3 - 7 * R1 @BT@[ [ 1, 2, 3 ], [ 0, -3, -6 ], [ 0, -6, -12 ] ]@BT@ -> R3 - 2 * R1 @BT@[ [ 1, 2, 3 ], [ 0, -3, -6 ], [ 0, 0, 0 ] ]@BT@ Determinant = 1 * -3 * 0 = 0

-- Sep 6 In-Class Exercise Thread

Input Matrix: @BT@[[1,2,3],[4,5,6],[7,8,9]]@BT@ . Input Matrix: The matrix being subtraced has deteriminant of zero since it has two rows equal: . @BT@A^{(1)}=[[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]]@BT@ Using @BT@A^{(1)}@BT@: Subtract 7*Row1 matrix you get the matrix below: . @BT@A^{(2)} = [[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]]@BT@ . Using @BT@A^{(2)}@BT@: Subtract -2*Row2 matrix you get the matrix below: . @BT@A^{(3)} = [[1,2,3],[0,-3,-6],[0,-6,-12]] - 2*[[0,0,0],[0,0,0],[0,-3,-6]]=[[1,2,3],[0,-3,-6],[0,0,0]]@BT@ . @BT@A^{(3)}@BT@ is upper diagonal so its determinant is: . @BT@det(A^{(3)})=(1)(-3)(0)=0@BT@

-- Sep 6 In-Class Exercise Thread

= 1*(5*9-6*8) - 2*(4*9-6*7) + 3*(4*8 -5*7) = 0

-- Sep 6 In-Class Exercise Thread

R2 -> R2 - 4R1; R3 -> R3 - 7R1 - 2R2; This will give upper diagonal matrix like below: |1 2 3| |0 -3 -6| |0 0 0| Det(A) along diagonal = (1)* (-3) * (0) = 0

-- Sep 6 In-Class Exercise Thread

Determinant is 0 Adding the last row to the first row, we get the first row as a scalar multiple of the second row. So, the Determinant is 0

-- Sep 6 In-Class Exercise Thread

Step 1 : Row 3 = Row 3 - 7*Row 1. Last row transforms to [0 ,-6, -12] Step 2 : Row 2 = Row 2 - 4*Row 1. Row 2 transforms to [0, -3, -6] The matrix is an upper triangle. ''Determinant = 1* -3 * 0 = 0'''

-- Sep 6 In-Class Exercise Thread

R2=R3-R2 R3=R3-2R2 R3=R3-R1 Matrix=@BT@[[1,2,3][3,3,3][0,0,0]]@BT@ Determinant=0

-- Sep 6 In-Class Exercise Thread

[[1,2,3],[4,5,6],[7,8,9]] -> [[1,2,3],[3,3,3],[7,8,9]] -> [[1,2,3],[3,3,3],[3,3,3]] R2 = R2-R1; R3 = R3-R2 R3 = R2 => determinant = 0