Post your solutions to the Sep 6 In-Class Exercise to this thread. Best, Chris

-- Sep 6 In-Class Exercise Thread

@BT@[ [1,2,3] , [4,5,6] , [7,8,9] ]@BT@ After applying the operations to the given matrix: R1->R1+R3 @BT@[ [8,10,12] , [4,5,6] , [7,8,9] ]@BT@ R2->2*R2 @BT@[ [8,10,12] , [8,10,12] , [7,8,9] ]@BT@ Now if we see, R1 and R2 are the same, therefore the determinant value would be 0.

-- Sep 6 In-Class Exercise Thread

Answer : 0

-- Sep 6 In-Class Exercise Thread

@BT@[[1,2,3],[4,5,6],[7,8,9]]@BT@ row2 - 4 x row1 and row3 - 7 x row1 generates below matrix @BT@[[1,2,3],[0,-3,-6],[0,-6,-12]]@BT@ row3 - 2 x row2 generates a new upper triangle matrix below @BT@[[1,2,3],[0,-3,-6],[0,0,0]]@BT@ Then the determinant is 1x(-3)x0=0

-- Sep 6 In-Class Exercise Thread

Row 1 = Row 1 + Row 2 @BT@[ [8,10,12], [4,5,6], [7,8,9]]@BT@ Row 2 = 2 * Row 2 @BT@[ [8,10,12], [8,10,12], [7,8,9]]@BT@ Two same rows, hence determinant = 0

-- Sep 6 In-Class Exercise Thread

@BT@[[1,2,3],[4,5,6],[7,8,9]]@BT@ Row 1 plus row 3 into row 3 @BT@[[1,2,3],[4,5,6],[8,10,12]]@BT@ -2 * row 2 plus row 3 into row 3 @BT@[[1,2,3],[4,5,6],[0,0,0]]@BT@ -4 * row 1 plus row 2 into row 2 @BT@[[1,2,3],[0,-3,-6],[0,0,0]]@BT@ Upper triangular, multiply diagonal.. @BT@1*-3*0=0@BT@

-- Sep 6 In-Class Exercise Thread

The matrix in upper triangular form is: @BT@[[1,2,3],[0,-3,-6],[0,0,0]]@BT@ By multiplying along the diagonal, we get det = 0

-- Sep 6 In-Class Exercise Thread

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

-- Sep 6 In-Class Exercise Thread

+1[[5 6],[8 9]] -2 [[4 6],[7 9]] + 3[[4 5],[7 8]] = 1(45-48) - 2(36-42) + 3(32-35) = -3 -2(-6) + 3(-3) = 0

-- Sep 6 In-Class Exercise Thread

@BT@[[1,2,3],[4,5,6],[7,8,9]]@BT@ 3 x row1 + row2 1,2,3 7,11,15 7,8,9 row3-row2 1,2,3 7,11,15 0,-3,-6 row2 = -7*row1+row2 1,2,3 0,-3,-6 0,-3,-6 now 2 rows are same so det(A) = 0