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

Best, Chris

[ [1,2,3] , [4,5,6] , [7,8,9] ]

After applying the operations to the given matrix:

R1->R1+R3

[ [8,10,12] , [4,5,6] , [7,8,9] ]

R2->2*R2

[ [8,10,12] , [8,10,12] , [7,8,9] ]

Now if we see, R1 and R2 are the same, therefore the determinant value would be 0.

Answer : 0

[[1,2,3],[4,5,6],[7,8,9]]

row2 - 4 x row1 and row3 - 7 x row1 generates below matrix

[[1,2,3],[0,-3,-6],[0,-6,-12]]

row3 - 2 x row2 generates a new upper triangle matrix below

[[1,2,3],[0,-3,-6],[0,0,0]]

Then the determinant is 1x(-3)x0=0

Row 1 = Row 1 + Row 2

[ [8,10,12], [4,5,6], [7,8,9]]

Row 2 = 2 * Row 2

[ [8,10,12], [8,10,12], [7,8,9]]

Two same rows, hence determinant = 0

[[1,2,3],[4,5,6],[7,8,9]]

Row 1 plus row 3 into row 3

[[1,2,3],[4,5,6],[8,10,12]]

-2 * row 2 plus row 3 into row 3

[[1,2,3],[4,5,6],[0,0,0]]

-4 * row 1 plus row 2 into row 2

[[1,2,3],[0,-3,-6],[0,0,0]]

Upper triangular, multiply diagonal..

1*-3*0=0

The matrix in upper triangular form is:

[[1,2,3],[0,-3,-6],[0,0,0]]

By multiplying along the diagonal, we get det = 0

1 * (59-68) = -3 -2 * (49 - 67) = 12 3 * (48-57) = -9

-3 + 12 - 9 = 0

+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

[[1,2,3],[4,5,6],[7,8,9]]

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