Solve Laplace Expansion Determinant -2a,-2b,-2c,2p+x,2q+y,2r+z,3x,3y,3z

[matrix_204]

how can i solve this determinant
if detA= a b c
p q r = -1
x y z

compute det B -2a -2b -2c
2p+x 2q+y 2r+z
3x 3y 3z

i want to kno what i should do to reach to the point of multiplying the two det,( ie, det A(-1) x det b)

 

[matrix_204]

I GOT IT, pretty easy question, lol, so dumb

 

[M98Ranger]

To solve this determinant, you can use the Laplace Expansion method. This method involves expanding the determinant along a row or column and breaking it down into smaller determinants until you reach a 2x2 determinant, which can be easily solved.

In this case, you can expand along the first row. This would give you the following expression:

detA = -2a * det(-2b, 2q+y, 2r+z, 3y, 3z) + 2p * det(2p+x, 2q+y, 2r+z, 3x, 3z) - 3x * det(2p+x, 2q+y, -2c, 3x, 3y)

Next, you can expand each of these smaller determinants along the first column to get rid of the variables in the first row. This would give you the following expressions:

detA = -2a * (-2b * det(2q+y, 2r+z, 3z) + 2q * det(2p+x, 2r+z, 3z) - 3z * det(2p+x, 2q+y, -2c)) + 2p * (2p * det(2q+y, 2r+z, 3z) - 2q * det(2p+x, 2r+z, 3z) + 3z * det(2p+x, 2q+y, -2c)) - 3x * (2p * det(2q+y, 2r+z, 3y) - 2q * det(2p+x, 2r+z, 3y) + 3y * det(2p+x, 2q+y, -2c))

Now, you can solve each of these smaller determinants using the formula for 2x2 determinants, which is ad-bc. For example, the first determinant would be:

det(2q+y, 2r+z, 3z) = (2q * 3z) - (3z * 2r+z) = 6q - 6z

Similarly, you can solve the other smaller determinants and substitute them back into the original expression. This would give you:

detA = -2a * (-2