Does $\det(A + B) = \det(A) + \det(B)$ hold?

No, in general $\det(A + B) ≠ \det(A) + \det(B)$.

The determinant of a matrix is a scalar value that can be calculated for any square matrix. It is denoted as det(A) or |A|. The determinant of a matrix is equal to the sum of the products of its elements along its main diagonal, multiplied by the alternating signs +1 and -1.

For example, the determinant of the 2 x 2 matrix:

[a, b] [c, d]

is calculated as follows:

|A| = ad – bc

The determinant of a matrix has some important properties, including the following:

• The determinant of a matrix is equal to the product of its eigenvalues.
• The determinant of a matrix is invariant under invertible linear transformations.
• The determinant of a matrix is used to calculate the inverse of a matrix.

However, the determinant does not follow the addition property that you mentioned. In other words, $\det(A + B) ≠ \det(A) + \det(B)$ in general.

Here are some examples to illustrate the fact that $\det(A + B) ≠ \det(A) + \det(B)$:

Example 1:

Let’s consider the following 2 x 2 matrices:

A = [1, 2] [3, 4]

B = [5, 6] [7, 8]

Then,

$\det(A) = 14 – 23 = -2$

$\det(B) = 58 – 67 = -2$

$\det(A + B) = (1+5)(4+8) – (2+6)(3+7) = -14$

Therefore, $\det(A + B) ≠ \det(A) + \det(B)$.

Example 2:

Let’s consider the following 3 x 3 matrices:

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

B = [10, 11, 12] [13, 14, 15] [16, 17, 18]

Then,

$\det(A) = 1*(59 – 68) – 2*(49 – 67) + 3*(48 – 57) = 0$

$\det(B) = 10*(1418 – 1517) – 11*(1318 – 1516) + 12*(1317 – 1416) = 0$

$\det(A + B) = (1+10)((5+14)(9+18) – (6+15)(8+17)) – (2+11)((4+14)(9+18) – (6+15)(7+16)) + (3+12)((4+14)(8+17) – (5+15)*(7+16)) = -28$

Therefore, $\det(A + B) ≠ \det(A) + \det(B)$.