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# 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)\$.