S6 Flashcards
If A and B are both n×n, then det(A+B) = det(A) + det(B).
False
Take for instance A and B.
Then det(A) + det(B) = 0 + 0 = 0, but det(A + B) = 1.
![](https://s3.amazonaws.com/brainscape-prod/system/cm/120/960/665/a_image_thumb.png?1659440373)
If A and B are row equivalent, then they have the same determinant.
False
If B is obtained from A by a row interchange, or by multiplying a row, then they are row equivalent, but have different determinant.
The linear transformation with matrix A is injective if and only if det(A) ≠ 0.
True
In class we saw that if det(A) ≠ 0, then A is invertible, and by the Invertible Matrix Theorem that we saw earlier, this is equivalent to the transformation defined by A being injective (if A is square).
The determinant of the partitioned matrix is det(A) · det(D).
![](https://s3.amazonaws.com/brainscape-prod/system/cm/120/960/799/q_image_thumb.png?1659494408)
True
We can get this matrix into echelon form by separately putting A and D into echelon forms Aech and Dech.
Then this is an upper triangular matrix, so the determinant is the product of the diagonal entries. But we can split this product into the diagonal entries from Aech and those from Dech, which are exactly det(A) and det(D).
![](https://s3.amazonaws.com/brainscape-prod/system/cm/120/960/799/a_image_thumb.png?1659440373)
The determinant of this partitioned matrix is det(A) · det(D)−det(B) · det(C).
![](https://s3.amazonaws.com/brainscape-prod/system/cm/120/960/914/q_image_thumb.png?1659494408)
False
Take for instance A = C = D = I2. Then we can take basically anything for B and the equation will fail:
![](https://s3.amazonaws.com/brainscape-prod/system/cm/120/960/914/a_image_thumb.png?1659440374)