determinants in matrices

Question

I was given the following question:

If $A$ is a $2 \times 2$ matrix, then $\det(2A)=2\det(A)$.

I realize that this is false because it's $4\det(A)$ ($\det$ standing for determinant btw)

I wanted to expand that question a bit. Does that mean if $A$ is a $4 \times 4$ matrix, then the $\det(4A)=16\det(A)$?

Answer 1

Actually, if A is 2 by 2, then $\text{det}(2A) = 4 \cdot \text{det}(A)$.

This is easy to see for the 2 by 2 case: write the determinant of A as $a*d - b*c$, where $a, b, c, d$ are the entries from left to right, top to bottom of A. Then the determinant of 2A is $$2a\cdot2d - 2b\cdot2c = 4ad - 4bc = 4(ad - bc) = 4 \, \text{det}(A)$$

In general, the determinant is linear in each row. This means if you multiply all the entries in a single row of a matrix, then the determinant is also multiplied by that amount. Because the example above has 2 rows, you multiply by 2 twice (multiplying the matrix A by 2 is the same as multiplying its first row by 2, and then multiplying its second row by 2). If A was 3 by 3, then $\text{det}(2A) = 2^3 \cdot \text{det}(A)$ instead, because you can factor out a 2 from each row. Again if A is 3 by 3, then $\text{det}(1.5A) = 1.5 ^ 3 * \text{det}(A)$.

Try seeing if you can generalize this to figure out what det(4A) would look like for A being a 4 by 4 matrix.

Answer 2

Since $\det(AB) = \det(A)\det(B)$ and $cA = c(I_nA) = (cI_n) A$, it follows that $$\det(cA) = \det(c(I_nA)) = \det((cI_n) A) = \det(cI_n)\det(A) = c^n\det(A).$$