# Is $\rm\det (A+B) = \det (A) + \det (B)$? [duplicate]

Is there a general rule for the determinant of the sum if we know the determinants? For example:

If $\rm \det A = 1$ and $\rm\det B = 2$ and there is $\rm A+B$, is $\rm\det (A+B) = \det (A) + \det (B) = 3$ ? Thank you.

• No${{{{{{}}}}}}$. – Angina Seng Sep 7 '17 at 18:45
• This can be easily broken with $2\times 2$ examples like the identity matrix and the matrix full of ones. – User8128 Sep 7 '17 at 18:45
• If you don't want to mess with specific examples, think about that if the statement was correct, then it would follow that: $2\det(A) = \det(A) + \det(A) = \det(A + A) = \det(2A) = 2^n\det(A)$, and for $n>1$ you get a contradiction – AsafHaas Sep 7 '17 at 18:48
• You might be interested in the following: math.stackexchange.com/questions/329972/… – projectilemotion Sep 7 '17 at 19:42
• You should have used Google for this question. – A---B Sep 7 '17 at 19:47

For a diagonal matrix with diagonal entries $a_{11},\cdots, a_{nn}$, the determinant is the product of said entries. So, even in easy cases such as $A$ and $B$ being both diagonal, there is less of a correlation between $\det A$, $\det B$ and $\det (A+B)$ than there is between the product of $n$ numbers, the product of other $n$ numbers and the product of their sums (in some order).
Counterexample of if det A = 1 and det B = 2 and there is A+B, is det (A+B) = det (A) + det (B) = 3 ?
Let $A= \left[ {\begin{array}{cc} 1 & 1\\ 0 & 1\\ \end{array} } \right]$ and $B= \left[ {\begin{array}{cc} 2 & 0\\ 2 & 1\\ \end{array} } \right]$, then $\det A=1$ and $\det B=2$
Now $A+B= \left[ {\begin{array}{cc} 3 & 1\\ 2 & 2\\ \end{array} } \right]$, then $\det (A+B)=4\neq \det A+\det B$