For this discussion topic, I decided to answer Section D question M30
Construct an example to show that the following statement is not true for all square matrices A and B of the same size: det (A + B) = det (A) + det (B).
Let’s construct two 2x2 matrices using variables to get the general solution
\[A=\begin{bmatrix} { a }_{ 1 } & { a }_{ 2 } \\ { a }_{ 3 } & { a }_{ 4 } \end{bmatrix}\quad and\quad B=\begin{bmatrix} { b }_{ 1 } & { b }_{ 2 } \\ b_{ 3 } & { b }_{ 4 } \end{bmatrix}\] Now, let’s solve each side, first det(A + B):
\[det\left( \begin{bmatrix} { a }_{ 1 } & { a }_{ 2 } \\ { a }_{ 3 } & { a }_{ 4 } \end{bmatrix}+\begin{bmatrix} { b }_{ 1 } & { b }_{ 2 } \\ b_{ 3 } & { b }_{ 4 } \end{bmatrix} \right)= \\ det\left( \begin{bmatrix} { a }_{ 1 }+b_{ 1 } & { a }_{ 2 }+b_{ 2 } \\ { a }_{ 3 }+{ b }_{ 3 } & { a }_{ 4 }+b_{ 4 } \end{bmatrix} \right)= \\ \left( { a }_{ 1 }+b_{ 1 } \right) \left( { a }_{ 4 }+b_{ 4 } \right) -\left( { a }_{ 2 }+b_{ 2 } \right) \left( { a }_{ 3 }+b_{ 3 } \right)= \\ { a }_{ 1 }{ a }_{ 4 }+{ a }_{ 1 }{ b }_{ 4 }+b_{ 1 }{ a }_{ 4 }+{ b }_{ 1 }{ b }_{ 4 }-{ a }_{ 2 }{ a }_{ 3 }-{ a }_{ 2 }{ b }_{ 3 }-b_{ 2 }{ a }_{ 3 }-{ b }_{ 2 }{ b }_{ 3 }\]
Then det(A) + det(B):
\[det\left( \begin{bmatrix} { a }_{ 1 } & { a }_{ 2 } \\ { a }_{ 3 } & { a }_{ 4 } \end{bmatrix} \right) \quad +\quad det\left( \begin{bmatrix} { b }_{ 1 } & b_{ 2 } \\ b_{ 3 } & b_{ 4 } \end{bmatrix} \right)= \\ { a }_{ 1 }{ a }_{ 4 }-{ a }_{ 2 }{ a }_{ 3 }+b_{ 1 }{ b }_{ 4 }-{ b }_{ 2 }{ b }_{ 3 }\]
so, you can see generally the two equations are not equal; however, lets find when they would be by setting them equal to each other:
\[{ a }_{ 1 }{ a }_{ 4 }+{ a }_{ 1 }{ b }_{ 4 }+b_{ 1 }{ a }_{ 4 }+{ b }_{ 1 }{ b }_{ 4 }-{ a }_{ 2 }{ a }_{ 3 }-{ a }_{ 2 }{ b }_{ 3 }-b_{ 2 }{ a }_{ 3 }-{ b }_{ 2 }{ b }_{ 3 }={ a }_{ 1 }{ a }_{ 4 }-{ a }_{ 2 }{ a }_{ 3 }+b_{ 1 }{ b }_{ 4 }-{ b }_{ 2 }{ b }_{ 3 }\\ \]
\[{ a }_{ 1 }{ b }_{ 4 }+b_{ 1 }{ a }_{ 4 }-{ a }_{ 2 }{ b }_{ 3 }-b_{ 2 }{ a }_{ 3 }=0\]
Here, we find that, for a 2x2 matrix, if you construct a matrix consistent with the formula above, then det (A + B) = det (A) + det (B). Let’s do an example:
Non-equivalent example
A <- matrix(c(1, 2, 3, 4), nrow = 2, byrow = TRUE)
B <- matrix(c(5, 6, 7, 8), nrow = 2, byrow = TRUE)
ls <- det(A + B)
rs <- det(A) + det(B)
all.equal(ls, rs, tolerance = 1e-06)## [1] "Mean relative difference: 0.5"Equivalent example
A <- matrix(c(1, 2, 3, 4), nrow = 2, byrow = TRUE)
B <- matrix(c(1/4, 1/3, 1/2, 1), nrow = 2, byrow = TRUE)
ls <- det(A + B)
rs <- det(A) + det(B)
all.equal(ls, rs, tolerance = 1e-06)## [1] TRUE